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-Dominant- [34]
1 year ago
12

A motorboat starting from rest travels in a straight line on a lake. If the boat reahes a speed of 8.0m/s in 10s what is the boa

t's average acceleration?
Physics
1 answer:
victus00 [196]1 year ago
5 0

The boat's average acceleration is  0.8 m/s².

<h3>What is acceleration?</h3>

Acceleration is defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

A motorboat starting from rest travels in a straight line on a lake. If the boat reaches a speed of 8.0m/s in 10s.

The final velocity v = 8 m/s and it starts from rest, then its initial velocity v = 0 m/s.

Time taken for the change in speed t = 10 s.

The acceleration is given by

a = (8 - 0)/ 10

a = 0.8 m/s²

Thus, the boat's average acceleration is 0.8 m/s².

Learn more about acceleration.

brainly.com/question/12550364

#SPJ1

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Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

<u>Part A)</u>

Let's use the parabolic motion equation to solve it. Let's define the variables:

  • y(i) is the initial height, it is a constant.
  • y(f) is the final height, in our case is 0
  • v(i) is the initial velocity (v(i)=16 m/s)
  • θ1 is the first angle, 30°
  • θ2 is the first angle, -30°

For the first stone

y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}              

0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}

0=y_{i1}+8t_{1}-4.905*t_{1}^{2} (1)  

For the second stone  

0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}    

0=y_{i2}-8t_{2}-4.905t_{2}^{2} (2)            

 

If we solve the equation (1) we will have:

t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}  

We can do the same procedure for the equation (2)

t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

<u>Part B)</u>

We can use the equation of the horizontal position here.

<u>First stone</u>

x_{f1}=x_{i1}+vcos(30)t_{1}

x_{f1}=0+13.86*t_{1}

x_{f1}=13.86*t_{1}

<u>Second stone</u>

x_{2}=x_{i2}+vcos(-30)t_{2}

x_{1}=0+13.86*t_{1}

x_{1}=13.86*t_{2}

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

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The change in state of every matter is accompanied by lost or gained of energy.

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The solid state of water is ice. The motion of particles of the water is relatively zero because the molecules are held at a fixed position.

The liquid state of water occurs when the temperature of the ice is increased above zero degree Celsius. The speed of the particles of water in liquid state is greater than solid state.

The gaseous state of water occurs when the temperature of the liquid water is increased beyond 100 degree Celsius. The speed of water in gaseous state is greater than liquid state.

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