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fiasKO [112]
3 years ago
7

A crossbow is fired horizontally off a cliff with an initial velocity of 15 m/s. If the arrow takes 4s to hit the ground, what i

s the range of the projectile?
Physics
1 answer:
marysya [2.9K]3 years ago
8 0

Answer:

<em>The range of the projectile is 60 m</em>

Explanation:

<u>Horizontal Motion</u>

When an object is thrown horizontally with a speed vo from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.

The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:

v_x=v_o

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

v_y=g.t

The horizontal distance is calculated as a constant speed motion:

x = v_x.t

Knowing the crossbow is fired horizontally at vo=vx=15 m/s and it takes t=4 s to hit the ground, thus the range of the projectile is:

x = 15*4 = 60

The range of the projectile is 60 m

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Read 2 more answers
A train is moving along a track with constant speed v1 relative to the ground. A person on the train holds a ball of mass m and
scoundrel [369]

Answer:

(a)0.5m(2v1v2+v2^{2})\\(b)0.5m(v2^{2}-v1^{2})\\(c)0.5m(2v1v2+v2^{2})\\(d)0.5m(v2^{2}-v1^{2})

Explanation:

velocity of ball in train reference = v2

velocity of ball in earth reference = v1+v2

(a)

Kinetic energy is given by 0.5mv^{2} where m and v are the mass and velocity of object respectively.

Change in kinetic energy is given by subtracting initial kinetic energy from the final kinetic energy. In this case

Initial kinetic energy= 0.5mv1^{2}

Final kinetic energy= 0.5m (v1+v2)^{2}=0.5m(v1^{2}+2v1v2+v2^{2})

Change in kinetic energy=0.5m (v1+v2)^{2}-0.5mv1^{2}=0.5m((v1+v2)^{2})-v1^{2}=0.5m(2v1v2+v2^{2}

(b)

Change in velocity in train reference will be

Initial kinetic energy= 0.5mv1^{2}

Final kinetic energy= 0.5mv2^{2}

Change in kinetic energy=0.5m(v2^{2}-v1^{2})

(c)

Work done, W = change in kinetic energy=0.5m (v1+v2)^{2}-0.5mv1^{2}=0.5m((v1+v2)^{2})-v1^{2}=0.5m(2v1v2+v2^{2}

(d)

Work done, W = change in kinetic energy=0.5m(v2^{2}-v1^{2})

4 0
3 years ago
A metal wire is bent into a square and carries auniform current throughout it. The net magneticfield at the center of this squar
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Answer:

The statement that the net magnetic field at the center of this square is zero is false.

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Explanation:

The net magnetic field at the center of this square is not equal to  zero.

The net magnetic field at the center of this square is given by the equation below:

B = 2√2μ₀I/πₐ

Where a = the side of the loop, and I is the current.

Thus, the statement that the net magnetic field at the center of this square is zero is false.

The net magnetic field inside a conductor must be zero - This is a true statement because the total charge on the conductor must be equal to zero.

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