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lys-0071 [83]
3 years ago
14

What is the action force and the reaction force when you sit down on a chair

Physics
1 answer:
Licemer1 [7]3 years ago
4 0
The action force is when you are putting your weight on the chair when you are sitting down. the reaction force is the force exerted by the chair that pushes up on your body and is equal to your weight. 
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Sonbull [250]

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Convection is what keeps the hurricane going in till the storm hits land.

Explanation:

Convection is the warm, moist current that rises above sea level and as it cools into water vapor it releases energy which a hurricane runs on. Once the hurricane hits land it continues to lose energy to keep it going because there's no convection from the sea currents rising above sea level to release energy.

Hope this helps.

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Are there any steps of the rock cycle that do not connect directly to one another
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The three main rock types are igneous, metamorphic and sedimentary.

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3 years ago
Why do intermolecular forces exist between water molecules?
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I think the answer is B

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7 0
3 years ago
Name the planets to which these moons belong
yuradex [85]
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4 0
3 years ago
Automobile which is being towed at constant velocity up the incline using the cable at c. The automobile has a mass of 5 mg and
Simora [160]

The value of the normal reaction force will be 4.65 kN and 31.76 kN.

<h3>What is force?</h3>

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The normal force on the car is,Na= Nb =1 kN

The force of the cable on the car is,F=1 kN

The force of the gravity is,W=mg

The balancing equation of force;

\rm N_A+N_B+F sin \theta_2 -mg cos \theta_2 =0 \\\\ \rm -Fcos \theta_2 +mg sin \theta_1 =0 \\\\ Fcos \theta_2 \times a -F sin \theta_2 \times b -mg cos \theta_1 \times c -mg sin \theta_1 \times e +N_B (c+d)=0 \\\\ N_A+N_B+F sin 30^0 - 5 \times 9.81 cos 30^0 =0 \\\\ -

Putting the value further;

\rm N_A+N_B+F sin 30^0 - 5 \times 9.81 cos 30^0 =0 \\\\  -F cos 30^0 + 9.81 sin 20^0 =0 \\\\ F cos 30^ 0 (0.6)-F sin30^0 (1.5)-(5 \times 10^6) \times 9.81 \times cos 20^0 (1)-(5)(9.81)sin 20^0 +N_B \times 1.75 =0

Solving the equation, we get;

\rm  F_A = 19.37 \ kN \\\\ N_A=4.65 \ kN \\\\ N_B=31.76 \ kN

Hence, the value of the normal force swill be 4.65 kN and 31.76 kN.

To learn more about the force, refer to the link;

brainly.com/question/26115859

#SPJ4

4 0
2 years ago
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