Answer:
7 minutes
Explanation:
From the question, the flow of the river applies equally to both the hat and the canoe. The best way to approach this problem is to look at it from a perspective that follows the water, and ignores the bank.
Hence, the water and the hat are stationary, and the man paddles away from the hat and then paddles back.
Now, the hat is at rest and relative to the water and the canoe always travels at a speed of 4 m/s relative to the water.
We are told that it takes 7 minutes for the man to travel away from the hat.
Since he is travelling at a constant speed, it will therefore take same time of 7 minutes for him to return.
So, it will take 7 minutes for the man to row back upriver to reclaim his hat.
A body of mass m=1kg is moving straight-line and its path as a function of time is given by the following
function: s= A - Bt+Ct - Dr, here C=10m/s2 and D=1m/s3 are some constants. What is the
magnitude of the force acting on the body at instant t=1s?
Answer: The force at t = 1s is 14 N
Explanation:
I guess that the position equation actually is: (by looking at the units of C and D)
S = A - B*t + C*t^2 - D*t^3
As you may know by Newton's third law, if we want to find the force, we first need to find the acceleration, and before that, the velocity.
the velocity can be found by integrating over time, the velocity is:
V = 0 - B + 2*C*T - 3*D*t^2
For the acceleration we integrate again:
A = 0 + 2*C - 6D*T
and we know that F = m*A
then:
Force(t) = 1kg*(2*10m/s^2 - 6*1m/s^3*t)
We want the force at t = 1s, so we replace t by 1s.
F(1s) = 1kg*(20m/s^2 - 6m/s^2) = 14 N
What is unique about visible
light in the electromagnetic
spectrum?
Answer:
The visible light spectrum is the segment of the electromagnetic spectrum that the human eye can view. More simply, this range of wavelengths is called visible light. Typically, the human eye can detect wavelengths from 380 to 700 nanometers.
Explanation:
Answer:
Visible light, which travels at a dizzying 186,282 miles per second through space, is just one part of light's broad spectrum, which encompasses all electromagnetic radiation. We can detect visible light because of cone-shaped cells in our eyes that are sensitive to the wavelengths of some forms of light
Explanation:
Consider the boundary-value problem introduced in the construction of the mathematical model for the shape of a rotating string: T d2y dx2 + rhoω2y = 0, y(0) = 0, y(L) = 0. For constants T and rho, define the critical speeds of angular rotation ωn as the values of ω for which the boundary-value problem has nontrivial solutions. Find the critical speeds ωn and the corresponding deflections yn(x). (Give your answers in terms of n, making sure that each value of n corresponds to a unique critical speed.)
Answer:
[tex]y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x[/tex]
[tex]y_n(x) = C_n \sin \frac{n \pi x}{L}[/tex]
Explanation:
The given differential equation is
[tex]T\frac{d^2y}{dx^2} + \rho w ^2y=0[/tex] and y(0) = 0, y(L) =0
where T and ρ are constants
The given rewrite as
[tex]\frac{d^2y}{dx^2} + \frac{\rho w^2}{T} y=0[/tex]
auxiliary equation is
[tex]m^2+ \frac{\rho w^2}{T} =0\\\\m= \pm\sqrt{\frac{\rho}{T} } wi[/tex]
Solution of this de is
[tex]y(x)=C_1 \cos \sqrt{\frac{\rho}{t} } wx + C_2 \sin \sqrt{\frac{\rho}{T} } wx[/tex]
y(0)=0 ⇒ C₁ = 0
[tex]y(x) = C_2 \sin \sqrt{\frac{\rho}{T} } wx[/tex]
y(L) = 0 ⇒
[tex]C_2 \sin \sqrt{\frac{\rho}{T} } wL=0[/tex]
we need non zero solution
⇒ C₂ ≠ 0 and
[tex]\sin \sqrt{\frac{\rho}{T} } wL=0[/tex]
[tex]\sin \sqrt{\frac{\rho}{T} } wL=0 \rightarrow \sqrt{\frac{\rho}{T} } wL=n \pi[/tex]
[tex]w_n = \sqrt{\frac{T}{\rho} } \frac{n \pi}{L}[/tex]
solution corresponding these [tex]w_n[/tex] values
[tex]y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x[/tex]
[tex]y_n(x) = C_n \sin \frac{n \pi x}{L}[/tex]
A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 3.5 s for the boat to travel from its highest point to its lowest, a total distance of 0.69 m. The fisherman sees that the wave crests are spaced 6.6 m apart. Part A How fast are the waves traveling
Answer:
v= 0.94 m/s
Explanation:
In order to calculate the speed of the waves, you use the following formula for the speed of the waves:
[tex]v=\lambda f[/tex] (1)
v: speed of the wave
λ: wavelength of the wave
f: frequency of the wave
The frequency is calculated by using the information about the time that boat takes to travel from its highest point to its lowest point. This time is a half of a period:
[tex]\frac{T}{2}=3.5s\\\\T=7.0s[/tex]
Then, the frequency is:
[tex]f=\frac{1}{T}=\frac{1}{7s}=0.142Hz[/tex]
The wavelength of the wave is the distance between crests of the wave
[tex]\lambda=6.6m[/tex]
With the values of the frequency and the wavelength, you can find the speed of the wave by using the equation (1):
[tex]v=(6.6m)(0.142Hz)=0.94\frac{m}{s}[/tex]
The speed of the wave is 0.94m/s
adeen says that you can increase the resistance of a copper wire by hammering the wire to make it narrower and longer. Arnell says that you can increase its resistance by heating the wire. Which one, if either, is correct, and why
Answer:
Both of Arnell and Adeen are right.
Explanation:
For Adeen, hammering the wire makes it flat, effectively reducing the cross sectional area, and increasing the length of the wire. Recall that the resistivity of a metal conductor increases with length and decreases with cross sectional area. From this, we can see that the resistance of the wire will increase due to the hammering. This means Adeen is correct.
For Arnell, heating a metal causes the atoms along with their electrons to vibrate in a random manner. These random motions is not organised and can be in any direction. For electricity to flow smoothly, the electrons must travel in the same direction in an orderly fashion. The randomly vibrating atoms will continuously bump into the flowing electron, resisting the smooth motion of the electrons. This collisions results in an additional resistance, hence, the resistance is increased. This means Arnell is correct.
Think of something from everyday life that follows a two-dimensional path. It could be a kicked football, a bus that's turning a corner, or a person jogging around a track, etc. Describe your scenario in detail, and then identify the acceleration at each point. When is the acceleration vector not aligned with the direction of travel
Answer:
Let us consider the case of a bus turning around a corner with a constant velocity, as the bus approaches the corner, the velocity at say point A is Va, and is tangential to the curve with direction pointing away from the curve. Also, the velocity at another point say point B is Vb and is also tangential to the curve with direction pointing away from the curve. Although the velocity at point A and the velocity at point B have the same magnitude, their directions are different (velocity is a vector quantity), and hence we have a change in velocity. By definition, an acceleration occurs when we have a change in velocity, so the bus experiences an acceleration at the corner whose direction is away from the center of the corner.
The acceleration is not aligned with the direction of travel because the change in velocity is at a tangent (directed away) to the direction of travel of the bus.
A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. The hoist weighs 350 N. The ropes, fastened at different heights, make angles of 50° and 38° with the horizontal. Find the tension in each rope and the magnitude of each tension.
Answer:
276 N and 225 N
Explanation:
Draw a free body diagram. There are three forces on the hoist:
Weight force 350 N pulling down,
Tension force T₁ pulling up and left 50° from the horizontal,
Tension force T₂ pulling up and right 38° from the horizontal.
Sum of forces in the x direction:
∑F = ma
T₂ cos 38° − T₁ cos 50° = 0
T₂ cos 38° = T₁ cos 50°
T₂ = T₁ cos 50° / cos 38°
Sum of forces in the y direction:
∑F = ma
T₂ sin 38° + T₁ sin 50° − 350 = 0
T₂ sin 38° + T₁ sin 50° = 350
Substitute:
(T₁ cos 50° / cos 38°) sin 38° + T₁ sin 50° = 350
T₁ cos 50° tan 38° + T₁ sin 50° = 350
T₁ (cos 50° tan 38° + sin 50°) = 350
T₁ = 350 / (cos 50° tan 38° + sin 50°)
T₁ = 276 N
T₂ = T₁ cos 50° / cos 38°
T₂ = 225 N
A point charge is located at the center of a thin spherical conducting shell of inner and outer radii r1 and r2, respectively. A second thin spherical conducting shell of inner and outer radii R1 and R2, respectively, is concentric with the first shell. The flux is as follows for the different regions of this arrangement.
Φ = −10.3 ✕ 103 N · m2/C for r < r1Φ = 0 for r1 < r < r2Φ = 36.8 ✕ 103 N · m2/C for r2 < r < R1Φ = 0 for R1 < r < R2Φ = −36.8 ✕ 103 N · m2/C for r > R2
Complete Question
The complete question is shown on the first uploaded image
Answer:
The point charge is [tex]Q_z = -0.0912 \ \mu C[/tex]
The inner shell is [tex]Q_t = 0.4168 \ \mu C[/tex]
The outer shell is [tex]Q_w = -0.6514 \ \mu C[/tex]
Explanation:
From the question we are told that
The inner radius of thin first spherical conducting shell is [tex]r_1[/tex]
The outer radius of thin first spherical conducting shell is [tex]r_2[/tex]
The inner radius of second thin spherical conducting shell is [tex]R_1[/tex]
The outer radius of second thin spherical conducting shell is [tex]R_2[/tex]
The magnetic flux for different region is [tex]\phi = -10.3 *10^3 N\cdot m^2 /C \ for \ r < r_1[/tex]
The magnetic flux for first shell is [tex]\phi = 36 * 10^3 N \cdot m^2 /C \ for \ r_2 < r <R_1[/tex]
The magnetic flux for second shell is [tex]\phi = -36 * 10^3 N \cdot m^2 /C \ for \ r <R_1[/tex]
The magnitude of the point charge is mathematically represented as
[tex]Q_z = \ \phi_z * \epsilon _o[/tex]
[tex]Q_z = -10.3*10^{3} * 8.85 *10^{-12}[/tex]
[tex]Q_z = -9.115*10^{-8} \ C[/tex]
[tex]Q_z = -0.0912 \ \mu C[/tex]
Considering the inner shell
[tex]Q_a = \phi_a * \epsilon _o[/tex]
=> [tex]Q_a = 36 .8 * 10^3 * 8.85*10^{-12}[/tex]
[tex]Q_a = 32.56*10^{-8} \ C[/tex]
[tex]Q_a =0.326} \ \mu C[/tex]
Charge on the inner shell is
[tex]Q_t = Q_a - Q_z[/tex]
[tex]Q_t = 0.326} \ \mu - ( -0.0912 \ \mu)[/tex]
[tex]Q_t = 0.4168 \ \mu C[/tex]
Considering the outer shell
[tex]Q_y = \phi_y * \epsilon_o[/tex]
=> [tex]Q_y = -36.8 *10^{3} * 8.85*10^{-12}[/tex]
[tex]Q_y = -32.56*10^{-8} \ C[/tex]
[tex]Q_y = - 0.326} \ \mu C[/tex]
Charge on the outer shell is
[tex]Q_w = Q_y - Q_z[/tex]
[tex]Q_w =- 0.326} \ \mu - ( -0.0912 \ \mu)[/tex]
[tex]Q_w = -0.6514 \ \mu C[/tex]
The motor on a helicopter turns at an angular speed of 6.2 x 102 revolutions per minute. (a) Express this angular speed in radians per second.
(b) If the rotor has a radius of 3.0 m, what arc length (linear displacement) tip of the blade trace out in 2.0 x 102 s?
Answer:
64.93 rad/s38.956 kmExplanation:
(a)
[tex]\dfrac{6.2\cdot 10^2\,\text{rev}}{60\,\text{s}}\times\dfrac{2\pi\,\text{rad}}{\text{rev}}=\dfrac{62\pi\,\text{rad}}{3\,\text{s}}\approx\boxed{64.93\,\text{rad/s}}[/tex]
__
(b)
[tex]d=r\theta=(3.0\,\text{m})(2.0\cdot 10^2\,\text{s})\left(\dfrac{62\pi\,\text{rad}}{3\,\text{s}}\right)=124\pi\cdot 10^2\,\text{m}=\boxed{38\,956\,\text{m}}[/tex]
Two bars are conducting heat from a region of higher temperature to one of lower temperature. The bars have identical lengths and cross-sectional areas, but are made from different materials. In the drawing they are placed in "parallel" between the two temperature regions arrangement A, while they are placed end-to-end in arrangement B. In which arrangement is the heat conducted the greatest?
A) Arrangement A
B) The heat conducted is the same in both arrangements.
C) Arrangement B
D) It is not possible to determine which arrangement conducts more heat
Answer:
A) Arrangement A
Explanation:
The rate of heat conduction is given by Fourier's Law of Heat Conduction. It is given as follows:
Q = KAΔT/L
where,
Q = Rate of Heat Transfer or Conduction
K = Thermal Conductivity
A = Cross-Sectional Area
ΔT = Difference in Temperature
L = Thickness
So, it is clear from the formula that for a constant temperature difference and value of thermal conductivity, the rate of heat transfer is directly proportional to the cross-sectional area and it is inversely proportional to the thickness.
Therefore, the arrangement with larger cross-sectional area and smaller thickness will be the one with the greatest heat transfer rate and as a result greatest heat shall be conducted through that arrangement.
It is clear that the parallel arrangement that is arrangement A, has higher cross-sectional area and smaller thickness. Therefore, the correct option is:
A) Arrangement A
A An ____ occurs when a very low resistance circuit is formed and can easily start a fire.
parallel circuit
series circuit
short circuit
voltage divider
Question 9 (4 points)
Which device measures the current in a circuit?
Answer:
1- Short circuit
2- ammeter
A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling horizontally with a speed of 250 km/hr (69.4 m/s) how far in advance of the recipients (horizontal distance) must the goods be dropped
Answer:
481 m
Explanation:
To fall 235 m, the time required is
t = √(2H/g)
t= √(2[tex]\times[/tex]235/9.8)
t=6.92 seconds.
The supplies will travel forward
6.92 [tex]\times[/tex] 69.4 ≈ 481 m
Therefore, the goods must be dropped 481 m in advance of the recipients.
A seesaw consisting of a uniform board of mass M and length L, supports at rest a father and daughter with masses "mf " and "md", respectively.
The support ( called the fulcrum ) is under the centre of gravity of the board , the father is the distance "d" from the center and the daughter is a distance "L/2" from the center.
A) determine the magnitude of the upward force "n" exerted by the support on the board ?
B) determine where the father should sit to balance the system at rest ?
Answer:
A) N = (M + mf + md) g
B) d = md L / (2 mf)
Explanation:
Draw a free body diagram. There are four forces acting on the board.
Weight force M g pulling down at the center of the board.
Normal force N pushing up at the center of the board.
Weight force mf g pulling down a distance d from the center.
Weight force md g pulling down at the end of the board.
Sum of forces in the y direction:
∑F = ma
N − M g − mf g − md g = 0
N = (M + mf + md) g
Sum of moments about the center of the board:
∑τ = Iα
md (L/2) − mf d = 0
d = md L / (2 mf)
The magnitude of the upward force n exerted by the support on the board will be the same with the magnitude of the weight of the seesaw. Which is
W = 9.8M Newton
The father should sit at L/4 at the other end in order to balance the system at rest
Since the seesaw is of a uniform board, the center of gravity will act at the center. If it is of length L, the support called the fulcrum will be positioned at L/2.
If the daughter is a distance L/2 from the center, Then, she must be positioned at L/4 from the one edge of the board.
A. The magnitude of the upward force n exerted by the support on the board will be the same with the magnitude of the weight of the seesaw. Which is
W = mg
W = 9.8M Newton
The weight of both the father and the daughter will act downward. so, they will not be included.
The father should sit at L/4 at the other end in order to balance the system at rest
Learn more here: https://brainly.com/question/15651723
An aluminium bar 600mm long with diameter 40mm has a hole drilled in the centre of the bar.The hole is 3omm in diameter and is 100mm long.If the modulus of elasticity for the aluminium is 85GN/m^2. Calculate the total contraction on the bar due to a compresive load of 180kn
Given that,
Length of bar = 600 mm
Diameter of bar = 40 mm
Diameter of hole = 30 mm
Length of hole = 100 mm
Modulus of elasticity = 85 GN/m²
Load = 180 kN
We need to calculate the area of cross section without hole
Using formula of area
[tex]A=\dfrac{\pi\times d^2}{4}[/tex]
Put the value into the formula
[tex]A=\dfrac{\pi\times40^2}{4}[/tex]
[tex]A=1256.6\ mm^2[/tex]
We need to calculate the area of cross section with hole
Using formula of area
[tex]A=\pi\times\dfrac{(d_{b}^2-d_{h}^{2})}{4}[/tex]
Put the value into the formula
[tex]A=\pi\times\dfrac{(40^2-30^2)}{4}[/tex]
[tex]A=549.77\ mm^2[/tex]
We need to calculate the total contraction on the bar
Using formula of total contraction
Total contraction = contraction in bar without hole part + contraction in bar with hole part
[tex]Total\ contraction = \dfrac{F\times L_{1}}{A_{1}\times E}+\dfrac{F\times L_{2}}{A_{2}\times E}[/tex]
Where, F = load
L = length
A = area of cross section
E = modulus of elasticity
Put the value into the formula
[tex]Total\ contraction=\dfrac{180\times10^3}{85\times10^{3}}(\dfrac{500}{1256.6}+\dfrac{100}{549.77})[/tex]
[tex]Total\ contraction = 1.227\ mm^2[/tex]
Hence, The total contraction on the bar is 1.227 mm²
Assume the earth to be a nonrotating sphere with mass MEME and radius RERE. If an astronaut weights WW on the ground, what is his weight when he is 2RE2RE above the surface of the earth
Answer:
The weight at a distance 2 RE from surface of earth is W/9
Explanation:
For the value of acceleration due to gravity (g), we have a formula, that is:
g = (G)(ME)/(RE)² ----- equation (1)
where,
G = Gravitational Constant
ME = Mass of Earth
RE = Radius of Earth
g = Acceleration due to gravity on surface of earth = 9.8 ms²
When the person goes 2RE, distance above earth's surface. Then the total distance from center of earth becomes: 2RE + RE = 3RE.
Therefore, equation (1) becomes:
gh = (G)(ME)/(3RE)²
where,
gh = acceleration due to gravity at height
gh = (G)(ME)/(RE)²9
using equation (1), we get:
gh = g/9
Now, he weight is given by formula:
W = mg ------- equation (2)
At height 2RE
Wh = (m)(gh)
where,
Wh = Weight at height = ?
m = mass of astronaut
Therefore, using vale of gh, we get:
Wh = mg/9
Using equation (2), we get:
Wh = W/9
What structures can be found in the axial region of the body?
Answer:
The axial region of the body consists of the bones of the head, trunk of a vertebrate, skull, vertebral column, and thoracic cage. The human skeleton consists of 80 bones.
Explanation:
The axial region of the body consists of the bones of the head, trunk of a vertebrate, skull, vertebral column, and thoracic cage. The human skeleton consists of 80 bones.
It is composed of the following six parts:
1. Skull (22 bones)
2. Ossicles of the middle ear
3. Hyoid bone
4. Rib cage
5. Sternum
6. Vertebral column
The axial region of the body forms the vertical axis of the body as the axial skeleton supports the head, neck, back, and chest.
Exercise 2: A 0.6 kg particle has a speed of 2 m / s at point A and kinetic energy of 7.5 J at point B. What is
(a) its kinetic energy at A?
(b) its speed at B?
(c) the total work done on the particle as it moves from A to B?
Explanation:
We have,
Mass of a particle is 0.6 kg
Speed at A is 2 m/s
Kinetic energy at B is 7.5 J
(a) The kinetic energy at A is given by :
[tex]K_A=\dfrac{1}{2}mv^2\\\\K_A=\dfrac{1}{2}\times 0.6\times 2^2\\\\K_A=1.2\ J[/tex]
(b) Kinetic energy at B is given by
[tex]K_B=\dfrac{1}{2}mV^2\\\\V=\sqrt{\dfrac{2K_B}{m}} \\\\V=\sqrt{\dfrac{2\times 7.5}{0.6}} \\\\V=5\ m/s[/tex]
(c) The work done on the particle as it moves form A to B is given by work energy theorem as :
[tex]W=\dfrac{1}{2}m(V^2-v^2)\\\\W=\dfrac{1}{2}\times 0.6\times (5^2-2^2)\\\\W=6.3\ J[/tex]
A candle is placed 14 cm in front of a concave mirror. The image of the candle is focused on a sheet of paper that is exactly 21 cm in front of the mirror. what is the magnification of the image
Answer:
The magnification of the image is [tex]M = -1.5[/tex]
Explanation:
From the question we are told that
The object distance is [tex]u = 14 cm[/tex]
The image distance is [tex]v = 21 \ cm[/tex]
The magnification of the image is mathematically represented as
[tex]M = - \frac{v}{u}[/tex]
substituting values
[tex]M = - \frac{21}{14}[/tex]
[tex]M = -1.5[/tex]
The negative value means that the image is real , inverted and enlarged
During normal beating, a heart creates a maximum 3.95-mV potential across 0.305 m of a person’s chest, creating a 0.75-Hz electromagnetic wave. During normal beating of the heart, the maximum value of the oscillating potential difference across 0.305 m of a person’s chest is 3.95 mV. This oscillating potential difference produces a 0.75-Hz electromagnetic wave.
What is the maximum electric field created?
Answer:
E = 0.0130 V/m.
Explanation:
The electric field is related to the potential difference as follows:
[tex] E = \frac{\Delta V}{d} [/tex]
Where:
E: is electric field
ΔV: is the potential difference = 3.95 mV
d: is the distance of a person's chest = 0.305 m
Then, the electric field is:
[tex]E = \frac{\Delta V}{d} = \frac{3.95 \cdot 10^{-3} V}{0.305 m} = 0.0130 V/m[/tex]
Therefore, the maximum electric field created is 0.0130 V/m.
I hope it helps you!
Now suppose that the Earth had the same mass and radius as it currently does, but all of the mass was concentrated into a thin (1 ft. thick) hollow spherical shell below your feet. Also suppose there was a small hole in the shell, just big enough four you to fit through. Compare the force of gravity on you outside the shell to the force of gravity if you stepped inside the shell.
Answer:
The force of gravity at the shell will be extremely great on me due to the huge mass collapsed into the small radius.
At the center of the shell, the gravitational forces all around should cancel out, giving me a feeling of weightlessness; which will be a lesser force compared to that felt while standing on the shell.
Explanation:
For the collapsed earth:
mass = 5.972 × 10^24 kg
radius = 1 ft
according to Newton's gravitation law, the force of gravity due to two body with mass is given as
Fg = GMm/[tex]R^{2}[/tex]
Where Fg is the gravitational force between the two bodies.
G is the gravitational constant
M is the mass of the earth
m is my own mass
R is the distance between me and the center of the earths in each case
For the case where I stand on the shell:
radius R will be 1 ft
Fg = GMm/[tex]1^{2}[/tex]
Fg = GMm
For the case where I stand stand inside the shell, lets say I'm positioned at the center of the shell. The force of gravity due to my mass will be balanced out by all other masses around due to the shell of the hollow earth. This cancelling will produce a weightless feeling on me.
Two large parallel conducting plates separated by 6 cm carry equal and opposite surface charge densities such that the electric field between them is uniform. The difference in potential between the plates is 400 V. An electron is released from rest at the negatively charged plate.
(a) What is the magnitude of the electric field between the plates?
(b) Which plate is at the higher potential?
1. The negatively charged plate
2. the positively charged plat
Answer:
a. 6666.67 V/m
b. 2. the positively charged plat
Explanation:
a. The computation of the magnitude of the electric field between the plates is shown below:
As we know that
[tex]E = \frac{V}{d}[/tex]
[tex]= \frac{400 V}{0.06 m}[/tex]
= 6666.67 V/m
hence, the magnitude of the electric field is 6666.67 V/m
b. Based on this the higher potential is positively charged plate as the flow of the current goes from positive to negative and it is inverse in case of th electron
We simply applied the above formula
The sun generates both mechanical and electromagnetic waves. Which statement about those waves is true?
ОА
The mechanical waves reach Earth, while the electromagnetic waves do not.
ОВ.
The electromagnetic waves reach Earth, while the mechanical waves do not.
Ос. Both the mechanical waves and the electromagnetic waves reach Earth.
OD. Neither the mechanical waves nor the electromagnetic waves reach Earth.
Answer:
B: The electromagnetic waves reach earth, while the mechanical waves do not.
correct Answer:
The electromagnetic waves reach Earth, while the mechanical waves do not
Enterprising students set an enormous slip-n-slide (a plastic sheetcovered in water to reduce friction) on flat ground. If the slip-n-slideis 250 m long, how small does the average acceleration have to be fora student starting at 5 m/s to slide to the end
Answer:
a = -0.05 m/s² (negative sign shows deceleration)
Explanation:
In order, to find out the minimum average acceleration for a student starting at 5 m/s to slide to the end, we can use 3rd equation of motion. 3rd equation of motion is given as follows:
2as = Vf² - Vi²
where,
a = minimum acceleration required = ?
s = minimum distance covered = 250 m
Vf = Final Speed = 0 m/s (for minimum acceleration the student will barely cover 250 m and then stop)
Vi = Initial Velocity = 5 m/s
Therefore,
2a(250 m) = (0 m/s)² - (5 m/s)²
a = - (25 m²/s²)/(500 m)
a = -0.05 m/s² (negative sign shows deceleration)
Attempting to impress the skeptical patrol officer with your physics knowledge, you claim that you were traveling so fast that the red light (693 nm) appeared yellow (582 nm) to you. How fast would you have been traveling (in mi/hr) if that had been the case?
Answer:
v_r = 1.268 × 10⁸ mi/hr
Explanation:
We are given;
wavelength of the red light; λr = 693 nm = 693 × 10^(-9) m
wavelength of the yellow light; λy = 582 nm = 582 × 10^(-9) m
Frequency is given by the formula;
f = v/λ
Where v is speed of light = 3 x 10^(8) m
Frequency of red light; f_o = [3 x 10^(8)]/(693 × 10^(-9)) = 4.33 x 10¹⁴ Hz
Similarly, Frequency of yellow light;
f = [3 x 10⁸]/(582 × 10^(-9)) = 5.15 x 10¹⁴ Hz
To find the speed of the car, we will use the formula;
f = f_o[(c + v_r)/c)]
Where c is speed of light and v_r is speed of car.
Making v_r the subject;
cf/f_o = c + v_r
v_r = c(f/f_o - 1)
So, plugging in the relevant values, we have;
v_r = 3 × 10⁸[((5.15 x 10¹⁴)/(4.33 x 10¹⁴)) - 1]
v_r = 3 × 10⁸(0.189)
v_r = 5.67 x 10⁷ m/s
Converting to mi/hr, 1 m/s = 2.23694 mile/hr
So, v_r = 5.67 × 10⁷ × 2.23694
v_r = 1.268 × 10⁸ mi/hr
A toy car's movements is measured using photogates.
Answer:
a) the velocity increases then decreases.
how you would teach the habit to your students.
Answer:
Probably just enforcing it little by little. (If you mean a habit. If you mean a certain topic it's different.)
Explanation:
If you are trying to teach your students a habit-
Try to correct them everytime they don't follow it. Expect them to do so and give a reward when they earn it.
If you mean a certain topic-
Make a slideshow or a presentation with an interactive game. Making it not boring and having them more attentive.
I gave a more general answer, so if you would like to go deeper into the subject I would be more happy to.
(If this helped, please let me know. If you have any questions do not be afriad to ask!)
momentum is closely related to
Answer:
Momentum is mainly related to forces
You are on an interstellar mission from the Earth to the 8.7 light-years distant star Sirius. Your spaceship can travel with 70% the speed of light and has a cylindrical shape with a diameter of 6 m at the front surface and a length of 25 m. You have to cross the interstellar medium with an approximated density of 1 hydrogen atom/m3.
(a) Calculate the time it takes your spaceship to reach Sirius.
(b) Determine the mass of interstellar gas that collides with your spaceship during the mission.
Note: Use 1.673 ×10−27 kg as proton mass.
Answer:
Explanation:
TJVFKBFVN
A positive test charge q is released from rest at distance r away from a charge of +Q and a distance 2r away from a charge of +2Q. 1)How will the test charge move immediately after being released?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct option is the second option
Explanation:
Generally the electric force exerted by the charge Q on the charge (q) is mathematically represented as
[tex]F_Q = \frac{kqQ}{r^2}[/tex]
Generally the electric force exerted by the charge 2Q on the charge (q) is mathematically represented as
[tex]F_{2Q} = \frac{kq2Q}{2r^2}[/tex]
Now the net force exerted on q is
[tex]F_{net} = \frac{kqQ}{r^2} - \frac{2k q Q}{4r^2}[/tex]
[tex]F_{net} = \frac{4kqQ- 2kqQ}{4r^2}[/tex]
[tex]F_{net} = \frac{kqQ}{2r^2}[/tex]
Looking at the resulting equation we see that [tex]F_{net} > 0[/tex]
This implies that the charge q would move to the right
A very small source of light that radiates uniformly in all directions produces an electric field with an amplitude of ܧ at a distance R from the source. What is the amplitude of the magnetic field at a point 2R from the source?
A very small source of light that radiates uniformly in all directions produces an electric field with an amplitude of ܧ at a distance R from the source. What is the amplitude of the magnetic field at a point 2R from the source?
If the distance from the source is doubled. The amplitude of the magnetic field is smaller 4 times.