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vazorg [7]
3 years ago
10

What is the frequency of a wave with a period of 8s?

Physics
1 answer:
PSYCHO15rus [73]3 years ago
3 0
I don’t know the answer
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The barrel of a rifle has a length of 0.855 m. A bullet leaves the muzzle of a rifle with a speed of 553 m/s. What is the accele
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Answer:

Acceleration of the bullet will be 1778835.6 m/sec^2      

Explanation:

We have given length of the barrel refile s= 0.855 m

When the bullet leaves the muzzle its velocity is 553 m/sec

So final velocity v = 553 m/sec

Initial velocity will be 0 that is u = 0 m/sec

According to third equation of motion v^2=u^2+2as

553^2=0^2+2\times a\times 0.855

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A ball is kicked horizontally at 4.6 m/s off of a cliff 13.4 m high. How far from the cliff will it land.
Mekhanik [1.2K]
<h3>Answer:</h3>

7.53 m

<h3>Explanation:</h3>

<u>We are given:</u>

Initial Horizontal Velocity of the Ball = 4.6 m/s

Initial Vertical Velocity of the Ball = 0 m/s

Height from which ball is kicked = 13.4 m

<u>Time taken by the ball to reach the ground:</u>

The ball has an initial vertical velocity of 0 m/s

it also has a downward acceleration of 10 m/s² due to gravity

<u>Solving for the time taken:</u>

s = ut + 1/2(at²)                 [second equation of motion]

replacing the values

13.4 = (0)(t) + 1/2 (10)(t²)

13.4 = 5t²

t² = 13.4/5                  [dividing both sides by 5]

t² = 2.68

t = 1.637 seconds     [taking the square root of both sides]

<u>Horizontal distance covered by the ball:</u>

Since there are no horizontal opposing forces on the ball,

the ball will more horizontally at a velocity of 4.6 m/s until it hits the ground

We calculated that the ball will hit the ground in 1.637 seconds

<u>Distance covered:</u>

s = ut + 1/2 (at²)                            [seconds equation of motion]

s = ut                                            [since a = 0m/s² in the horizontal plane]

replacing the values

s = 4.6 * 1.637

s = 7.53 m

Hence, the ball landed 7.53 m from the cliff

5 0
3 years ago
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