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3 years ago

10

A golfer hits a golf ball at an angle of 30 degrees from the ground, with an initial velocity of 120 ft/sec. It lands on the gro

und, after 5 seconds
How far did the golf ball travel horizontally?

1 answer:

Gelneren [198K]3 years ago

8 0

Data:

Vo = 120 ft / s

α = 30°

t = 5 s

x = ?

Formulas:

cos(α) = Vo,x / Vo => Vo,x = Vo * cos(α)

x = Vo,x * t

Calculations:

Vo,x = 120 ft / s * cos(30) = 103.92 ft /s

x = 103.92 ft/s * 5 s = 519.6 ft

Answer: 519.6 ft

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Unit time will decrease

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a) The work done by the tree is $-1.44\cdot 10^5 J$

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Explanation:

a)

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

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where

W is the work done on the car

m is the mass of the car

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v is its final speed

For the car in this problem, we have:

m = 2000 kg

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v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

$W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J$

The work is negative because it is done in the direction opposite to the direction of motion of the car.

b)

The work done by the tree on the car can also be rewritten as

$W=Fd$

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

$W=-1.44\cdot 10^5 J$ is the work done

$d=50.0 cm = 0.50 m$ is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

$F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N$

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

Learn more about work:

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brainly.com/question/6443626

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Here,

$\theta$ = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

$\epsilon= -N \frac{d\Phi }{dt}$

$\epsilon = -N \frac{(BAcos\theta)}{dt}$

$\epsilon = -NAcos\theta \frac{dB}{dt}$

$\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)$

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