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sergey [27]
3 years ago
8

An unknown material has a mass of 0.447 kg, and its temperature increases by 2.87°C when 943 J of heat are added. What is the sp

ecific heat of the material? (Unit=J(kg*C))
Physics
1 answer:
Sergeeva-Olga [200]3 years ago
4 0

Answer:

735 J/kg/C

Explanation:

Q = mcT

943 = (0.447)( c )(2.87)

1.28289c = 943

c = <u>7</u><u>3</u><u>5</u><u> </u><u>J</u><u>/</u><u>k</u><u>g</u><u>/</u><u>C</u><u> </u><u>(</u><u>3</u><u> </u><u>s</u><u>f</u><u>)</u>

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A batter swings at a baseball. The action force is the bat hitting the ball with a force of 5N. What is the reaction force?
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2 years ago
A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline
balandron [24]

Answer:

<em>2.78m/s²</em>

Explanation:

Complete question:

<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>

According to Newton's second law of motion:

\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\

Where:

\mu is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

gsin\theta - \mu g cos\theta = a_x\\

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>

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Answer:

You kinda left out the options you want us to choose from.

Resend the question with Full details

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