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Rom4ik [11]
3 years ago
9

A 75-g mass sits 75 cm from the center of a rotating platform undergoing a uniform angular acceleration of 0.125 rad/s^2. The co

efficient of static friction between the mass and the platform is 0.250.
What is the speed of the mass when it slides off?
a. 0.889 m/s
b. 1.36 m/s
c. 1.44 m/s
d. 1.58 m/s
e. It will never slide off.
Physics
1 answer:
jasenka [17]3 years ago
3 0

Answer:

velocity of mass when it slides of will be 1.36 m/sec

So option (b) will be correct option.

Explanation:

We have given mass of the slits m = 75 gram = 0.075 kg

Radius r = 75 cm = 0.75 m

Coefficient of kinetic friction \mu =0.250

Acceleration due to gravity g=9.8m/sec^2

When the mass slides of

\frac{mv^2}{r}=\mu mg

v=\sqrt{\mu rg}=\sqrt{0.250\times 0.75\times 9.8}=1.36m/sec

So velocity of mass when it slides of will be 1.36 m/sec

So option (b) will be correct option.

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The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

  • initial velocity of the sprinter, u = 18 km/h
  • final velocity of the sprinter, v = 27 km/h
  • time of motion of the sprinter, t = 3.5 x 10⁻⁴ h

Convert the velocity of the sprinter to m/s;

initial \ velocity, u = 18 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 5 \ m/s\\\\final \ velocity, v =27 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 7.5 \ m/s\\\\

The time of motion is seconds;

t = 3.5 \times 10^{-4} \ h \times \frac{3600 \ s}{1 \ h} = 1.26 \ s

The sprinter’s average acceleration is calculated as follows;

a = \frac{v- u}{t} \\\\a = \frac{7.5 \ m/s \ - \ 5 \ m/s}{1.26 \ s} \\\\a = 1.98 \ m/s^2

Thus, the sprinter’s average acceleration is 1.98 m/s²

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3 years ago
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Answer:

Explanation:

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I ∝ 1 /r²  where I is intensity and r is distance from source . If I₁ and I₂ be intensity at distance r₁ and r₂ .

I₁ /I₂ = r₂² /r₁²

If r₂ = 4r₁ ( given )

I₁ / I₂ = (4r₁ )² / r₁²

= 16 r₁² / r₁²

I₁ / I₂ = 16

I₂ = I₁ / 16

So intensity will become 16 times less bright .

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A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm
iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

m = m_{o}* m_{e}

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

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and me as

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m_{e} =26

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m = 14*26 = 364

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Kinetic energy depends on the mass and the speed of a moving object.

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