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Tcecarenko [31]
3 years ago
5

A disk, with a radius of 0.25 m, is to be rotated like a merry-go-round through 1000 rad, starting from rest, gaining angular sp

eed at the constant rate through the first 500 rad and then losing angular speed at the constant rate until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed 400 m/s².
(a) What is the least time required for the rotation?
(b) What is the corresponding value of α₁?

Physics
1 answer:
aniked [119]3 years ago
6 0

Answer:

(a) The least time required for the rotation is 50 seconds

(b) The corresponding value of α₁ is 1.6 rad/s²

Explanation: Please see the attachments below

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A horizontal desk surface measures 1.7 m by 1.0 m. If the Earth's magnetic field has magnitude 0.42 mT and is directed 68° below
AlexFokin [52]

Answer:

The magnetic flux through the desk surface is 6.6\times10^{-4}\ T-m^2.

Explanation:

Given that,

Magnetic field B = 0.42 T

Angle =68°

We need to calculate the magnetic flux

\phi=BA\costheta

Where, B = magnetic field

A = area

Put the value into the formula

\phi=0.42\times10^{-3}\times1.7\times1.0\cos22^{\circ}

\phi=0.42\times10^{-3}\times1.7\times1.0\times0.927

\phi=6.6\times10^{-4}\ T-m^2

Hence, The magnetic flux through the desk surface is 6.6\times10^{-4}\ T-m^2.

3 0
3 years ago
Why is it useful to calculate average speed?
Phantasy [73]
When you are driving
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3 years ago
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10
sashaice [31]

Answer:

Time interval;Δt ≈ 37 seconds

Explanation:

We are given;

Angular deceleration;α = -1.6 rad/s²

Initial angular velocity;ω_i = 59 rad/s

Final angular velocity;ω_f = 0 rad/s

Now, the formula to calculate the acceleration would be gotten from;

α = Change in angular velocity/time interval

Thus; α = Δω/Δt = (ω_f - ω_i)/Δt

So, α = (ω_f - ω_i)/Δt

Making Δt the subject, we have;

Δt = (ω_f - ω_i)/α

Plugging in the relevant values to obtain;

Δt = (0 - 59)/(-1.6)

Δt = -59/-1.6

Δt = 36.875 seconds ≈ 37 seconds

6 0
4 years ago
A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth
Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

\tan\theta=\dfrac{8}{3}

\thta=\tan^{-1}\dfrac{8}{3}

\theta=69.44^{\circ}

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

mg=2T\cos\theta

Put the value into the formula

3\times9.8=2T\times\cos69.44

T=\dfrac{3\times9.8}{2\times\cos69.44}

T=41.85\ N  

Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.

4 0
4 years ago
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6. Drop to one quarter of its original value
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