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eduard
3 years ago
7

Water from a fire hose is directed horizontally against a wall at a rate of 69.4 kg/s and a speed of 19.6 m/s. Calculate the mag

nitude of the force exerted on the wall, assuming the water comes to a complete stop when hitting the car, with no splash-back.
Physics
1 answer:
Gnoma [55]3 years ago
5 0

Answer:

Force, |F| = 1360.24 N

Explanation:

It is given that,

Water from a fire hose is directed horizontally against a wall at a rate of 69.4 kg/s, \dfrac{m}{t}=69.4\ kg/s

Initial speed of the water, u = 19.6 m/s

Finally water comes to stop, v = 0

To find,

The magnitude of the force exerted on the wall.

Solution,

Let F is the force exerted on the wall. The product of mass and acceleration is called the force exerted. Using second law of motion to find it as :

F=m\dfrac{v-u}{t}

F=\dfrac{-mu}{t}

F=-69.4\ kg/s\times 19.6\ m/s

|F| = 1360.24 N

So, the magnitude of the force exerted on the wall is 1360.24 N.

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Answer:

17.3 m

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Given that,

Mass of a hammer is 0.58 kg

Velocity with which the hammer slides is 6.69 m/s at constant speed.

The roof makes an angle of 18 ◦ with the horizontal, and its lowest point is 18.2 m from the ground. We need to find the horizontal distance traveled by the hammer between the time is leaves the roof of the house and the time it hits the ground. Firstly, we will find the time taken by the hammer when it reaches ground in vertical direction.

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Neglecting negative value,

To find horizontal distance, multiply 2.72 s with the horizontal component of velocity.

d=2.72\times 6.69\times \cos(18)\\\\d=17.3\ m

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