Answer:
Zero.
Explanation:
By definition, the displacement is the difference between the final position and the initial position, so independent from the distance traveled (in this case a full lap), since the initial and final position are the same, the net displacement is just Zero.If a rock is dropped from the top of a tower at the front of and it takes 3.6 seconds to hit the ground. Calculate the final velocity of the penny in m/s.
Answer:
36 m/s
Explanation:
t = 3.6s
u = 0m/s
a = +g = 10m/s²
v = ?
using,
v = u + at
v = 0 + 10(3.6)
v = 36 m/s
A ball is thrown 24 m/s into the air. How high does it go?
556.4 m
0 m
29.4 m
-556.4 m
Answer:
option c is correct
Explanation:
we know that
2as=vf^2-vi^2
vf=24 m/s
vi= 0 m/s
a=g= 9.8 m/s^2
s=vf^2-vi^2/2a
s=(24)²-(0)²/2*9.8
s=576/19.6
s=29.4 m
therefore option c is correct
Dolphins rely on echolocation to be able to survive in the ocean. In a 20 °C ocean, a dolphin produces an ultrasonic sound with a frequency of 125 kHz. What is the wavelength of this sound, in meters?While remaining stationary, the dolphin emits a sound pulse and receives an echo after 0.220 s. How far away, in meters, is the reflecting object from the dolphin?
Answer:
wavelength = 0.01 m
distance = 162.8 m
Explanation:
Given that;
Speed of sound in water = 1,480 meters per second
Frequency of ultrasound = 125KHZ
From=
v=λf
v= speed of sound
λ= wavelength of sound
f= frequency of sound
λ= 1,480 ms-1/125 * 10^3 Hz
λ= 0.01 m
From
v = 2x/t
where;
v= velocity of sound in water
x= distance traveled
t = time taken
x = vt/2
x = 1,480 ms-1 * 0.220 s/2
x= 162.8 m
. A car going initially with a velocity 15 m/s accelerates at a rate of 2 m/s2 for 10 seconds. It then accelerates at a rate of -1.5 m/s until stop. Find the car’s maximum speed. Calculate the total distance traveled by the car.
Answer:
The maximum speed of the car is 35 m/s
The total distance traveled by the car is 658.33 m
Explanation:
Given;
initial velocity of the car, u = 15 m/s
acceleration of the car, a = 2 m/s²
time of car motion, t = 10 s
(i)
Initial distance traveled by the car is given by;
d₁ = ut + ¹/₂at²
d₁ = (15 x 10) + ¹/₂(2)(10)²
d₁ = 150 + 100
d₁ = 250 m
The maximum speed of the car during this is given by;
v² = u² + 2ad₁
v² = (15)² + (2 x 2 x 250)
v² = 1225
v = √1225
v = 35 m/s
(ii)
The final distance cover by the car during the deceleration of 1.5 m/s².
Note: the final or maximum speed of the car becomes the initial velocity during deceleration.
v² = u² + 2ad₂
where;
v is the final speed of the car when it stops = 0
0 = u² + 2ad₂
0 = (35²) + (2 x - 1.5 x d₂)
0 = 1225 - 3d₂
3d₂ = 1225
d₂ = 1225 / 3
d₂ = 408.33 m
The total distance traveled by the car is given by;
d = d₁ + d₂
d = 250 m + 408.33 m
d = 658.33 m
An 7.40 kg block drops straight down from a height of 0.83 m, striking a platform spring having a force constant of 9.50 102 N/m. Find the maximum compression of the spring.
Answer:
0.25 m.
Explanation:
mass of the block = 7.40 kg, height = 0.83 m, force constant of the spring = 9.50 x [tex]10^{2}[/tex] N/m.
The maximum compression on the spring can be determined by;
Potential energy stored in the spring = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]
But, potential energy = mgh
So that,
mgh = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]
7.4 x 9.8 x 0.83 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]
60.1916 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]
[tex]x^{2}[/tex]= [tex]\frac{60.1916}{9.50*10^{2} }[/tex]
= 0.06336
x = 0.2517
x = 0.25 m
The maximum compression of the spring is 0.25 m.
A plane mirror is placed to the right of an object. The image formed by the mirror will be a
real image that appears to be on the right of the mirror.
real image that appears to be on the left of the mirror.
virtual image that appears to be on the right of the mirror.
virtual image that appears to be on the left of the mirror.
Hamish is studying what happens when he sends a sound wave through different mediums, and he records his data in a table.
A 2-column table with 4 rows titled Hamish's Waves. The first column labeled Wave has entries 1, 2, 3, 4. The second column labeled Information has entries liquid, solid, gas, liquid.
Which statement could made about the data collected in Hamish’s table?
Wave 1 will move the fastest.
Wave 2 will move the slowest.
Wave 3 will move the slowest.
Wave 4 will move the fastest.
What is common between transverse waves and longitudinal waves?
Both include an amplitude, crest, and rarefactions
Both move faster at higher temperatures
Both move slower through densely packed molecules
Both include a wavelength from compression to compression
An angle of refraction is the angle between the refracted ray and the
incident ray.
normal.
medium.
boundary.
Answer:
A plane mirror is placed to the right of an object. The image formed by the mirror will be a virtual image that appears to be on the left of the mirror.
Explanation:
A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s.
Complete Question
A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?
Answer:
The position of the object at t = 10s is [tex]X = 38.3 \ m[/tex]
Explanation:
From the question we are told that
The acceleration along the x axis is [tex]a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)[/tex]
The position of the object at t = 0 is x = -14.0 m
The velocity at t = 0 s is [tex]v_{0}x = 7.10 m/s[/tex]
Generally from the equation for acceleration along x axis we have that
[tex]a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)[/tex]
=> [tex]\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt[/tex]
=> [tex]V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]
At t =0 s and [tex]v_{0}x = 7.10 m/s[/tex]
=> [tex]7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1[/tex]
=> [tex]K_1 = 7.10[/tex]
So
[tex]\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]
=> [tex]\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}[/tex]
=> [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2[/tex]
At t =0 s and x = -14.0 m
[tex]-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2[/tex]
=> [tex]K_2 = -14[/tex]
So
[tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14[/tex]
At t = 10.0 s
[tex]X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14[/tex]
=> [tex]X = 38.3 \ m[/tex]
What would happen if there is more male hyenas than female hyenas in a population?
Choices:
Male hyenas will compete to mate with the females.
Some male hyenas will die.
Male hyenas for wait for more females to join the population.
Answer:
Option 1
Explanation:
I always see animals do that
A model of a helicopter rotor has four blades, each 3.4 m in length from the central shaft to the tip of the blade. The model is rotated in a wind tunnel at 550 rev/min. What is the radial acceleration of the blade tip, expressed as a multiple of the acceleration g due to gravity?
A. (5.72 × 104)g
B. (6.23 × 102)g
C. (1.15 × 103)g
D. (2.25 × 103)g
David Wetterman drops a 5 kg watermelon from the top of a 30 m building. What is the velocity of the watermelon as it smashes
into the ground (neglecting air resistance)?
-(1)
A)
24.25 m/s
B)
32.45 m/s
C)
60 m/s
D)
588 m/s
Answer:
A. 24.25 m/s
Explanation:
velocity = [tex]\sqrt{2 * g * d}[/tex]
velocity = sqr 2 * 9.8 * 30 = sqr 588 = 24.25 m/s
The velocity of the watermelon as it smashes into the ground will be 24.2 m/s
State the third equation of motion?
The third equation of motion is -
v² - u² = 2aS
Given David Wetterman drops a 5 kg watermelon from the top of a 30 m building.
Height of building [S] = 30 m
Mass of watermelon [M] = 5 Kg
Initial velocity [v] = 0 m/s
acceleration [g] = 9.8 m/s²
Using the third equation of motion -
v² - u² = 2aS
v² = 2aS
v² = 2 x 9.8 x 30
v² = 588
v = 24.2 m/s
Therefore, the velocity of the watermelon as it smashes into the ground will be 24.2 m/s.
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A radio wave transmits 38.5 W/m2 of power per unit area. A flat surface of area A is perpendicular to the direction of propagation of the wave. Assuming the surface is a perfect absorber, calculate the radiation pressure on it.
Answer:
[tex]P=2.57\times 10^{-7}\ N/m^2[/tex]
Explanation:
Given that,
A radio wave transmits 38.5 W/m² of power per unit area.
A flat surface of area A is perpendicular to the direction of propagation of the wave.
We need to find the radiation pressure on it. It is given by the formula as follows :
[tex]P=\dfrac{2I}{c}[/tex]
Where
c is speed of light
Putting all the values, we get :
[tex]P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2[/tex]
So, the radiation pressure is [tex]2.57\times 10^{-7}\ N/m^2[/tex].
A 5.3 kg block rests on a level surface. The coefficient of static friction is μ_s=0.67, and the coefficient of kinetic friction is μ_k= 0.48 A horizontal force, x is applied to the block. As x is increased, the block begins moving. Describe how the force of friction changes as x increases from the moment the block is at rest to when it begins moving. Show how you determined the force of friction at each of these times ― before the block starts moving, at the point it starts moving, and after it is moving. Show your work.
As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.
Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.
The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:
n + (-w) = 0
n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N
The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be
f = 0.67 (51.94 N) ≈ 35 N
so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.
The block starts moving as soon as x = 35 N, at which point f = 35 N.
At any point after the block starts moving, we have
f = 0.48 (51.94 N) ≈ 25 N
so that x = 25 N is the required force to keep the block moving at a constant speed.
As x is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force
Magnitude of normal force ( object at rest ); n = 51.94 N Required magnitude of x before the movement of object ; x = 35 NMagnitude of x after object start moving x = 25 NGiven data :
mass of block at rest ( m ) = 5.3 kg
Coefficient of static friction ( μ_s ) =0.67
Coefficient of kinetic friction is ( μ_k ) = 0.48
Horizontal force applied to block = x
First step : magnitude of normal force ( n ) when object is at rest
n = w where w = m*g
n - w = 0
n - ( 5.3 * 9.81 ) = 0 ∴ n = 51.94 N
Second step : Required magnitude of x before the movement of object
F = μ_s * n
F = 0.67 * 51.94 = 34.79 N ≈ 35 N
∴ The object will start moving once F and x = 35 N
Final step : Magnitude of x after object start moving
F = μ_k * n
= 0.48 * 51.94 = 24.93 N ≈ 25 N
∴ object will continue to accelerate at a constant speed once F and x = 25N
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A rolling ball moves from x1 = 8.0 cm to x2 = -4.1 cm during the time from t1 = 2.9 s to t2 = 6.0 s .
Complete Question
A rolling ball moves from [tex]x_1 = 8.0 \ cm[/tex] to [tex]x_2 = - 4.1 \ cm[/tex] during the time from [tex]t_1 = 2.9 s[/tex] to [tex]t_2 = 6.0s[/tex]
What is its average velocity over this time interval?
Answer:
The velocity is [tex]v = 3.903 \ m/s[/tex]
Explanation:
From the question we are told that
The first position of the ball is [tex]x_1 = 8.0 \ cm[/tex]
The second position of the ball is [tex]x_2 = - 4.1 \ cm[/tex]
Generally the average velocity is mathematically represented as
[tex]v = \frac{ x_1 - x_2}{t_2 - t_1}[/tex]
=> [tex]v = \frac{ 8 - -4.1 }{ 6 - 2.9 }[/tex]
=> [tex]v = 3.903 \ m/s[/tex]
A household refrigerator consumes electrical energy at the rate of 200 W. lf electricity costs 5 k per kWh, calculate the cost of operating the appliance for 30 days
Answer:
= 720000 [k]
Explanation:
The cost is equal to 5 [$/kW-h], kilowatt per hour, this value should be multiplied by the power, and then by the time.
[tex]5[\frac{k}{kw*h}]*200[w]*30[day]*24[\frac{h}{day} ][/tex]
= 720000 [k]
Acceleration is sometimes expressed in multiples of g, where g = 9.8 m/s^2 is the magnitude of the acceleration due to the earth's gravity. In a test crash, a car's velocity goes from 26 m/s to 0 m/s in 0.15 s. How many g's would be experienced by a driver under the same conditions?
Answer:
Acceleration = 18g
Explanation:
Given the following data;
Initial velocity, u = 26m/s
Final velocity, v = 0
Time = 0.15 secs
To find the acceleration;
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.
Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]
Substituting into the equation, we have;
[tex]a = \frac{0 - 26}{0.15}[/tex]
[tex]a = \frac{26}{0.15}[/tex]
Acceleration = 173.33m/s2
To express it in magnitude of g;
Acceleration = 173.33/9.8
Acceleration = 17.7 ≈ 18g
Acceleration = 18g
How many significant figures are in 0.0067?
Answer:
2
Explanation:
there are 2 significant figures in there
Need help ASAP..please help
Answer:
option 3
Explanation:
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it is the question 12 part okay
Answer:
Yeah it's ok I think. Also, I can't see the answer you gave so maybe updating the question would be nice.
A car’s brakes decelerate it at a rate of -2.40 m/s2. If the car is originally travelling at 13 m/s and comes to a stop, then how far, in meters, will the car travel during that time?
Answer:
Approximately [tex]35.2\; \rm m[/tex].
Explanation:
Given:
Initial velocity: [tex]u = 13\; \rm m \cdot s^{-1}[/tex].
Acceleration: [tex]a = -2.40\; \rm m \cdot s^{-2}[/tex] (negative because the car is slowing down.)
Implied:
Final velocity: [tex]v = 0\; \rm m \cdot s^{-1}[/tex] (because the car would come to a stop.)
Required:
Displacement, [tex]x[/tex].
Not required:
Time taken, [tex]t[/tex].
Because the time taken for this car to come to a full stop is not required, apply the SUVAT equation that does not involve time:
[tex]\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{{\left(0\; \rm m \cdot s^{-1}\right)}^2 - {\left(13\; \rm m \cdot s^{-1}\right)}^2}{2\times \left(-2.40\; \rm m\cdot s^{-2}\right)} \approx 35.2\; \rm m \end{aligned}[/tex].
In other words, this car would travel approximately [tex]35.2\; \rm m[/tex] before coming to a stop.
Measurements of the radioactivity of a certain isotope tell you that the decay rate decreases from 8255 decays per minute to 3110 decays per minute over a period of 4.50 days.
What is the half-life (T1/2) of this isotope?
Answer:half-life (T1/2) of this isotope =
Explanation:
The number of nuclei of any radioactive substance at a given time is expressed by
Nt = N0e⁻kt
Nt=decay of material at a time t, =3110 decays per minute
N=decays at t=0, 8255 decays per minute
k=constant
Nt=N0e−kt
3110= 8255 e⁻k(4.50)
3110/ 8255=e−k(4.50)
0.3767 = e−k(4.50)
In 0.3767 = -k (4.50)
0.976=-4.5k
k=0.976/4.5
=0.2159
Also we know that t 1/2= time that it takes half the original material to decay.it is related to the rate constant by
T₁/₂=ln 2 / k
Therefore half-life (T1/2) of this isotope
T₁/₂=ln 2/0.2159
T₁/₂=3.12 days
A block slides down an inclined plane from rest. Initially the block is at 4.5m above the ground. Find the speed of the block when it is 1.5m above the ground. 1) 7.7m/s 2) 9.4m/s 3) 5.4m/s 4) 3.2m/s
Since, no external force is acting , so the system is in equilibrium .
Initial total energy = Final total energy
[tex]mg(4.5) = mg(1.5) + \dfrac{mv^2}{2}\\\\\dfrac{v^2}{2}=3\times g \\\\v^2=3\times 9.8\times 2\\\\v = \sqrt{58.8}\ m/s\\\\v = 7.67 \ m/s[/tex] ( Here , g = acceleration due to gravity = 9.8 m/s² )
Therefore, option 1) is correct.
Hence, this is the required solution.
someone help please
waves disturb ____, but do not transmit it.
a. energy
b. matter
c. sound
d. none of the above
Answer:
b. matter
Explanation:
Waves disturb matter but do not transmit it.
Waves are disturbances that transmit energy from one point to another. Although they cause disturbances, they do not transfer the matters in the medium.
Energy is propagated by a wave. When for example, sound waves are produced, the disturbance is propagated via particle - particle interaction But after the wave train moves, the particles remain.Three moles of a monatomic ideal gas are heated at a constant volume of 1.20 m3. The amount of heat added is 5.22x10^3 J.(a) What is the change in the temperature of the gas?________ ? K(b) Find the change in its internal energy.________ ? J(c) Determine the change in pressure.________ ? Pa
Answer:
A) 140 k
b ) 5.22 *10^3 J
c) 2910 Pa
Explanation:
Volume of Monatomic ideal gas = 1.20 m^3
heat added ( Q ) = 5.22*10^3 J
number of moles (n) = 3
A ) calculate the change in temp of the gas
since the volume of gas is constant no work is said to be done
heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT
make ΔT subject of the equation
ΔT = Q / n.(3/2).R
= (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )
= 140 K
B) Calculate the change in its internal energy
ΔU = Q this is because no work is done
therefore the change in internal energy = 5.22 * 10^3 J
C ) calculate the change in pressure
applying ideal gas equation
P = nRT/V
therefore ; Δ P = ( n*R*ΔT/V )
= ( 3 * 8.3144 * 140 ) / 1.20
= 2910 Pa
A) The change in the temperature of the gas is; ΔT = 139.5 K
B) The change in internal energy of the gas is; ΔU = 5.22 × 10³ J
C) The change in pressure of the gas is; ΔP = 2899.5 Pa
We are given;
Volume of Monatomic ideal gas; V = 1.2 m³
Amount of heat added; Q = 5.22 × 10³ J
number of moles; n = 3
A) To calculate the change in temperature of the monatomic idea gas, we will use formula;
Q = ³/₂nRΔT
Where R is a constant = 8.314 J/mol.K
ΔT is the change in temperature
Making ΔT the subject of the formula;
ΔT = ²/₃(Q/(nR))
ΔT = ²/₃(5.22 × 10³)/(3 × 8.314)
ΔT = 139.5 K
B) Due to the fact that no work was done, then from first law of thermodynamics, we can say that;
ΔU = Q
Thus;
change in internal energy; ΔU = 5.22 × 10³ J
C) The change in pressure will be calculated from the formula;
ΔP = (n*R*ΔT)/V
ΔP = (3 * 8.314 * 139.5)/1.2
ΔP = 2899.5 Pa
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When a potential difference of 10 V is placed across a certain solid cylindrical resistor, the current through it is 2 A. If the diameter of this resistor is now tripled, the current will be:______.A) 18 A.
B) 2/3 A.
C) 3 A.
D) 2/9 A.
E) 2 A.
Answer:
sorry I wish I could it help you
In the winter sport of curling, players give a 20 kg stone a push across a sheet of ice. The Slone moves approximately 40 m before coming to rest. The final position of the stone, in principle, onlyndepends on the initial speed at which it is launched and the force of friction between the ice and the stone, but team members can use brooms to sweep the ice in front of the stone to adjust its speed and trajectory a bit; they must do this without touching the stone. Judicious sweeping can lengthen the travel of the stone by 3 m.1. A curler pushes a stone to a speed of 3.0 m/s over a time of 2.0 s. Ignoring the force of friction, how much force must the curler apply to the stone to bring it op to speed?A. 3.0 NB. 15 NC. 30 N
D. 150 N2The sweepers in a curling competition adjust the trajectory of the slope byA. Decreasing the coefficient of friction between the stone and the ice.
B. Increasing the coefficient of friction between the stone and the ice.C. Changing friction from kinetic to static.D. Changing friction from static to kinetic.3. Suppose the stone is launched with a speed of 3 m/s and travel s 40 m before coming to rest. What is the approximate magnitude of the friction force on the stone?A. 0 NB. 2 NC. 20 ND. 200 N4. Suppose the stone's mass is increased to 40 kg, but it is launched at the same 3 m/s. Which one of the following is true?A. The stone would now travel a longer distance before coming to rest.B. The stone would now travel a shorter distance before coming to rest.C. The coefficient of friction would now be greater.D. The force of friction would now be greater.
Answer:82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.
The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2
Fd = (1/2)mv^2
F = (1/2)mv^2/d.
Plug in m = 20 kg, v = 3 m/sec, d = 40 m.
83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.
Explanation:Hope I helped :)
Is Geothermal Energy renewable? Why or why not? Use in your own words.
Answer:
Yes, geothermal energy is a renewable energy resource because the water can be heated by pumping it through the rocks.
What is magnet made of
Answer:
metals like iron or nickel
Explanation:
Two identical wind-up cars A and B are released. Car B has a 2 kilogram weight strapped to the back of the car. Which will have the greatest average speed towards the end of the motion?
A)Car A
B)They will both have an average speed of zero.
C)They will have the same average speed.
D)Car B
E)There is not enough information to answer.
Answer:
Car A would have a better average speed
Explanation:
added weight to a object that is self propelled will be slower than a identical object with no added weight
What is the energy contained in a 0.950 m3 volume near the Earth's surface due to radiant energy from the Sun?
In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give? Let's find out by doing a calculation in a very artificial situation. Suppose a man standing on frictionless ice fires a rifle horizontally. The mass of the man together with the rifle is 70 kg, and the mass of the bullet is 10 g. If the bullet leaves the muzzle at a speed of 500 m/s, what is the final speed of the man?
Answer:
Explanation:
m1v1=m2v2
m1=70 kg
m2=10 g=0.01 kg
v2=500 m/s
m1v1=m2v2
v1=m2v2/m1
v1=0.01*500/70
v1=0.07