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Zolol
2 months ago
9

A tennis ball bounces on the floor three times. If each time it loses 22.0% of its energy due to heating, how high does it rise

after the third bounce, provided we released it 2.3 m from the floor?
Physics
1 answer:
Ostrovityanka [3.2K]2 months ago
6 0

Answer:

H = 109.14 cm

Explanation:

Given,                                                            

Assume that the total energy equals 1 unit.                                

Energy remaining after the first collision = 0.78 x 1 unit

Balance after the first impact = 0.78 units

Remaining energy after the second impact = 0.78 ^2 units

Balance after the second impact = 0.6084 units

Remaining energy after the third impact = 0.78 ^3 units

Balance after the third impact = 0.475 units

The height reached after the third collision is equivalent to the remaining energy.

Let H denote the height achieved after three bounces.

0.475 (m g h) = m g H                  

H = 0.475 x h                                    

H = 0.475 x 2.3 m                          

H = 1.0914 m                      

H = 109.14 cm                      

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Keith_Richards [3271]

Answer:

The resulting velocity for him will be 0.187 m/s in reverse direction.

Explanation:

Given:

The mass of the man is, M=75\ kg

The mass of the ball is, m=4\ kg

The initial velocity of the man is, u_m=0\ m/s(rest)

The initial velocity of the ball is, u_b=0\ m/s(rest)

The final velocity of the ball is, v_b=3.50\ m/s

The final velocity of the man is, v_m=?\ m/s

To determine this scenario, we employ the principle of momentum conservation.

This principle states that the total initial momentum equals the total final momentum.

Momentum is calculated by multiplying mass by velocity.

Initial momentum = Initial momentum of the man and the ball

Initial momentum = Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm

Final momentum = Final momentum of the man and the ball

Final momentum = Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14

Hence, the total initial momentum equals the total final momentum

0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s

The negative sign indicates that the man moves backward.

Thus, his final velocity ends up being 0.187 m/s backward.

3 0
1 month ago
A Federation starship (8.5 ✕ 106 kg) uses its tractor beam to pull a shuttlecraft (1.0 ✕ 104 kg) aboard from a distance of 14 km
Keith_Richards [3271]
Consider the diagram below.

m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ =  acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted

Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t²                  (1)
x = 0.5a₂t²                  (2)

F = m₁a₁ = m₂a₂         (3)
Additionally,
x + y = 14000 m          (4)

From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²

From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s

Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m

The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.

Result: The starship shifts by 16.5 m (to the nearest tenth)

3 0
2 months ago
An object is at rest on the ground. The object experiences a downward gravitational force from Earth. Which of the following pre
Yuliya22 [3333]

Answer:

A) and B) are valid.

Explanation:

When an object remains at rest, it is indicative that no net force acts upon it.

The downward gravitational force from Earth must be counterbalanced by an upward force of equal magnitude in order to maintain rest.

This upward force is provided by the normal force, which adjusts to satisfy Newton’s 2nd Law and is always perpendicular to the surface supporting the object (in this instance, the ground).

At the molecular level, this normal force comes from the ground's bonded molecules acting like tiny springs, compressed by the object’s molecules, providing an upward restorative force.

Thus, statements A) and B) are true.

6 0
2 months ago
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For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italia
serg [3582]

Answer:

Part a)

A = 0.0581 m

Part b)

T = 0.37 s

Explanation:

A slice is dropped onto the plate from a height of 0.250 m,

therefore the speed of the slice upon impact is calculated as

v = \sqrt{2gh}

We know that

v = \sqrt{2(9.81)(0.250)}

v = 2.21 m/s

Now applying the conservation of momentum:

mv = (m + M)v_f

m = 0.300 kg

M = 0.400 kg

From this equation, we find:

0.300 (2.21) = (0.300 + 0.400) v_f

v_f = 0.95 m/s

0.400 (9.81) = 200 x_1

When the slice rests on the plate, the new mean position can be expressed as

x_1 = 0.01962 m

(0.300 + 0.400)9.81 = 200 x_2

We also determine that the speed of SHM is represented as

x_2 = 0.0343 m

Here, we derive values from

v = \omega\sqrt{A^2 - x^2}

\omega = \sqrt{\frac{k}{m + M}}

\omega = \sqrt{\frac{200}{0.300 + 0.400}}

\omega = 16.9 rad/s

a = x_2 - x_1 = 0.0343 - 0.01962 = 0.0147 m

Using the previous formula gives:

0.95 = 16.9\sqrt{A^2 - 0.0147^2}

A = 0.0581 m

Part b)

The time period for the scale is computed as

T = \frac{2\pi}{\omega}

T = \frac{2\pi}{16.9}

T = 0.37 s

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2 months ago
Paula is studying two different animals. Both animals are classified within the same genus, but they are different species. Base
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