The velocity of a 150 kg cart changes from 6.0 m/s to 14.0 m/s. What is the magnitude of the impulse that acted on it?
2 answers:
Answer: The magnitude of impulse acted on the cart will be 1200 kg.m/s
Explanation:
Impulse is defined as the product of mass of the object and the change in its velocity. It is basically the change in momentum of the body.
Mathematically,
where,
J = Impulse
m = mass of the object
= change in the velocity
We are given:
Putting values in above equation, we get:
Hence, the magnitude of impulse acted on the cart will be 1200 kg.m/s
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The work done by Joe is 0 J.
<u>Explanation</u>:
When a force is applied to an object, there will be a movement because of the applied force to a certain distance. This transfer of energy when a force is applied to an object that tends to move the object is known as work done.
The energy is transferred from one state to another and the stored energy is equal to the work done.
W = F . D
where F represents the force in newton,
D represents the distance or displacement of an object.
Force = 0 N, D = 20 cm = 0.20 m
W = 0 0.20 = 0 J.
Hence the work done by Joe is 0 J.
Answer:
Part A:
Distance=864000 m=864 km
Part B:
Energy Used=ΔE=8638000 Joules
Part C:
Explanation:
Given Data:
v=20m/s
Time =t=12 hours
In Secs:
Time=12*60*60=43200 secs
Solution:
Part A:
Distance = Speed**Time
Distance=v*t
Distance= 20*43200
Distance=864000 m=864 km
Part B:
Energy Used=ΔE= Energy Required-Kinetic Energy of swans
Energy Required to move= Power Required*time
Energy Required to move=200*43200=8640000 Joules
Kinetic Energy=
Energy Used=ΔE=8640000 -2000
Energy Used=ΔE=8638000 Joules
Part C:
Fraction of Mass used=Δm/m
For This first calculate fraction of energy used:
Fraction of energy=ΔE/Energy required to move
ΔE is calculated in part B
Fraction of energy=8638000/8640000
Fraction of energy=0.99977
Kinetic Energy=
Now, the relation between energies ratio and masses is:
Answer:
a) 6 times farther. b) 6 times longer.
Explanation:
Once released, in the horizontal direction, no other forces act on the ball, so it continues moving at the same initial velocity, which is given by the projection of the velocity vector in the horizontal direction, as follows:
vₓ = v* cos (25º) = 23 m/s * 0.906 = 20.8 m/s
In the vertical direction, the initial velocity is the projection of the velocity vector along the vertical axis, as follows:
vy = v* sin (25º) = 23 m/s * 0.422 = 9.72 m/s
Assuming that the acceleration is constant, and equal to 1/6*g, we can calculate the total time of flight, with the following kinematic equation for the vertical displacement:
If the total displacement in the vertical direction is 0 (which means that the time if the total time of flight), we can solve for t, as follows:
On earth, this time could be calculated in the same way:
As the time is defined by the vertical movement, we can find the horizontal distance travelled on the moon, as follows:
Δx = v₀ₓ * t = 20.8 m/s * 11. 9 s = 248.1 m
On earth, the distance travelled had been as follows:
Δx = v₀ₓ * t = 20.8 m/s * 1.98 s = 41.3 m
⇒ Δx(moon) / Δx(earth) = 248.1 / 41.3 = 6.00
b) As we have just found, the time of flight on the moon and on the earth are as follows:
tmoon = 11. 9 s
tearth = 1.98 s
⇒ t(moon) / t(earth) = 11.9 / 1.98 = 6.0