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Black_prince [1.1K]
3 years ago
8

A rubber ball and a steel ball are dropped from the same height onto a concrete floor. They have the same mass, and lets assume

that they both rebound to the same height after hitting the ground. How would the average forces experienced by the balls compare. Group of answer choices The steel ball would experience the greater average force The average forces would be the same The rubber ball would experience the greater average force
Physics
1 answer:
Mkey [24]3 years ago
5 0

Answer:

The average forces would be the same

Explanation:

Both have the same velocity on impact as they fell from the same height.

Both have the same velocity after the bounce because they reach the same height.

Both have the same mass

Both will thus experience the same impulse because both have the same change in momentum.

Therefore both experience the same average force.

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Sodium is a highly reactive metal and chlorodyne is a poisonous gas which compound do they for a chemically combines
stellarik [79]
A metal and a non-metal will form salt. Na+ and Cl- will form NaCl or commonly known as table salt
5 0
3 years ago
A ball whose mass is 0.3 kg hits the floor with a speed of 5 m/s and rebounds upward with a speed of 2 m/s. If the ball was in c
Sonbull [250]

Answer:

1400 N

Explanation:

Change in momentum equals impulse which is a product of force and time

Change in momentum is given by m(v-u)

Equating this to impulse formula then

m(v-u)=Ft

Making F the subject of the formula then

F=\frac {m(v-u)}{t}

Take upward direction as positive then downwards is negative

Substituting m with 0.3 kg, v with 2 m/s, and u with -5 m/s and t with 0.0015 s then

F=\frac {0.3(2--5)}{0.0015}=1400N

4 0
3 years ago
A 200g air-track glider is attached to a spring. The glider is pushed in 10cm and released. A student with a stopwatch finds tha
cricket20 [7]

The spring constant will be k= 5.5N/m for a 200g air track glider attached to a spring.

<h3>What is spring constant?</h3>

The spring constant, k, is a measure of the stiffness of the spring. It is different for different springs and materials.

Calculation for What is the spring constant

First step is to calculate the time period

T = 12 second/10

T = 1.2 second

Now let calculate the spring constant using this formula

k=\dfrac{4\pi ^2m}{T^2}

Where,

m=0.2kg

T=1.2second

k represent spring constant=?

Let plug in the formula

k=\dfrac{4 \pi \times 0.2}{(1.2)^2}

k=\dfrac{39.48\times 0.2}{1.44}

k=\dfrac{7.90}{1.44}

k=5.48 N/m

k=5.5 N/m ( Approximately)

Therefore the spring constant will be 5.5 N/m

To know more about spring constant follow

brainly.com/question/1968517

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4 0
2 years ago
Clara rushes 30m to a truck then turns and walks back. Total travel is 120s what is her average velocity?
navik [9.2K]

Taking into account the definition of velocity, Clara's average velocity is 0.5 m/s.

<h3>Definition of velocity</h3>

Velocity is a physical magnitude that relates the displacement of an object, the time it takes to make this change in position and direction. So it is considered a vector magnitude.

In other words, the velocity can be defined as the amount of space traveled per unit of time with which a body moves, considering the direction, and can be calculated using the expression:

velocity= distance traveled÷ time

<h3>Average velocity of Clara</h3>

Clara rushes 30 m to a truck then turns and walks back. Total travel is 120s. Then, you know:

  • distance traveled= 30m rushing + 30m walking back= 60 m
  • time= 120 s

Replacing in the definition of velocity:

velocity= 60 m÷ 120 s

Solving:

<u><em>velocity= 0.5 m/s</em></u>

Finally, Clara's average velocity is 0.5 m/s.

Learn more about velocity:

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brainly.com/question/25288184

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5 0
2 years ago
If the angular velocity of the pully is -8.4rad/s at a given time, and its anglar acceleration is -2.8rad/s2, what is the angula
snow_lady [41]

Here we know that

\omega_i = - 8.4 rad/s

\alpha = - 2.8 rad/s^2

t = 1.5 s

now from kinematics we have

\omega_f = \omega_i + \alpha t

now from above all values we have

\omega_f = (-8.4 rad/s) + (-2.8 rad/s^2)(1.5)

\omega_f = -8.4 + (-4.2)

\omega_f = -12.6 rad/s

so final angular speed is -12.6 rad/s

6 0
3 years ago
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