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Gnoma [55]
4 years ago
7

A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min.

Physics
1 answer:
Anastaziya [24]4 years ago
3 0

Answer:

a) 482.5N

b) 17.4N

Explanation:

40 rev/min = 40 * 2π / 60 = 4.2 rad/s

3 rev/min = 3 * 2π / 60 = 0.314 rad/s

The centripetal acceleration that the child is subjected to

a = \omega^2r

And their centripetal force is

f = ma = m\omega^2r

From here we can plug the number in

(a) f_1 = m\omega_1^2r_1 = 22*4.2^2*1.25 = 482.5N

(b) f_2 = m\omega_2^2r_2 = 22*0.314^2*8 = 17.4N

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The magnetic force on a charged particle is given as: F = qvBsinФ
B is the magnetic field.
q is the charge on the particle.
v is the velocity of the charged particle.
 Ф is the angle between magnetic field (B) & the velocity of charge (v).


⇒ F = qvb sin Ф

Here, Ф = 90 degree.
and sin 90 = 1
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3 years ago
Each statement below represents a feature that has been claimed to exist on Mars. Sort the statements into the correct bin accor
Wittaler [7]

Answer:

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5 0
3 years ago
NEED ANSWER ASAP!!!
xenn [34]

Answer:

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3 0
3 years ago
A 2. 5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0.
OLga [1]

The speed of the spring when it is released is 3.5 m/s.

The given parameters:

  • <em>Mass of the block, m = 2.5 kg</em>
  • <em>Spring constant, k = 56 N/m</em>
  • <em>Extension of the spring, x = 0.75 m</em>

The speed of the spring when it is released is calculated by applying the principle of conservation of energy as follows;

K.E = U_x\\\\\frac{1}{2} mv^2 = \frac{1}{2} kx^2\\\\mv^2 = kx^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{56 \times 0.75^2}{2.5} } \\\\v = 3.5  \ m/s

Thus, the speed of the spring when it is released is 3.5 m/s.

Learn more about conservation of energy here:  brainly.com/question/166559

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2 years ago
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