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Serjik [45]
3 years ago
6

If the distance between two objects is decreased to 1 10 of the original distance, how will it change the force of attraction be

tween them? A. The new force will be 100 times more than the original. B. The new force will be 20 times more than the original. C. The new force will be 1 20 of the original. D. The new force will be 1 100 of the original. E. The new force will be 1 10 of the original.
Physics
1 answer:
aleksandr82 [10.1K]3 years ago
6 0

(A) It will 100 times larger than the original force.

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A golf club rotates 215 degrees and has a length (radius) equal to 29 inches. The time it took to swing the club was 0.8 seconds
vichka [17]

Answer:

The average linear velocity (inches/second) of the golf club is 136.01 inches/second

Explanation:

Given;

length of the club, L = 29 inches

rotation angle, θ = 215⁰

time of motion, t = 0.8 s

The angular speed of the club is calculated as follows;

\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega =  (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s

The average linear velocity (inches/second) of the golf club is calculated as;

v = ωr

v = 4.69 rad/s  x  29 inches

v = 136.01 inches/second

Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second

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3 years ago
Can you get cool powers.
Morgarella [4.7K]

Answer: Sadly no I wish

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3 0
2 years ago
What will the stopping distance be for a 2,000-kg car if -2,000 N of force are applied when the car is traveling 20 m/s?
astraxan [27]

Answer is B- 200 m

Given:

m (mass of the car) = 2000 Kg

F = -2000 N

u(initial velocity)= 20 m/s.

v(final velocity)= 0.

Now we know that

<u>F= ma</u>

Where F is the force exerted on the object

m is the mass of the object

a is the acceleration of the object

Substituting the given values

-2000 = 2000 × a

a = -1 m/s∧2

Consider the equation

<u>v=u +at</u>

where v is the initial velocity

u is the initial velocity

a is the acceleration

t is the time

0= 20 -t

t=20 secs


s = ut +1/2(at∧2)

where s is the displacement of the object

u is the initial velocity

t is the time

v is the final velocity

a is the acceleration

s= 20 ×20 +(-1×20×20)/2

<u>s= 200 m</u>


3 0
3 years ago
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7 0
4 years ago
A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. The mass is pull
sdas [7]

In spring mass system we know that angular frequency is given as

\omega = 2\pi f

f = 8.38 Hz

\omega = 2\pi(8.38)

\omega = 52.65  rad/s

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v = A\omega

here we know that

A = 17.5 cm

v = 0.175 (52.65)

v = 9.21 m/s

so maximum speed is 9.21 m/s

7 0
3 years ago
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