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Annette [7]
2 years ago
9

A 10.0-kg mass is placed on a 25.0o incline and friction keeps it from sliding. The coefficient of static friction in this case

is 0.580, and the coefficient of sliding friction is 0.520. The mass is given a shove causing it to slide down the incline. What is the frictional force while the mass is sliding
Physics
1 answer:
vovangra [49]2 years ago
5 0

The frictional force while the mass is sliding will be 46.2 N.

<h3>What is friction force?</h3>

Opposition forces on the surface cause heat loss during the motion of an object known as the friction force.

Given data:

m(mass)= 10.0-kg

Θ (Inclination angle)=25.0o

Coefficient of sliding friction,\rm \mu_k=0.520

Coefficient of static friction,\rm  \mu_s=0.520

The friction force, F=?

Resolve the force in the inclined plane;

\rm F=\mu_s mg cos25^0 \\\\ F=0.520 \times 10 \times 9.81 \times  cos 25 ^0 \\\\ F= 46.2 \ N

Hence, the frictional force while the mass is sliding will be 46.2 N.

To know more about friction force refer to the link;

brainly.com/question/1714663

#SPJ1

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(c) velocity in mph is 115.0mph

Explanation:

(a).

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\boxed{P = 64,680J}

(b).

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\boxed{v = 51.43m/s}

(c).

The velocity in mph is

\dfrac{51.43m}{s} * \dfrac{3600s}{hr} * \dfrac{1mile}{1609.34m}

\boxed{v= 115.0mph}

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3 years ago
The lowest possible temperature in outer space is 3.13 K. What is the rms speed of hydrogen molecules at this temperature? (The
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Solution:

To calculate the root mean squarer or rms speed of hydrogen molecule, we use the given formula:

v_{rms} = \sqrt{\frac{3TR}{m}}

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Putting the values in the above formula:

v_{rms} = \sqrt{\frac{3\times 3.13\times 8.314}{2.02\times 10^{-3}}}

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0.0034 sec

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T = initial time period = 2.51 s

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L = 1.56392 m

L' = new length

ΔT = Rise in temperature = 142 °C

α = coefficient of linear expansion = 19 x 10⁻⁶ °C

New length due to rise of temperature is given as

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L' = 1.56392 + (1.56392) (19 x 10⁻⁶) (142)

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But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

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a = -gμ.............. Equation 3

From the question,

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Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

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From the question,

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Substitute these values into equation 4

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