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Firdavs [7]
3 years ago
10

A bullet is fired horizontally with an initial velocity of 144.7 m/s from a tower 11 m high. if air resistance is negligible, wh

at is the horizontal distance the bullet travels before hitting the ground?
Physics
1 answer:
scoundrel [369]3 years ago
6 0
H = 11 m, the vertical distance that the bullet falls.
Initial vertical velocity  = 0
Horizontal velocity  = 144.7 m/s

The time, t, taken to fall 11 m is given by
(1/2)*(9.8 m/s²)*(t s)² = (11 m)
4.9t² = 11
t = 1.4983 s

If aerodynamic resistance is ignored, the horizontal distance traveled before the bullet hits the ground is
d = (144.7 m/s)*(1.4983 s) = 216.8 m

Answer: 216.8 m
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Define average velocity.Immersive Reader
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The slope of a graph of position vs time

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An angry physics student releases a wrecking ball as shown. The wrecking ball is just about to hit the building at the final tim
daser333 [38]

Answer:

the force between the building and the ball is non-conservative (friction-type force)

Explanation

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6 0
3 years ago
The two conducting rails in the drawing are tilted upwards so they each make an angle of 30.0° with respect to the ground. The v
dolphi86 [110]

Answer:

The current flows through the rod is 14.9 A.

Explanation:

Given that,

Magnetic field = 0.045 T

Mass of aluminum rod  = 0.19 kg

Length = 1.6 m

Angle = 30.0°

We need to calculate the force

Using resolving force

F\cos\theta=mg\sin\theta

F=mg\tan\theta

Put the value into the formula

F=0.19\times9.8\times\tan30

F=1.075\ N

We need to calculate the current flows through the rod

Using formula of magnetic force

F=iLB

i=\dfrac{F}{LB}

Put the value into the formula

i=\dfrac{1.075}{1.6\times0.045}

i=14.9\ A

Hence, The current flows through the rod is 14.9 A.

8 0
3 years ago
A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 340 N is appl
Taya2010 [7]

Answer:

The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

Explanation:

Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.

Given that,

Mass of block = 60 kg

Acceleration = 2.0 m/s²

Mass = 100 kg

Horizontal force = 340 N

Let the frictional force be f.

We need to calculate the frictional force

Using balance equation

F-f=ma

Put the value into the formula

340-f=60\times2.0

f=340-60\times2.0

f=220\ N

We need to calculate the coefficient of friction

Using formula of friction force

f= \mu mg

\mu=\dfrac{f}{mg}

\mu=\dfrac{220}{60\times9.8}

\mu =0.37

We need to calculate the acceleration of the 100 kg block

Using formula of newton's law

F = ma

a=\dfrac{F}{m}

a=\dfrac{220}{100}

a=2.2\ m/s^2

Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

3 0
3 years ago
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