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liraira [26]
3 years ago
12

a stone is vertically thrown upward with the velocity of 72km/hr find the maximum height reached the height​

Physics
1 answer:
erik [133]3 years ago
3 0

Answer:

y = 20.38 [m]

Explanation:

In order to solve these problems, we must use the following kinematics equation.

v_{f} ^{2} =v_{i} ^{2}-(2*g*y)

where:

Vf = final velocity = 0

Vi = initial velocity = 72 [km/h]

g = gravity acceleration = 9.81 [m/s^2]

y = vertical elevation [m]

We need to convert [km/h] to [m/s]

72[\frac{km}{h}]*[\frac{1h}{3600s}]*[\frac{1000m}{1km} ] = 20 [m/s]

Note: the negative sign of the equation means that the acceleration acts in the opposite direction to the movement of the body. And the final speed is zero, because when the body reaches the maximum height, the Stone does not move its speed has been reduced to its entirety.

0 = (20)^2 - (2*9.81*y)

20^2 = 2*9.81*y

y = 20.38 [m]

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A 600 g model rocket is on a cart that is rolling to the right at a speed of 4.0 m/s. The rocket engine, when it is fired, exert
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Answer:

The rocket should be launched when the cart is 13.48m away from a point directly below the hoop.

Explanation:

Step 1: Data given

mass of the rocket = 600 grams

speed = 4.0 m/s

Step 2: Calculate weight

Fw = mg

with Fw = the weight (in Newton)

with m = the mass (in kg)

with g = the acceleration due to gravity (9.81 m/s²).

Fw = (0.600 kg)(9.81 m/s²)  = 5.886 N

Step 3: Calculate force available to provide acceleration

The rocket engine, when it is fired, exerts a 8.0 AND vertical thrust on the rocket.

5.886 N of that force will be used to counter the rocket's weight, leaving 2.114 N of force available to provide acceleration.  

Step 4: Calculate the rocket's upward acceleration:

Fnet = m*a

With Fnet = the net force (the force that remains after the rocket's weight is compensated)

with a = the rocket's acceleration (in m/s²)

2.114 N = (0.600 kg)*a

a = 3.52 m/s²  = the rocket's upward acceleration

Step 5: Calculate how long it will take to rise 20 meters into the air.

Δy = v0*t + 1/2 at²

with v0 = 0m/s

Δy = 1/2 at²

20 m = 1/2(3.52)t²

20 m = (1.76 m/s²)t²

11.36 = t²

t = 3.37 s

This means the rocket will take 3.37 seconds to reach the hoop.  It should be launched when the cart is 3.37 seconds away from being directly beneath the hoop.  

Step 6: Calculate the distance

v = Δx / t

4.0 m/s = Δx / 3.37 s

Δx = 13.48 m

The rocket should be launched when the cart is 13.48m away from a point directly below the hoop.

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5 0
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B. Electrical energy

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8 0
3 years ago
an object travels a distance of 6 m in 2 seconds if average speed is found using the equation distance traveled/ elapsed time wh
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