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3241004551 [841]
2 years ago
8

A 0.5kg ball falls from a building that Is 50m high. how much kinetic energy will it have when it has fallen half way to the gro

und?
Physics
1 answer:
castortr0y [4]2 years ago
4 0
Calculate velocity at halfway to the ground.

vfinal = root 2ad

v = root (2*9.81m/s^2*25)

v = 69.367175234 m/s

Kinetic energy = 1/2mv^2

Kinetic energy = 1/2 * (69.367175234 m/s^s^2

Ek = 2405.9025 Joules
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Discuss the limitations of using the Doppler shift to determine an object's speed.
pantera1 [17]

Answer and Explanation:

Limitation of Doppler shift :

The Doppler impact is relevant when the speeds of the wellspring of sound and spectator are considerably less than the speed of sound. The movement of both the spectator and the source is along a similar straight line.When movement is not in straight line or velocity is not much less than speed of light then we can not use Doppler shift

This is the limitation of Doppler shift to determine the object distance

3 0
3 years ago
Without steering, a car (mass 1501 kg) can travel with a certain speed around a banked frictionless corner angled at 25.0° to th
natima [27]

Answer:

Its tricky but am still working on it

6 0
3 years ago
A car tire rotates with an average angular speed of 32 rad/s. In what time interval will the tire rotate 3.5 times? Answer in un
stiks02 [169]

Answer:

\Delta t = 0.687\ s

Explanation:

given,

Angular speed of the tire = 32 rad/s

Displacement of the wheel = 3.5 rev

Δ θ = 3.5 x 2 π

        = 7 π rad

now,

Time interval of the car to rotate 7π rad

using equation

\omega_{avg} =\dfrac{\Delta \theta}{\Delta t}

\Delta t=\dfrac{7\pi}{32}

\Delta t = 0.687\ s

Time taken to rotate 3.5 times is equal to 0.687 s.

3 0
2 years ago
A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at
pav-90 [236]

Answer:

Explanation:

Given

mass of drop m=2 kg

height of fall h=1 m

ball leaves the foot with a speed of 18 m/s at an angle of 55^{\circ}

Velocity of ball just before the collision with the floor

u^=2gh

u=\sqrt{2gh}

u=\sqrt{2\times 9.8\times 1}=4.42 m/s

Impulse delivered in Y direction

J_y=m(v\sin (55)-(-u))

J_y=2(18\sin (55)+4.42)

J_y=38.32 kg-m/s

Impulse in x direction

J_x=m\times v\cos (55)

J_x=2\times 4.42\cos (55)=20.646

J_{net}=\sqrt{J_x^2+J_y^2}

J_{net}=\sqrt{(38.32)^2+(20.64)^2}

J_{net}=43.52 kg-m/s

at an angle of \tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}

\phi =tan^{-1}(1.856)

\phi =61.7^{\circ}  

7 0
2 years ago
Two people are pushing a car to the right. The mass of the car is 1000.0 kg. One person applies a force of 275 N to the car, whi
givi [52]

Explanation:

Below is an attachment containing the solution.

7 0
2 years ago
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