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laila [671]
4 years ago
15

If a rock is skipped into a lake at 24 m/s2, with that what force was the rock thrown if it was 1.75kg?

Physics
1 answer:
erma4kov [3.2K]4 years ago
6 0

Answer: f= M×A

1.75kg×24= 42N

Explanation:

Because to find force you do Mass times acceleration so I did 1.75 kg times 24 would equal 42 Newtons!

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Alberta is going to have dinner at her grandmother's house, but she is running a bit behind schedule. As she gets onto the highw
viktelen [127]

Complete Question

Alberta is going to have dinner at her grandmother's house, but she is running a bit behind schedule. As she gets onto the highway, she knows that she must exit the highway within 55 min if she is not going to arrive late. Her exit is 43 mi away. How much time would it take at the posted 60 mph speed?

Answer:

The time it would take at the given speed is  x = 43.00 \ minutes

Explanation:

From the question we are told that

      The time taken to exist the highway is  t = 55 min

      The distance to the exist is  d =  43\  mi

       Alberta speed is v =  60 mph

The time it would take travelling at the given speed is mathematically represented as

        t_z = \frac{d}{v}

  substituting values

        t_z = \frac{43}{60}

        t_z = 0.71667\ hrs

Converting to minutes

         1  hour =  60 minutes

So      0.71667 hours = x minutes

   Therefore

               x = 0.71667 * 60

                x = 43.00 \ minutes

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3 years ago
Scientists have learned about the structure of Earth's interior by
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C) studying seismic waves

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Because they are different they all show different traits.
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During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau
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Answer:

T=6.75s

Explanation:

We must separate the motion into two parts, the first when the rocket's engines is on  and the second when the rocket's engines is off. So, we need to know the height (h_1) that the rocket reaches while its engine is on and we need to know the distance (h_2) that it travels while its engine is off.

For solving this we use the kinematic equations:

In the first part we have:

h_1=v_0T+\frac{1}{2}aT^2\\h_1=0*T+\frac{1}{2}(16\frac{m}{s^2})T^2\\h_1=8\frac{m}{s^2}T^2\\

and the final speed is:

v_f=v_0+aT\\v_f=0+16\frac{m}{s^2}T\\v_f=16\frac{m}{s^2}T

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.

So, we have:

v_f^2=v_0^2+2gh_2\\2gh_2=v_f^2-v_0^2\\h_2=\frac{v_f^2-v_0^2}{2g}\\h_2=\frac{0^2-(16\frac{m}{s^2}T)^2}{2(-9.8\frac{m}{s^2})}\\h_2=\frac{-256\frac{m^2}{s^4}T^2}{-19.6\frac{m}{s^2}}\\h_2=13.06\frac{m}{s^2}T^2

The sum of these heights will give us the total height, which is known:

h=h_1+h_2\\960m=8\frac{m}{s^2}T^2+13.06\frac{m}{s^2}T^2\\960m=21.06\frac{m}{s^2}T^2\\T^2=\frac{960m}{21.06\frac{m}{s^2}}\\T^2=45.58s^2\\T=\sqrt{45.58s^2}\\T=6.75s

This is the time that its needed in order for the rocket to reach the required altitude.

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Answer:

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