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Georgia [21]
3 years ago
7

3.A carnival ride has a radius of 5.00 m and exerts an applied force of 987 N on an 80 kg rider.

Physics
1 answer:
Llana [10]3 years ago
5 0

Answer:

gggn×2

lol

Explanation:

hujhyhhhgyhngfg

ojui

ukk

lol

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Calculate the frequency if the number of revolutions is 300 and the paired poles are 50.
Andrei [34K]

Answer: A

Explanation: We know that f=p*n

f=50*300=15000 Hz = 15kHz.

Have a great day! <3

3 0
2 years ago
What is the mass of an object that weighs 500 newtons on earth?
Eva8 [605]
The mass would be 51 kg. So b is the right answer.
6 0
4 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

     v_s =  \frac{d }{t}

substituting values

   v_s =  \frac{72000 * 3.0*10^{8} }{90000}

    v_s =  2.40*10^{8} \ m/s

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

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

substituting value

       T  =  90000 *  \sqrt{ 1 -  \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }

        T = 54000 \ s

=>    T  = 15 \ hours

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

       

7 0
3 years ago
12.
vazorg [7]

Answer: A

Explanation: How large a parachute is (in other words, the parachute's surface area) affects its air resistance, or drag force. ... In the case of these parachutes, the drag force is opposite to the force of gravity, so the drag force slows the parachutes down as they fall.

7 0
4 years ago
Read 2 more answers
Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a sp
Alex787 [66]

Answer:

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.

Explanation:

i want the answer i don't know

7 0
3 years ago
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