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12345 [234]
3 years ago
9

A rotating merry go-round makes one complete revolution in 4.2 s. (a) what is the linear speed of a child seated 1.3 m from the

center?
Physics
2 answers:
gregori [183]3 years ago
6 0
The radius, r, of the child from the center of the wheel is
r = 1.3 m

The wheel makes one revolution in 4.2 s. Its angular velocity is
ω = (2π rad)/(4.2 s) = 1.496 rad/s

The linear speed of the child is the tangential velocity, given by
v = rω
   = (1.3 m)*(1.496 rad/s)
   = 1.945 m/s

Answer: 1.95 m/s  (nearest hundredth)

Lera25 [3.4K]3 years ago
3 0
Remember in solving any Physics problem it is important to write all the given parameter. Given are 1 complete revolution of merry go-round or can be written as 1rev/s, this should be converted to radians, since it makes one complete revolution. Radius given equal to 1.3 m from the center. Simply follow the process,  ω=1rev/s x 2πrads/rev  ω=answer will in "rad/s" or cancel the unit "rev". The formula to follow is Speed=rω. just multiply radius and the angular speed. 
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stiv31 [10]

Answer:

Explanation:

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If a person touches the inner surface of the sphere then he will not be harmed as there is no charge on the inner surface of the sphere.

If a person carries the charge of the opposite sign of the same magnitude then the sphere and person get neutralized upon touching the sphere.

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3 years ago
A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi
Mamont248 [21]

Answer:

a=368.97\ m/s^2

Explanation:

Given that,

Initial angular velocity, \omega=0

Acceleration of the wheel, \alpha =7\ rad/s^2

Rotation, \theta=14\ rotation=14\times 2\pi =87.96\ rad

Let t is the time. Using second equation of kinematics can be calculated using time.

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s

Let \omega_f is the final angular velocity and a is the radial component of acceleration.

\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s

Radial component of acceleration,

a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2

So, the required acceleration on the edge of the wheel is 368.97\ m/s^2.

6 0
3 years ago
If you have 80g of a radioactive substance whose half life is 2 days, how long will it take for the substance to decay to the po
Ber [7]

Answer:

6 days.

Explanation:

From radioactivity, The expression for half life is given as,

R/R' = 2⁽ᵃ/ᵇ)................... Equation 1

Where R = original mass of the radioactive substance, R' = Remaining mass of the radioactive substance after decay, a = Total time taken to decay, b = half life.

Given: R = 80 g, R' = 10 g, b = 2 days.

Substitute into equation 1

80/10 = 2⁽ᵃ/²⁾

8 = 2⁽ᵃ/²⁾

2³ = 2⁽ᵃ/²)

Equating the base and solving for a

3 = a/2

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a = 6 days.

5 0
3 years ago
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Elis [28]

Answer:

Each object exerts a force on the other, and the two forces are equal and in opposite directions

Explanation:

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