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Yuki888 [10]
3 years ago
13

If a ball is dropped from rest, what is it’s velocity after 4 seconds

Physics
1 answer:
Ghella [55]3 years ago
4 0

Answer: -39.2 m/s directed downwards

Explanation:

We can solve this problem with the following equation:

V=V_{o}+gt

Where:

V is the ball's velocity at 4 s

V_{o}=0m/s is the ball's intial velocity, since it was dropped from rest

g=-9.8 m/s^{2} is the acceleration due gravity, aways directed downwards

t=4 s is the time

Solving:

V=0m/s-9.8 m/s^{2}(4 s)

Finally:

V=-39.2 m/s <u>The negative sign indicates the velocity is downwards</u>

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Jim strikes a 0.058 kg golf ball with a force of 272 N, giving it a velocity of 62.0 m/s. How long was the ball in contact with
Mekhanik [1.2K]

Answer: 0.013 seconds

Explanation:

Given that

Mass of golf ball = 0.058 kg

Force = 272 N

Velocity = 62.0 m/s

Time taken = ?

Recall that force is the rate of change of momentum per unit time

i.e Force = Change in momentum / Time

i.e Force = (Mass x velocity) / Time

272N = (0.058 kg x 62.0 m/s) / Time

272N = 3.6kgm/s / Time

Time = (3.6kgm/s / 272N)

Time = 0.013 seconds

Thus, the ball was in contact with the club for 0.013 seconds

8 0
4 years ago
A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with
gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s

The block's speed is 31.422 cm/s

4 0
3 years ago
A pitcher releases a fastball that moves toward home plate.
Vsevolod [243]

The two additional forces that act on the ball as it travels between the pitcher and the home plate are air resistance and gravity.

<h3>What are the forces that affect object in motion;</h3>
  • Air resistance: this is the force that oppose the motion of an object in air due to frictional force
  • Gravity: this is the force due to weight of the object and acts downwards.

The two additional forces that act on the ball as it travels between the pitcher and the home plate include:

  1. Air resistance and
  2. Gravitational force  

<h3>How the forces affect the motion of the ball</h3>
  • Air resistance oppose the motion of the ball as it travels in air.
  • Gravity is the force due to weight of the ball and acts downwards.

Learn more about forces on object in motion here: brainly.com/question/10454047

5 0
3 years ago
50 kg of water at 75o C is cooled to 25o C. How much heat was given off?
attashe74 [19]

Answer:

b the answer is b

Explanation:

b is the awnser because it cools after the heat on the water witch lets the steam out

8 0
2 years ago
Find the period of a wave with a frequency 40 kHz​
Burka [1]

We know that frequency is an inverse value of time f=\dfrac{1}{t} this implies that time is the inverse value of frequency t=\dfrac{1}{f}.

Now since <em>f</em> is <em>40kHz</em><em>,</em><em> </em>we can calculate period or duration for that matter time.

t=\dfrac{1}{40\mathbf{kHz}}=0.000025\mathbf{s}=25\mathbf{\mu s}

Hope this helps.

r3t40

4 0
3 years ago
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