Answer:
a) a = 2.35 m/s^2
Explanation:
(a) In order to calculate the magnitude of the acceleration of the ball, you use the following formula, for the position of the ball:
[tex]x=v_ot+\frac{1}{2}at^2[/tex] (1)
x: position of the ball after t seconds = 87 m
t: time = 8.6 s
a: acceleration of the ball = ?
vo: initial velocity of the ball = 0 m/s
You solve the equation (1) for a:
[tex]x=0+\frac{1}{2}at^2\\\\a=\frac{2x}{t^2}[/tex]
You replace the values of the parameters in the previous equation:
[tex]a=\frac{2(87m)}{(8.6s)^2}=2.35\frac{m}{s^2}[/tex]
The acceleration of the ball is 2.35 m/s^2
A certain radio wave has a wavelength of 6.0 × 10-2m. What is its frequency in hertz?
Answer:
The frequency of the wave is 5 x 10⁹ Hz
Explanation:
Given;
wavelength of the radio wave, λ = 6.0 × 10⁻²m
radio wave is an example of electromagnetic wave, and electromagnetic waves travel with speed of light, which is equal to 3 x 10⁸ m/s².
Applying wave equation;
V = F λ
where;
V is the speed of the wave
F is the frequency of the wave
λ is the wavelength
Make F the subject of the formula
F = V / λ
F = (3 x 10⁸) / (6.0 × 10⁻²)
F = 5 x 10⁹ Hz
Therefore, the frequency of the wave is 5 x 10⁹ Hz
A neutron star has about one and a half times the mass of our Sun but has collapsed to a radius of 10 kmkm . Part A What is the acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface?
Answer:
gₓ = 1.36 x 10¹³ g
Explanation:
The value of acceleration due to gravity at a certain place is given by the following formula:
gₓ = GM/R²
where,
gₓ = acceleration due to gravity on the surface of neutron star
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of the star = 10 * Mass of sun = (10)(2 x 10³⁰ kg) = 2 x 10³¹ kg
R = 10 km = 10⁴ m
Therefore,
gₓ = (6.67 x 10⁻¹¹ N.m²/kg²)(2 x 10³¹)/(10⁴)²
gₓ = 1.334 x 10¹⁴ m/s²
Hence, comparing it with the free-fall acceleration at Earth's Surface:
gₓ/g = (1.334 x 10¹⁴)/9.8
gₓ = 1.36 x 10¹³ g
The acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is [tex]1.35 \times 10^5 \ g_E[/tex].
The given parameters:
Mass of the neutron star, m = 1.5 MRadius of the neutron star, R = 10 kmkmThe acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is calculated as follows;
[tex]F = mg = \frac{GM_sm}{R^2} \\\\(1.5 M_s)g = \frac{GM_s(1.5 M_s)}{R^2} \\\\g = \frac{GM_s}{R^2} \\\\[/tex]
where;
[tex]M_s[/tex] is the mass of the Sun = 1.989 x 10³⁰ kg.
[tex]g = \frac{6.67 \times 10^{-11} \times 1.989 \times 10^{30} }{(10,000,000)^2} \\\\g = 1.326 \times 10^{6} \ m/s^2[/tex]
In terms of gravity of Earth [tex](g_E)[/tex];
[tex]= \frac{1.326 \times 10^6}{9.81} = 1.35 \times 10^5 \\\\= 1.35 \times 10^5 \ g_E[/tex]
Thus, the acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is [tex]1.35 \times 10^5 \ g_E[/tex].
Learn more about acceleration due to gravity here: https://brainly.com/question/88039
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
A horizontal wire is hung from the ceiling of a room by two massless strings. The wire has a length of 0.11 m and a mass of 0.010 kg. A uniform magnetic field of magnitude 0.055 T is directed from the ceiling to the floor. When a current of I = 29 A exists in the wire, the wire swings upward and, at equilibrium, makes an angle φ with respect to the vertical, as the drawing shows. Find (a) the angle and (b) the tension in each of the two strings.
Answer:
Explanation:
The magnetic force acting horizontally will deflect the wire by angle φ from the vertical
Let T be the tension
T cosφ = mg
Tsinφ = Magnetic force
Tsinφ = BiL , where B is magnetic field , i is current and L is length of wire
Dividing
Tanφ = BiL / mg
= .055 x 29 x .11 / .010 x 9.8
= 1.79
φ = 61° .
Tension T = mg / cosφ
= .01 x 9.8 / cos61
= .2 N .
A gas is collected from a radioactive material; upon inspection, the gas is identified as helium. The presence of the helium indicates the radioactive sample is most likely decaying by: A). alpha B). beta+ C). beta- D). gamma
Answer:
option (a) alpha I have doublt
Consider a system of an 85.0 kg man, his 14.5-kg dog, and the earth. The gravitational potential energy of the system increases by 1.85 103 J when the man climbs a spiral staircase from the first to the second floor of an apartment building. If his dog climbs a normal staircase from the same first floor to the second floor, by how much does the potential energy of the system increase (in J)
Answer:
Explanation:
Increase in gravitational potential energy = m x g x h
where m is mass , g is gravitational acceleration and h is height
In the first case when man climbs
increase in potential = 85 x g x h = 1.85 x 10³ J
gh = 21.7647
when dog climbs
increase in potential = 14.5 x g x h J
= 14.5 x 21.7647
= 315.6 J
Two identical metal balls of radii 2.50
cm are at a center to center distance of
1.00 m from each other. Each ball is
charged so that a point at the surface of
the first ball has an electric potential of
+1.20 x 103 V and a point at the surface
of the other ball has an electric
potential of -1.20 x 103 V. What is the total charge on each ball?
Answer:
+1.33 × [tex]10^{-7}[/tex] C and -1.33 × [tex]10^{-7}[/tex] C respectively.
Explanation:
Electric potential (V) is the work done in moving a unit positive charge from infinity to a reference point within an electric field. It is measured in volts.
V = [tex]\frac{kq}{r}[/tex] ............. 1
where: k is a constant = 9 × [tex]10^{9}[/tex] N[tex]m^{2} C^{-2}[/tex], q is the charge and r is the distance between the charges.
From equation 1,
q = [tex]\frac{Vr}{k}[/tex] ............... 2
The charge on each ball can be determined as;
given that; V = 1.2 × [tex]10^{3}[/tex], k = 9 × [tex]10^{9}[/tex] N[tex]m^{2} C^{-2}[/tex] and r = 1.00 m.
From equation 2,
q = [tex]\frac{1.2*10^{3} * 1.0}{9*10^{9} }[/tex]
= 1.33 × [tex]10^{-7}[/tex] C
Thus, the charge on the first ball is +1.33 × [tex]10^{-7}[/tex] C, while the charge on the second ball is -1.33 × [tex]10^{-7}[/tex] C.
A baton twirler in a marching band competition grabs one end of her 1.2 kg, 1.0 meter long baton. She throws her baton into the air such that it rises to a height of 5.0 meters while spinning end over end at a rate of 3.5 revolutions per second. How much work did she do on the baton?
Answer:
349 J
Explanation:
Length L of baton = 1.0 m
Mass m of baton = 1.2 kg
Weight W of baton = 1.2 kg x 9.81 m/[tex]s^{2}[/tex] = 11.772 N
Height h reached = 5.0 m
Angular speed ω = 3.5 rev/s = 2π x 3.5 (rad/s) = 21.99 rad/s
Total work done on baton will be the work done in taking it to a height of 5.0 m and the kinetic energy with which the baton rolls.
Work done to bringing it to the height of 5.0 m = weight x height above ground
W x h = 11.772 x 5 = 58.86 J
Velocity v of spinning baton = ω x L = 21.99 x 1 = 21.99 m/s
Kinetic energy = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex] =
Total work done on baton = 58.86 + 290.14 = 349 J
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.
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
list and discuss how the nature of a rural settlements affect the type and expanse of agricultural activities
Answer:
Availability of land for Agricultural activities- The rural areas are known for a lesser degree of development which means lesser factories and other work buildings. The area has undeveloped lands which are usually used for a commercial type of agricultural activities.
Bad road networks: Bad road networks are mainly associated with rural settlements. This hinders to an extent the agricultural activities of planting and harvesting of crops due to difficulties in moving of the crops.
An excited hydrogen atom releases an electromagnetic wave to return to its normal state. You use your futuristic dual electric/magnetic field tester on the electromagnetic wave to find the directions of the electric field and magnetic field. Your device tells you that the electric field is pointing in the negative x direction and the magnetic field is pointing in the negative y direction. In which direction does the released electromagnetic wave travel
Answer: the magnetic wave will travel out of the screen.
Explanation:
Electric field direction is perpendicular to the magnetic field direction. Both are also perpendicular to the direction of the particles.
Using right hand rule to solve this problem,
This pointed finger depicts the electric field direction which the curly fingers depict the direction of the magnetic field. The pointed thumb will depict the direction in which the wave travel. Which is out of the screen.
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
Is light one dimensional?
Answer: No.
Explanation: Light exists in 3+1 dimensional space (3 space, 1 time).
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
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.
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
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
A 12.0-kg block is pushed across a rough horizontal surface by a force that is angled 30.0◦ below the horizontal. The magnitude of the force is 75.0 N and the acceleration of the block as it is pushed is 3.20 m/s2. What is the magnitude of the contact force exerted on the block by the surface?
Answer:
157.36 N
Explanation:
Contact force is the force which is created due to contact and it is applied on the contact point . The force applied by body on the surface is its weight .
If R be the reaction force of the ground
R = mg + F son30
= 12 x 9.8 + 75 sin 30
= 117.6 + 37.5
= 155.10 N .
friction force = f
Net force in forward direction = F cos 30 - f = ma
75cos 30 - f = 12 x 3.2
f = 65 - 38.4
= 26.6 N
Total force on the surface =√( f² + R² )
√ (26.6² + 155.1²)
= √707.56 + 24056²
=√ 24763.57
= 157.36 N.
contact force = 157.36 N .
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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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]
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
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
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.
A human expedition lands on an alien moon. One of the explorers is able to jump a maximum distance of 16.0 m with an initial speed of 2.90 m/s. Find the gravitational acceleration on the surface of the alien moon. Assume the planet has a negligible atmosphere. (Enter the magnitude in m/s2.)
Answer:
Gravitational acceleration (g) = 0.4205 m/s²
Explanation:
Given:
Distance (R) = 20 m
Initial speed (u) = 2.90 m/s
Find:
Gravitational acceleration (g)
Computation:
⇒ Distance (R) = [Initial speed (u)]²/ Gravitational acceleration (g)
⇒ Gravitational acceleration (g) = [Initial speed (u)]² / Distance (R)
⇒ Gravitational acceleration (g) = 2.90 ² / 20
⇒ Gravitational acceleration (g) = 8.41 / 20
⇒ Gravitational acceleration (g) = 0.4205 m/s²
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:
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.
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.
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
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
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