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Keith_Richards [23]
2 years ago
12

A 12.0 kg object is stationary on a surface that is inclined 30.0° relative to the horizontal. The coefficient of friction is 0.

65. Use a free body diagram to determine the magnitude of the force of static friction on the block.
Help please?
Physics
1 answer:
aniked [119]2 years ago
7 0

Answer:

hope you can understand the calculations

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Why might we expect venus and earth to be similar?
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They have similar distances from the sun therefore receive similar amounts of solar energy but Venus has no atmosphere to protect it so greater variation in temp.
Both are rock planets unlike those further away from the sun which, because of their distance receive less solor energy are cold, are ice planets
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What happens to heat energy after an object is cooled down lolololol asking for my bff
alisha [4.7K]

Answer:

Heat energy does not cool. Objects cool. Energy is not an object. Now, what happens when an object cools is that it gives off some of its thermal energy by one or more mechanisms: radiation, conduction, or convection. In radiation the energy escaped the object as electromagnetic waves - you see the red hot poker slowly dim as it cools, in conduction the energy is transferred Mechanically by the atoms of the hot object being in physical contact with those of the cool object and in convection the hot material moves to a cooler region where it gives up its heat energy by either radiation or conduction.

7 0
2 years ago
Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o
spin [16.1K]

Answer:

The true weight of the aluminium is m_{alu} = 4.5021 kg

Explanation:

Given data

m_{app} = 4.5 kg

\rho_{air} = 1.29 \frac{kg}{m^{3} }

\rho_{al} = 2.7× 10^{3} \frac{kg}{m^{3} }

The true mass of the aluminium is given by

m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }

Put all the values in above equation we get

m_{alu} = \frac{(2700)(4.5)}{2700-1.29}

m_{alu} = 4.5021 kg

Therefore the true weight of the aluminium is m_{alu} = 4.5021 kg

6 0
2 years ago
An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

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2 years ago
What is the explanation for the fact that the desert sand is very hot in the day and very cool at night? 1. Sand reflects light
zepelin [54]
3. Sand has a low specific heat compared to air
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