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LekaFEV [45]
3 years ago
13

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

Physics
1 answer:
aliina [53]3 years ago
3 0

Answer:

9.43 m/s

Explanation:

First of all, we calculate the final kinetic energy of the car.

According to the work-energy theorem, the work done on the car is equal to its change in kinetic energy:

W=K_f - K_i

where

W = -36.733 J is the work done on the car (negative because the car is slowing down, so the work is done in the direction opposite to the motion of the car)

K_f is the final kinetic energy

K_i = 66,120 J is the initial kinetic energy

Solving,

K_f = K_i + W = 66,120 + (-36,733)=29,387 J

Now we can find the final speed of the car by using the formula for kinetic energy

K_f = \frac{1}{2}mv^2

where

m = 661 kg is the mass of the car

v is its final speed

Solving for v, we find

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.43 m/s

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The boiling point of sulfur is 444.6 celsius .sulfur melting point is 586.1 fahrenheit lower than its boiling?
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(a) Melting point is 136.8°C

(b) Melting point is 278.24°F

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a ship travels a port p and travels 30 km due north. then it changes course and travels 20 km in a direction  30° east of north
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When we represent what is given to us on a coordinate plane, we have a figure as shown in the attachment.

To find the distance between P and R, we have to find the Net Displacement of the ship (brown arrow in the figure).

For that, we use the rules for Vector addition.

We see that the first displacement D_{1} = 30 km (blue arrow) is along the y-axis, but the second part of the ship's journey D_{2} = 20 km (red arrow) is at an angle with reference to y-axis.

So, we first find the components of the red arrow along X and Y.

Component of D_{2} along X-axis is given by  D_{2x}  = D_{2} Sin 30 = 10 km

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We now add all the vectors along X and along Y separately.

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Now that we have the components of the net displacement along X and Y, we make use of Pythagorean Theorem to calculate the D_{net}

D_{net}  = \sqrt{D_{netX} ^{2} + D_{netY} ^{2}}

Therefore, [tex]D_{net} = 48.37 km.

Hence, the distance between the ports P and R is 48 km.

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