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Crank
3 years ago
15

A 0.500 kg ball is dropped from rest at a point 1.20 m above the floor. The ball rebounds straight upward to a height of 0.700 m

a) What is the velocity one instant before the collision?b) what is the velocity one instant after the collision with the floor?c) what is the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?
Physics
1 answer:
KatRina [158]3 years ago
6 0

To solve this problem we will apply the concepts related to energy conservation. With this we will find the speed before the impact. Through the kinematic equations of linear motion we will find the velocity after the impact.

Since the momentum is given as the product between mass and velocity difference, we will proceed with the velocities found to calculate it.

Part A) Conservation of the energy

mgh = \frac{1}{2} mv^2

v_1 = \sqrt{2gh}

v_1 = \sqrt{2(9.8)(1.20)}

v_1 = 4.84m/s

\therefore v_1 = -4.84/s \rightarrow \text{Negative direction downward}

Part B)  Kinematic equation of linear motion,

v_2^2 = u_0^2 +2a\Delta y

Here

v= 0 Because at 1.5m reaches highest point, so v=0

0 = u_2^2 +2(-9.8)(0.7)

u_2 = 3.7m/s

Therefore the velocity after the collision with the floor is 3.7m/s

PART C) Total change of impulse is given as,

J = P_2 -P_1

J = mU_2-mV_1

J = m(U_2-V_1)

J = (0.5)(3.7-(-4.84))

J = 4.27kg \cdot m/s \rightarrow \text{Upward because is positive}

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Why are some substances dissolve both polar and nonpolar substances?
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<em></em>

<em></em>

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8 0
3 years ago
A circular radar antenna on a Coast Guard ship has a diameter of 2.10 m and radiates at a frequency of 16.0 GHz. Two small boats
Anna35 [415]

Answer:

d = 76.5 m

Explanation:

To find the distance at which the boats will be detected as two objects, we need to use the following equation:

\theta = \frac{1.22 \lambda}{D} = \frac{d}{L}

<u>Where:</u>

θ: is the angle of resolution of a circular aperture

λ: is the wavelength

D: is the diameter of the antenna = 2.10 m

d: is the separation of the two boats = ?

L: is the distance of the two boats from the ship = 7.00 km = 7000 m

To find λ we can use the following equation:

\lambda = \frac{c}{f}

<u>Where:</u>

c: is the speed of light = 3.00x10⁸ m/s

f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz

\lambda = \frac{c}{f} = \frac{3.00 \cdot 10^{8} m/s}{16.0 \cdot 10^{9} s^{-1}} = 0.0188 m            

Hence, the distance is:

d = \frac{1.22 \lambda L}{D} = \frac{1.22*0.0188 m*7000 m}{2.10 m} = 76.5 m

Therefore, the boats could be at 76.5 m close together to be detected as two objects.

 

I hope it helps you!

7 0
2 years ago
Timed plz hurry
puteri [66]

Answer:

b

Explanation:

emphasizes self-importance

4 0
3 years ago
Read 2 more answers
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