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meriva
2 years ago
6

Help? Will give thanks if right (thank you)

Physics
1 answer:
qwelly [4]2 years ago
8 0

Hello there.

1.986*10^6

10^6=1,000,000

1.986*1,000,000

<u><em>Answer=1,986,000</em></u>

Explanation: First you had to ten to the six power of the ten with six times. It gave us 1000000 should be have six zeros at any time. Then you can multiply by 1.986*1000000=1,986,000 is the right answer. Hope this helps! Thank you for posting your question at here on Brainly. -Charlie

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The angular speed of the rotor in a centrifuge increases from 420 to 1420 rad/s in a time of 5.00 s. (a) Obtain the angle throug
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Answer:

a) θ = 2500 radians

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Explanation:

Using equations of motion,

θ = (w - w₀)t/2

θ = angle turned through = ?

w = final angular velocity = 1420 rad/s

w₀ = initial angular velocity = 420

t = time taken = 5s

θ = (1420 - 420) × 5/2 = 2500 rads

Again,

w = w₀ + αt

α = angular accelaration = ?

1420 = 420 + 5α

α = 1000/5 = 200 rad/s²

6 0
3 years ago
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The waves become longer but slower
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A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?
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When a positive charge moves in the direction of the electric field, what happens to the electrical potential energy associated
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6 0
3 years ago
A yet-to-be-built spacecraft starts from Earth moving at constant speed to the yet-tobe-discovered planet Retah, which is 20 lig
Gre4nikov [31]

Answer:

The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

Explanation:

From the question we are told that

The distance between earth and Retah is  d = 20 \ light \ hours =  20 * 3600 *  c =  72000c \ m

Here c is the peed of light with value c =  3.0*10^8 m/s

The time taken to reach Retah from earth is  t =  25 \ hours  =  25 * 3600 =90000 \ sec

The velocity of the spacecraft is mathematically evaluated  as

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substituting values

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The time elapsed in the spacecraft’s frame is mathematically evaluated as

      T  =  t *  \sqrt{ 1 -  \frac{v^2}{c^2} }

substituting value

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So  The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

       

7 0
3 years ago
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