Answer:
physiology is the study of mechanisms and functions in a living system
Explanation:
it's basically a part of biology that works with how a living organism or body part functions
A box is being pulled to the right. What is the magnitude of the Kinect frictional force?
The so-called "force" of friction always acts exactly opposite to the direction in which an object is moving. So for this question, since the box is moving to the right, the "force" of friction is acting toward the left.
There's not enough information given in the question to determine its magnitude.
You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a horizontal shelf of 8 m, then another drop of 3 m to the bottom of the canyon, which has a horizontal floor. You kick a 0.87 kg rock, giving it an initial horizontal velocity that barely clears the shelf below.
1. What initial horizontal velocity v will be required to barely clear the edge of the shelf below you? The acceleration of gravity is 9.8 m/s2 . Consider air friction to be negligible. Answer in units of m/s.
2. How far from the bottom of the second cliff will the projectile land? Answer in units of m.
Answer:
1. v = 6.67 m/s
2. d = 9.54 m
Explanation:
1. To find the horizontal velocity of the rock we need to use the following equation:
[tex] d = v*t \rightarrow v = \frac{d}{t} [/tex]
Where:
d: is the distance traveled by the rock
t: is the time
The time can be calculated as follows:
[tex] t = \sqrt{\frac{2d}{g}} [/tex]
Where:
g: is gravity = 9.8 m/s²
[tex] t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s [/tex]
Now, the horizontal velocity of the rock is:
[tex] v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s [/tex]
Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.
2. To calculate the distance at which the projectile will land, first, we need to find the time:
[tex] t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s [/tex]
So, the distance is:
[tex] d = v*t = 6.67 m/s*1.43 s = 9.54 m [/tex]
Therefore, the projectile will land at 9.54 m of the second cliff.
I hope it helps you!
The animation shows a ball which has been kicked upward at an angle. Run the animation to watch the motion of the ball. Click initialize to set up the animation and start to run it.
Ghosts are left by the ball once per second. The animation can also be paused and moved forward in single frame mode using the step button. The cursor can be used to read the (x,y) coordinates of a position in the grid by holding down the left mouse button. Assume the grid coordinates read out in meters. When entering components, presume that x is positive to the right and y is positive upwards. Note that this ball is NOT being kicked on Earth. Do not expect an acceleration of 9.80 m/s2 downward, though you can presume that gravity is acting straight down. Use this animation to answer the following questions. Note that there are a number of different ways to go about each of the following questions. Your answer needs to be within 5% of the correct answer for credit. Please enter your answer to 3 significant digits.
What is the maximum height which the ball reaches? 42.24 m
What is the horizontal component of the initial velocity of the ball? 5.57 m/s
What is the vertical component of the initial velocity of the ball? 16.18 m/s
What is the vertical component of the acceleration of the ball? _____????
Answer:
The acceleration of the ball is [tex]a_y = - 0.3672 \ m/s^2[/tex]
Explanation:
From the question we are told that
The maximum height the ball reachs is [tex]H_{max} = 42.24 \ m[/tex]
The horizontal component of the initial velocity of the ball is [tex]v_{ix} = 5.57 \ m/s[/tex]
The vertical component of the initial velocity of the ball is [tex]v_{iy} = = 16.18 m/s[/tex]
The vertically motion of the ball can be mathematically represented as
[tex]v_{fy}^2 = v_{iy} ^2 + 2 a_{y} H_{max}[/tex]
Here the final velocity at the maximum height is zero so [tex]v_{fy} = 0 \ m/s[/tex]
Making the acceleration [tex]a_y[/tex] the subject we have
[tex]a_y = \frac{v_{iy} ^2}{2H_{max}}[/tex]
substituting values
[tex]a_y = - \frac{5.57^2}{2* 42.24}[/tex]
[tex]a_y = - 0.3672 \ m/s^2[/tex]
The negative sign shows that the direction of the acceleration is in the negative y-axis
Select the correct answer. mega M 10 6 1,000,000 kilo k 103 1,000 hecto h 102 100 deka da 10 1 10 deci d 10–1 0.1 centi c 10–2 0.01 milli m 10–3 0.001 micro µ 10–6 0.000001 nano n 10–9 0.000000001 pico p 10–12 0.000000000001 One nanometer is equal to how many centimeters? A. 109 mm B. 10–6 cm C. 10–7 cm D. 10–9 mm
Answer:
C. 10⁻⁷ cm
Explanation:
One nanometer = 10⁻⁹ meter
1 meter = 10² cm
one nanometer = 10⁻⁹ x 10² cm
= 10⁻⁹⁺² cm
= 10⁻⁷ cm .
Answer:
The answer is C
Explanation:
I got it right on my quiz
A bike with tires of radius
0.330 m speeds up from rest
to 5.33 m/s in 6.27's. Through
what angle do the wheels turn
in that time?
(Unit = rad)
Answer:
The wheels turn 101.27 radians in that time.
Explanation:
First, we need to find the angular velocity of the tires. We use the following formula for this purpose:
Δv = rω
ω = Δv/r
where,
ω = Angular Velocity of Tires = ?
Δv = change in linear velocity = 5.33 m/s - 0 m/s = 5.33 m/s
r = radius of tire = 0.33 m
Therefore,
ω = (5.33 m/s)/(0.33 m)
ω = 16.15 rad/s
Now, the angular displacement covered by tires can be found out by using the general formula of angular velocity:
ω = θ/t
θ = ωt
where,
θ = angular displacement = ?
t = time = 6.27 s
Therefore,
θ = (16.15 rad/s)(6.27 s)
θ = 101.27 radians
This problem concerns the properties of circular orbits for a satellite of mass m orbiting a planet of mass M in an almost circular orbit of radius r. In doing this problem, you are to assume that the planet has an atmosphere that causes a small drag due to air resistance. "Small" means that there is little change during each orbit so that the orbit remains nearly circular, but the radius can change slowly with time. The following questions will ask about the net effects of drag and gravity on the satellite's motion, under the assumption that the satellite's orbit stays nearly circular. Use G if necessary for the universal gravitational constant.
What is the potential energy U of the satellite?Express your answer in terms ofm, M, G, and r.What is the kinetic energy K of the satellite?Express the kinetic energy in termsof m, M, G, and r.
Answer:
A) U = - GMm/r
B) K = 0.5 mGM/r
Explanation:
A) The potential energy U of the satellite
U = - GMm/r
G = universal gravitational constant which is ( 6.67e-11 Nm^2/c^2 )
M = mass of the planet
m = mass
r = distance ( radius )
B) Kinetic energy
kinetic energy expressed as K = 0.5 m Vo^2
NOTE : Vo^2 = GM / r
hence kinetic energy will be expressed as
K = 0.5 mGM/r
Linear charge density 4.00×10−12 C/m surrounds an infinitely long line charge. A positively charged elementary particle (mass 1.67×10−27 kg, charge +1.60×10−19 C) is 15.0 cm from this line charge. Consider that this elementary particle is moving at speed 3.20×103 m/s directly toward the line charge.
Part A- Find the initial kinetic energy of this elementary particle.
Part B- Find the closest distance that the elementary particle get to the line charge?
Answer:
A)Kopya
B)YASAK
Explanation:
kopya yasak dostum adın da belli. Başın belaya girmesin
Two loudspeakers are located 2.65 m apart on an outdoor stage. A listener is 19.1 m from one and 20.1 m from the other. During the sound check, a signal generator drives the two speakers in phase with the same amplitude and frequency. The transmitted frequency is swept through the audible range (20 Hz–20 kHz). The speed of sound in the air is 343 m/s. What are the three lowest frequencies that give minimum signal (destructive interference) at the listener's location?
Answer:
171.5Hz,514.5Hz and 857.5Hz
Explanation:
We are given that
Distance between two loudspeaker,d=2.65 m
Distance of listener from one end=19.1 m
Distance of listener from other end=20.1 m
[tex]\Delta L=20.1-19.1=1[/tex]m
Speed of sound,v=343m/s
For destructive interference
[tex]f_{min,n}=\frac{(n-0.5)v}{\Delta L}[/tex]
Using the formula and substitute n=1
[tex]f_{min,1}=\frac{(1-0.5)\times 343}{1}=171.5Hz[/tex]
For n=2
[tex]f_{min,2}=\frac{(2-0.5)\times 343}{1}=514.5Hz[/tex]
For n=3
[tex]f_{min,3}=(3-0.5)\times 343=857.5Hz[/tex]
Hence, the three lowest frequencies that give minimum signal (destructive interference) at the listener's location is given by
171.5Hz,514.5Hz and 857.5Hz
A 350-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,010 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable? (Use 3.156 107 for the number of seconds in a year.)
Answer:
t = 47 years
Explanation:
To find the number of years in which the electrons cross the complete transmission, you first calculate the drift velocity of the electrons in the transmission line, by using the following formula:
[tex]v_d=\frac{I}{nAq}[/tex] (1)
I: current = 1,010A
A: cross sectional area of the transmission line = π(d/2)^2
d: diameter of the transmission line = 2.00cm = 0.02 m
n: free charge density = 8.50*10^28 electrons/m^3
q: electron's charge = 1.6*10^-19 C
You replace the values of all parameters in the equation (1):
[tex]v_d=\frac{1010A}{(8.50*10^{28}m^{-3})(\pi(0.02m/2)^2)(1.6*10^{-19}C)}\\\\v_d=2.36*10^{-4}\frac{m}{s}[/tex]
with this value of the drift velocity you can calculate the time that electrons take in crossing the complete transmission line:
[tex]t=\frac{d}{v_d}=\frac{350km}{2.36*10^{-4}m/s}=\frac{350000m}{2.36*10^{-4}m/s}\\\\t=1,483,050,847\ s[/tex]
Finally, you convert this value of the time to years:
[tex]t=1,483,050,847s*\frac{1\ year}{3.154*10^7s}=47\ years[/tex]
hence, the electrons take around 47 years to cross the complete transmission line.
14. High-voltage power lines are a familiar sight throughout the country. The aluminum wire used for some of these lines has a cross-sectional area of 4.9 x 10-4 m2 . What is the resistance of ten kilometers of this wire
Answer:
The resistance of the aluminium wire is 0.541 ohms
Explanation:
Given:
Cross sectional area of the aluminum wire, A = 4.9 x 10-4 m2
Length of the wire, L = 10 km = 10,000 m
Resistance of the wire = (pL) / A
Where:
p is resistivity of aluminium wire = 2.65 x 10^-8 ohm-m
L is length of the wire
A is cross sectional area of the wire
Substitute these values and solve for resistance of the wire.
Resistance = (2.65 x 10^-8 x 10,000) / (4.9 x 10^-4)
Resistance = 0.541 ohms
Therefore, the resistance of the aluminium wire is 0.541 ohms
A ball is thrown eastward into the air from the origin (in the direction of the positive x-axis). The initial velocity is 60 i + 64 k, with speed measured in feet per second. The spin of the ball results in a southward acceleration of 6 ft/s2, so the acceleration vector is a = −6 j − 32 k. Where does the ball land? (Round your answers to one decimal place.) ft from the origin at an angle of ° from the eastern direction toward the south. With what speed does the ball hit the ground? (Round your answer to one decimal place.)
The ball's position in the air at time t is given by the vector,
p(t) = (60 i + 64 k) t + 1/2 (-6 j - 32 k) t ²
and its velocity is given by
v(t) = (60 i + 64 k) + (-6 j - 32 k) t
The ball is in the air for as long as it takes for the vertical (k) component of the position vector to reach 0, so we solve,
64 t - 32/2 t ² = 0 ==> t = 0 OR t = 4
and so the ball is in the air for 4 s.
After this time, the ball has position vector
p(4) = (60 i + 64 k) t + 1/2 (-6 j - 32 k) t ² = 240 i - 48 j
which has magnitude
||p(4)|| = √(240² + (-48)²) = 48 √26 ≈ 244.8 ft
in a direction θ in the x,y plane from the positive x axis such that
tanθ = -48/240 = -1/5 ==> θ = -arctan(1/5) ≈ -11.3º
or 11.3º South of East.
The ball hits the ground with speed
||v(4)|| = ||60 i - 24 j - 64 k|| = √(60² + (-24)² + (-64)²) = 4 √517 ≈ 91.0 ft/s
kinematic relationships can be used in all dimensions and finding the position and velocity when reaching the ground is
r = 244.8 ft, tea = 21.8º from East to South
v = 91.0 ft / s
given parameters
Initial velocity v = (60 i + 64 k) ft / sBody acceleration a = (-6 j - 32 k) ft / s²
to find
where it reaches the groundground speed
Kinematics allows finding the position, velocity and acceleration of the body, in this case we have a problem in three dimensions, where they establish a Cartesian coordinate system, a method to solve this exercise is to solve each component independently
a) The acceleration of gravity acts on the z axis, so let's find the time it takes to reach the ground, if the initial vertical velocity is v_{oz} = 64 ft/s and the acceleration is a_z = g = -32 ft / s², we assume that the ball leaves the ground (z₀ = 0)
z = z₀ + v_{oz} t + ½ a_z t²
when reaching the ground its height of zero and
0 = 0 + v_{oz} t + ½ a_z t²
t (v_{oz) + ½ a_z t) = 0
t (64 - 16 t) = 0
the solusion of this squadron is
y = 0
t = 4 s
the first time is when it leaves and the second time is for when it reaches the ground, therefore the flight time is t = 4s
with this time we find the displacement is each exercise
X axis
in this axis there is no acceleration, so we use the uniform motion relationships
vₓ = x / t
x = vₓ t
x = 60 4
x = 240 ft
Y Axis
on this axis there is an acceleration of a_y = -6 ft/s and an initial velocity v_{oy} = 0
we use the kinematic relation
y = v_{oy} t + ½ a_y t²
y = 0 - ½ 6 4²
y = - 48 ft
let's use the Pythagoras theorem to find the position
r² = (x -x₀) ² + (y -y₀) ² + (z -z₀) ²
r² = (240-0) ² + (-48-0) ² + (0-0) ²
r = 244.8 ft
We use trigonometry to find the direction
tan θ = y / x
θ = tan⁻¹ [tex]\frac{y}{x}[/tex]
θ = tan⁻¹ [tex]\frac{96}{240}[/tex]
θ = -21.8º
This angle is measured clockwise from the x axis, it can also be read
θ = 21.8º from East to South
b) Let's look for the speed when we hit the ground
X axis
vₓ = v_{ox} + aₓ t
vₓ = 60 - 0
vₓ = 60 ft / s
Y Axis
v_y = v_{oy} + a_y t
v_y = 0 - 6 4
v_y = -24 ft / s²
Z axis
v_z = v_{oz} + a_z t
v_z = 64 -32 4
v_z = -64 ft / s
With the Pytagoras theorem find the modulus of this speed is
v² = vx² + vy² + vz²
v² = 60² + 24 ² + 64²
v = 91.0 ft / s
In conclusion, the kinematic relationships can be used in all dimensions and finding the position and velocity when reaching the ground is
a) r = 244.8 ft, θ = 21.8º from East to South
b) v = 91.0 ft / s
learn more about kinematics here:
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A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper cable makes angles of 41° and 63° with the horizontal. Calculate the tension in each of the three cables.
Answer:
T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N
Explanation:
This is a balance exercise where we must apply the expressions for translational balance in the two axes
∑ F = 0
Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º
decompose the tension of the two upper cables
cos 41 = T₁ₓ / T1
sin 41 = T₁y / T1
T₁ₓ = T₁ cos 41
T₁y= T₁ sin 41
for cable gold
cos 63 = T₂ / T₂
sin 63 = [tex]T_{2y}[/tex] / T₂
We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.
Let's start by analyzing the point where the traffic light meets the vertical cable
T₃ - W = 0
T₃ = W
T₃ = 200 N
now let's write the equations for the single point of the three wires
X axis
- T₁ₓ + T₂ₓ = 0
T₁ₓ = T₂ₓ
T1 cos 41 = T2 cos 63
T1 = T2 cos 63 / cos 41 (1)
y Axis
[tex]T_{1y}[/tex] + T_{2y} - T3 = 0
T₁ sin 41 + T₂ sin 63 = T₃ (2)
to solve the system we substitute equation 1 in 2
T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W
T₂ (cos 63 tan 41 + sin 63) = W
T₂ = W / (cos 63 tan 41 + sin 63)
We calculate
T₂ = 200 / (cos 63 tan 41 + sin 63)
T₂ = 200 / 1,2856
T₂ = 155.6 N
we substitute in 1
T₁ = T₂ cos 63 / cos 41
T₁ = 155.6 cos63 / cos 41
T₁ = 93.6 N
therefore the tension in each cable is
T₁ = 93.6 N
T₂ = 155.6 N
T₃ = 200 N
Adiabatic Equation of State Derive the adiabatic equation of state (2.3.19) using particle conservation (2.3.7) and energy conservation (2.3.21), by assuming that the heat flow vector q and all collision terms in these equations are zero.
p=Cn^{\gamma } Equation 19
\frac{\partial n}{\partial t}+\bigtriangledown (nu)=G-L Equation 7
\bigtriangledown (\frac{3}{2}pu)=\frac{\partial }{\partial t}(\frac{3}{2}p)\mid c Equation 22
Answer:
Explanation:
jhgfhmhvmghmm
A 35 grams bullet travels with a velocity of magnitude 126 km/h. What is the bullet's linear momentum?
The linear momentum of the bullet, given the data from the question is 1.225 Kg.m/s
What is momentum?Momentum is defined as the product of mass and velocity. It is expressed as
Momentum = mass × velocity
With the above formula, we can obtain the momentum of the bullet. Details below.
The following data were obtained from the question:
Mass of bullet = 35 g = 35 / 1000 = 0.035 KgVelocity = 126 Km/h = 126 / 3.6 = 35 m/sMomentum =?Momentum = mass × velocity
Momentum = 0.035 Kg × 35 m/s
Momentum = 1.225 Kg.m/s
From the calculation made above, we can conclude that the linear momentum of the bullet is 1.225 Kg.m/s
Learn more about momentum:
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An electric field of 2.09 kV/m and a magnetic field of 0.358 T act on a moving electron to produce no net force. If the fields are perpendicular to each other, what is the electron's speed?
Answer:
The velocity is [tex]v = 5838 \ m/s[/tex]
Explanation:
From the question we are told that
The electric field is [tex]E = 2.09 kV/m = 2.09 *10^{3} \ V/m[/tex]
The magnetic field is [tex]B = 0.358 \ T[/tex]
Generally the force experienced by the electron due to the magnetic field is
[tex]F_m = qvB[/tex]
Generally the force experienced by the electron due to the electric field is
[tex]F_e = qE[/tex]
Since from the question the net force is zero then
[tex]F_e = F_m[/tex]
=> [tex]v = \frac{E}{B}[/tex]
Substituting values
[tex]v = \frac{2.09*10^{3}}{0.358 }[/tex]
[tex]v = 5838 \ m/s[/tex]
A spring stretches by 0.0190 m when a 3.36-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Answer:
m = 4.87 kg
Explanation:
In order to find the required mass you first calculate the spring constant of the spring. When the system reaches the equilibrium you obtain the following equation:
[tex]Mg=kx[/tex] (1)
That is, the weight of the object is equal to the restoring force of the spring.
M: mass of the object = 3.36 kg
g: gravitational constant = 9.8m/s^2
k: spring constant = ?
x: elongation of the spring = 0.0190m
You solve the equation (1) for k:
[tex]k=\frac{Mg}{x}=\frac{(3.36kg)(9.8m/s^2)}{0.0190m}=1733.05\frac{N}{m}[/tex]
Next, to obtain a frequency of 3.0Hz you can use the following formula, in order to calculate the required mass:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex] (2)
You solve the equation (2) for m:
[tex]m=\frac{1}{4\pi^2}\frac{k}{f^2}\\\\m=\frac{1}{4\pi^2}\frac{1733.05N/m}{(3.0Hz)^2}=4.87kg[/tex]
The required mass to obtain a frequency of 3.0Hz is 4.87 kg
A rectangular tank is filled to a depth of 10m with freshwater and open to air at atmospheric pressure.
It is fitted with two drain plugs on the bottom of the tank, one at the left side and one at the right side.
Both plugs have chains attached and are removed by lifting the chains.
The left-side plug is a flat circular disk of 50cm diameter and 1cm height.
The right-side plug is a hemisphere and is also 50cm in diameter.
Q1. What is the upwards force required to lift the left- side circular disk plug?
Q2. What is the upwards force required to lift the right-side hemisphere plug?
Answer:
1. 39068.07 N
2. 19534.036 N
Explanation:
depth of water h = 10 m
atmospheric pressure Patm = 101325 Pa
density of water p = 1000 kg/m^3
acceleration due to gravity g = 9.81 m/s^2
pressure due to depth of water = pgh
P = 1000 x 9.81 x 10 = 98100 Pa
total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa
Left plug has diameter = 50 cm = 0.5 m
radius = 0.5/2 = 0.25 m
height = 1 cm = 0.01 m
height below tank surface = 10 - 0.01 = 9.99
pressure at this depth = 1000 x 9.81 x 9.99 = 98001.9 Pa
total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa
surface area of plug = π[tex]r^{2}[/tex] = 3.142 x [tex]0.25^{2}[/tex] = 0.196 [tex]m^{2}[/tex]
force required to lift left plug = pressure x area
F = 199326.9 x 0.196 = 39068.07 N
The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug
A = 0.196/2 = 0.098 [tex]m^{2}[/tex]
force F required to lift right plug = 199326.9 x 0.098 = 19534.036 N
Chris and Sue are wearing harnesses and are hanging from the ceiling by means of ropes attached to them. They are face to face and push off against each other. Sue has a mass of 52 kg and Chris has a mass of 78 kg. Following the push, Sue reaches a height of 0.65 meters above her starting point. How high does Chris reach?
Answer:
h = 0.288m
Explanation:
Assume
[tex]v_1[/tex] = Speed of Sue
[tex]v_2[/tex] = Speed of Chris immediately after the push
Sue's KE = [tex]\frac{1}{2} mv_1\ ^2 = 26 v_1 \ ^2[/tex]
now she swings this is converted into gravitation at PE of
mg n = 52 × 9.8 × 0.65
= 331.24
[tex]26v_{1}\ ^2[/tex] = 331.24
So, [tex]v_1 = 3.569[/tex]
They started at rest by conservation of momentum in case of push off the magnitude of sue momentum and it is equal to the magnitude of Chris momentum in the opposite or inverse direction
[tex]m_1v_1 = m_2v_2[/tex]
[tex]52 \times 3.569 = 78 \times v_2[/tex]
[tex]v_2 = 2.380[/tex]
Chris kt = [tex]\frac{1}{2} \times 78 \times 2.380^2[/tex]
= 220.827
220.827 = mgh
So, h = 0.288m
A projectile is fired from ground level at an angle above the horizontal on an airless planet where g = 10.0 m/s2. The initial x and y components of its velocity are 86.6 m/s and 50.0 m/s respectively. How long after firing does it take before the projectile hits the level ground?
Answer:
10 s
Explanation:
We are given that
[tex]g=10.0m/s^2[/tex]
Initially
[tex]v_x=86.6m/s,y=50.0m/s[/tex]
We have to find the time after firing taken by projectile before it hits the level ground.
v=[tex]\sqrt{v^2_x+v^2_y}[/tex]
[tex]v=\sqrt{(86.6)^2+(50)^2}=99.99 m/s[/tex]
[tex]\theta=tan^{-1}(\frac{v_x}{v_y})[/tex]
[tex]\theta=tan^{-1}(\frac{50}{86.6})=30^{\circ}[/tex]
Now,
[tex]t=\frac{vsin\theta}{g}[/tex]
Using the formula
[tex]t=\frac{99.99sin30}{10}[/tex]
[tex]t=4.99\approx 5 s[/tex]
Now, total time,T=2t=[tex]2\times 5=10s[/tex]
Hence, after firing it takes 10 s before the projectile hits the level ground.
Two boats start together and race across a 50-km-wide lake and back. Boat A goes across at 50 km/h and returns at 50 km/h. Boat B goes across at 25 km/h, and its crew, realizing how far behind it is getting, returns at 75 km/h. Turnaround times are negligible, and the boat that completes the round-trip first wins.
Required:
a. Which boat wins?
b. What is the average velocity of the winning boat?
Answer:
Boat A wins and 50km/hr
Explanation:
To solve this problem we need to calculate the time required for both boats to make the round trip.
For Boat A.
The time required for the first trip =distance/ speed = 50/50 = 1h
For the return trip the time is same since it's the same distance on the same speed.
Hence the time taken for boat A to make the round trip is 2hrs.
For boat B,
The time required for the first trip =distance/ speed = 50/25 = 2h
The time for the return trip is;
=distance/ speed = 50/75 = 2/3h= 2/3×60= 40mins
By comparing both time.
Boat A requires less time hence Boat A wins.
2.The average velocity of the winning boat is the velocity of boat A.
Which is total distance/ total time for the trip.
100km/2hrs = 50km/hr
boat A would win the race and the average velocity of the winning boat would be 50 km/h
What is Velocity?The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object.
As given in the problem Two boats start together and race across a 50-km-wide lake and back. Boat A goes across at 50 km/h and returns at 50 km/h. Boat B goes across at 25 km/h, and its crew, realizing how far behind it is getting, returns at 75 km/h.
Turnaround times are negligible, and the boat that completes the round-trip first wins
Average velocity of the boat A = ( 50 + 50 ) / 2
= 50 km /h
Thus, boat A would win the race and the average velocity of the winning boat would be 50 km/h
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As the space shuttle orbits the Earth, the shuttle and the astronauts
accelerate towards the Earth with the same acceleration. What effect does
this create?
A. The 'g' forces astronauts feel.
B. Weightlessness
C. Astronauts feeling dizzy when they land.
D. Astronauts losing weight while in space.
Is light one dimensional?
Answer: No.
Explanation: Light exists in 3+1 dimensional space (3 space, 1 time).
Describe briefly with used of practical knowledge the effect of the voltage stores across and current
flows through the resistors when connected in:
i. Series
ii. Parallel
(circuit diagram is necessary)
Answer:
Answer in explanation
Explanation:
The path of flow of circuit is called electric circuit. In an electric circuit, resistances are connected in two different ways, which are:
1- Series Combination of Resistances
2- Parallel Combination of Resistances
Series combination of resistances
In series combination resistors are connected end to end, and there is only one path for the flow of current.
Characteristics Of Series Circuit:
1. In series circuit there is only one path for the flow of current.
2. Same amount of current flows through each resistor.
I = I₁ = I₂ = I₃ = In
3. Total voltage of the battery is equal to the sum of voltages across each resistor.
V = V₁ +V₂ + V₃ + ... + Vn
4. In series combination the combined resistance of the resistors can be obtained by adding the value of each resistor.
R = R₁ + R₂ + R₃ + … + Rn
Parallel combination of resistances
Resistances are set to be connected in parallel, when each of them is connected directly from the terminals of electric source.
Characteristics Of Series Circuit:
1. There are more than one path for the flow of current.
2. The value of potential difference remains constant on each resistor.
V = V₁ = V₂ = V₃ = Vn
3. Total current is equal to the sum of current passing through each resistor.
I = I₁ + I₂ + I₃ +...+ In
4. Reciprocal of equivalent resistance is equal to the sum of reciprocals of resistances connected in parallel.
1/R = 1/R₁ + 1/R₂ + 1/R₃ + … + 1/Rn
The circuit diagrams are in attachment.
Which action is due to field forces?
A. an apple falling from a tree
B. a moving car stopping when the brakes are applied
C. the rowing of a boat
D. pushing a chair against the wall
Answer:
a
an apple falling from a tree
Answer an apple falling from a tree
Explanation:
A circular loop of flexible iron wire has an initial circumference of 164cm , but its circumference is decreasing at a constant rate of 11.0cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 1.00T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop. Find the magnitude of the emf induced in the loop after exactly time 4.00s has passed since the circumference of the loop started to decrease AND find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
Answer:
emf = 0.02525 V
induced current with a counterclockwise direction
Explanation:
The emf is given by the following formula:
[tex]emf=-\frac{\Delta \Phi_B}{\Delta t}=-B\frac{\Delta A}{\Delta t}[/tex][tex]\ \ =-B\frac{A_2-A_1}{t_2-t_1}[/tex] (1)
ФB: magnetic flux = BA
B: magnitude of the magnetic field = 1.00T
A2: final area of the loop; A1: initial area
t2: final time, t1: initial time
You first calculate the final A2, by taking into account that the circumference of loop decreases at 11.0cm/s.
In t = 4 s the final circumference will be:
[tex]c_2=c_1-(11.0cm/s)t=164cm-(11.0cm/s)(4s)=120cm[/tex]
To find the areas A1 and A2 you calculate the radius:
[tex]r_1=\frac{164cm}{2\pi}=26.101cm\\\\r_2=\frac{120cm}{2\pi}=19.098cm[/tex]
r1 = 0.261 m
r2 = 0.190 m
Then, the areas A1 and A2 are:
[tex]A_1=\pi r_1^2=\pi (0.261m)^2=0.214m^2\\\\A_2=\pi r_2^2=\pi (0.190m)^2=0.113m^2[/tex]
Finally, the emf induced, by using the equation (1), is:
[tex]emf=-(1.00T)\frac{(0.113m^2)-(0.214m^2)}{4s-0s}=0.0252V=25.25mV[/tex]
The induced current has counterclockwise direction, because the induced magneitc field generated by the induced current must be opposite to the constant magnetic field B.
Silver and Copper rods of equal areas are placed end to end with the free end of the silver rod in ice at 0.00 degrees Celsius and the free end of the copper rod in steam at 100. degrees Celsius. The Silver rod is 15.0 cm in length and the copper rod is 25.0 cm in length.
a) What is the temperature of the junction between copper and sliver when they have come to equilibrium?
b) How much ice (in grams) melts per second?
Answer:
A.) The temperature of the junction between copper and sliver when they have come to equilibrium is 35 degree Celsius
B.) ice (in grams) melts per second = 0.078 kg/s
Explanation:
A.) Given that the two material are of the same area.
The Silver rod is 15.0 cm in length and the copper rod is 25.0 cm in length.
Silver temperature = 0 degree Celsius
Copper temperature = 100 degree Celsius
Thermal conductivity k of silver = 429 W/m•K
Thermal conductivity k of copper = 385W/m.k
Rate of energy transferred P in the two materials can be expressed as
P = k.A.dT/L
dT = change in temperature
Since the rate and the area are the same
429 ( T -0 )/0.15 = 385( 100 - T )/0.25
2860T = 1540(100 - T)
Open the bracket
2860T = 154000 - 1540T
Collect the like terms
2860T + 1540T = 154000
4400T = 154000
T = 154000/4400
T = 35 degree Celsius
The temperature of the junction between copper and sliver when they have come to equilibrium is 35 degree Celsius
B.) Rate of energy transferred P will be
P = 2860 × 35 = 100100
P = Q/t ..... (1)
Where Q = energy transferred
But Q = mcØ .....(2)
And specific heat capacity c of water = 4182J/k.kg
Substitutes Q into formula 1.
P = mcØ/t
Make m/t the subject of formula
m/t = P/cØ
m/t = 100100/ 4182( 35 + 273 )
m/t = 100100/1288056
m/t = 0.078 kg/s
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t2. (a) What is the maximum height reached by the ball? ft (b) What is the velocity of the ball when it is 384 ft above the ground on its way up? (Consider up to be the positive direction.) ft/s What is the velocity of the ball when it is 384 ft above the ground on its way down?
Answer:
a) h_max = 400ft
b) v = 1024 ft/s
Explanation:
The general equation of a motion is:
[tex]s(t)=v_ot-\frac{1}{2}gt^2[/tex] (1)
You have the following equation of motion is given by:
[tex]s(t)=160t-16t^2[/tex] (2)
You compare both equations (1) and (2) and you obtain:
vo: initial velocity = 160ft/s
g: gravitational acceleration = 32ft/s
a) The maxim height reached by the ball is given by:
[tex]h_{max}=\frac{v_o^2}{2g}=\frac{(160ft/s)^2}{64ft/s^2}=400ft[/tex]
b) The velocity is given by:
[tex]v^2=v_o^2-2gh\\\\v=\sqrt{(160ft/s)^2-2(32ft/s^2)(384ft)}=1024ft/s[/tex]
Answer:
a) Smax = 400 ft
b) v = 32 ft/s
c) v = - 32 ft/s
Explanation:
(a)
The function given for the height of ball is:
s = 160 t - 16 t²
Therefore, in order to find the time to reach maximum height (in-flexion point), we must take the derivative with respect to t and set it equal to zero:
Therefore,
160 - 32 t = 0
32 t = 160
t = 160/32
t = 5 sec
Therefore, maximum distance is covered at a time interval of 5 sec.
Smax = (160)(5) - (16)(5)²
Smax = 800 - 400
Smax = 400 ft
b)
First we calculate the time at which ball covers 384 ft
384 = 160 t - 16 t²
16 t² - 160 t + 384 = 0
Solving Quadratic Equation:
Either:
t = 6 sec
Or:
t = 4 sec
Since, the time to reach maximum height is 5 sec. Therefore, t < 5 sec
Therefore,
t = 4 sec
Now, we find velocity at 4 sec by taking derivative od s with respect to t at 4 sec:
v = 160 - 32 t
v = 160 - (32)(4)
v = 32 ft/s
c)
Since, t = 6 s > 5 s
The second value of t = 6 sec must correspond to the instant when the ball is 384 ft above the ground while traveling downward.
Hence, velocity at that time will be:
v = 160 - (32)(6)
v = -32 ft/s
Negative sign due to downward motion.
A space probe on the surface of Mars sends a radio signal back to the Earth, a distance of 9.75 ✕ 107 km. Radio waves travel at the speed of light (3.00 ✕ 108 m/s). How many seconds does it take for the signal to reach the Earth?
Answer:
It takes 325 seconds for the signal to reach Earth.
Explanation:
First, you must make a unit change from m/s to km/s in order to make a comparison with the distance of the radio signal sent to Earth. For that, you know that 1 m is 0.001 km. So:
[tex]3*10^{8} \frac{m}{s} =3*10^{8}\frac{0.001 km}{s}=300,000\frac{km}{s}[/tex]
The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.
If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:
a ⇒ b
c ⇒ x
[tex]x=\frac{c*b}{a}[/tex]
In this case, the rule of three is applied as follows: if by definition of speed, 300,000 km of light are traveled in 1 second, 9.75 * 10⁷ km in how long are they traveled?
[tex]time=\frac{9.75*10^{7}km*1second }{300,000 km}[/tex]
time=325 seconds
It takes 325 seconds for the signal to reach Earth.
The number of seconds does it take for the signal to reach the Earth is 325 seconds.
The calculation is as follows;[tex]= 9.75 \times 10^7 \div 300,000 km[/tex]
= 325 seconds
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To Da
The radius of a circular track is 7 m. What will be the speed of a scooter, if it takes 20 seconds to complete one round of the
track?
published yet
04
published yet
A: 2.2 m/s
published yet
B: 22 m/s
02
published yet
C: 0.22 m/s
D: 022m/s
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Answer:
A
Explanation:
Firstly, recall the equation for the circumference of a circle (C=2*pi*r). Then apply the formula speed = distance/time.
Answer:
i see they did already i am just supporting them
Answer:
A
Explanation:
Firstly, recall the equation for the circumference of a circle (C=2*pi*r). Then apply the formula speed = distance/time.
Two resistors, A and B, are connected in parallel across a 8.0 V battery. The current through B is found to be 3.0 A. When the two resistors are connected in series to the 8.0 V battery, a voltmeter connected across resistor A measures a voltage of 2.4 V. Find the resistances A and B.
Answer:
R_A = 2.67 ohms
R_B = 1.14 ohms
Explanation:
When the resistors are connected in parallel, the voltage will be the same across both resistors A and B.
Thus, we now have the current and the voltage across B and so we can use Ohm's Law to find the resistance.
V/I = R
Thus, resistance of B; R_B = 8/3
R_B = 2.67 ohms
Now, when the resistors are connected in series, the voltage drop across B is;
V = 8V - 2.4V = 5.6V
Since we now have the resistance of B , we can find the current using Ohm's Law. Thus;
I = V/R
I = 5.6/2.67
I = 2.1 A
Now, current is the same for all resistances in a series circuit because this is the same current through resistors A and B. So, we can use Ohm's law again to find the resistance across A.
So, R = V/I
R_A = 2.4/2.1
R_A = 1.14 ohms