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2 years ago

Please help me and I’ll mark as brainliest i promise

Physics

1 answer:

irinina [24]2 years ago

6 0

Answer:

Explanation:

because C should be the correct one

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A uniform sphere (I = 2/5 MR 2 ) rolls down an incline. (a) What must be the incline angle if the linear acceleration of the cen

Liono4ka [1.6K]

Answer:

Part a)

$\theta = 8.05 degree$

Part b)

$a = 1.37 m/s^2$

Explanation:

As the uniform sphere is rolling down the inclined plane then the net force on the sphere is given as

$mg sin\theta - F_f = ma$

also we have torque equation on it

$F_f R = I\alpha$

for pure rolling

$a = R \alpha$

$F_f = \frac{Ia}{R^2}$

now we have

$mg sin\theta = ma + \frac{Ia}{R^2}$

now we have

$mg sin\theta = (m + \frac{2}{5}m)a$

$a = \frac{5}{7}g sin\theta$

now given that

$a = 0.10 g$

so we have

$0.10 g = \frac{5}{7} g sin\theta$

$sin\theta = 0.14$

$\theta = 8.05 degree$

Part b)

If the inclined plane is frictionless then the acceleration is given as

$a = g sin\theta$

$a = 9.8(0.14)$

$a = 1.37 m/s^2$

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4 years ago

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One peice of evidence that the universe is expanding

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3 years ago

I HAVE 5 MINUTES!!!!!! A block oscillating on the end of a spring moves from its position of maximum spring stretch to maximum s

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Therefore, if the block moves from its position of maximum spring stretch to maximum spring compression in 0.25 s, the time required for a full cycle is twice as much; T = 0.5 s.

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Why is it important to study electromagnetic waves?

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The study of EM is essential to understanding the properties of light, its propagation through tissue, scattering and absorption effects, and changes in the state of polarization. ... Since light travels much faster than sound, detection of the reflected EM radiation is performed with interferometry.

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3 years ago

A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.08190.0819 m, its frequency

Anvisha [2.4K]

Answer:

The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

Explanation:

Given that,

Amplitude = 0.08190 m

Frequency = 2.29 Hz

Wavelength = 1.87 m

(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave

Using formula of distance

$d=2A$

Where, d = distance

A = amplitude

Put the value into the formula

$d=2\times0.08190$

$d=0.1638\ m$

Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

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3 years ago