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devlian [24]
3 years ago
8

While standing in a low tunnel, you raise your arms and push against the ceiling with a force of 100 n. your mass is 70 kg. what

force does the ceiling exert on you?
Physics
1 answer:
alekssr [168]3 years ago
3 0
This is a manifestation of Newton's third law of motion called the Law of Action-Interaction. It states that when a body is at rest or not moving, for every force exerted on the object, an equal,opposite force is exerted also towards you. Thus, the ceiling also exerts an equal force of 100 Newtons.
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A 60 kg adult and a 30kg child are passengers on a rotor ride at an amusement park as shown in the diagram above. When the rotat
Ulleksa [173]

Answer:

 fr ’= ½ F

Explanation:

For this exercise we use the translational equilibrium equation, on the axis parallel to the wall

              fr - W = 0

              fr = W

for the adult man they indicate that the friction force is equal to F

              F = M g

we write the equilibrium equation for the child

             fr ’= w’

             fr ’= m g

in the statement they tell us that the mass of the adult is 2 times the mass of the child

             M = 2m

we substitute

            fr ’= M / 2 g

            fr ’= ½ Mg

we substitute

            fr ’= ½ F

therefore the force of friction in the child is half of the friction in the adult

8 0
3 years ago
Consider the following equilibrium at 979 K for the dissociation of molecular iodine into atoms of iodine. I2(g) equilibrium rea
GenaCL600 [577]

Answer:  I_2=  0.0050 M

I = 0.0155 M

Explanation:

Initial moles of  I_2 = 0.072 mole

Volume of container = 3.9 L

Initial concentration of I_2=\frac{moles}{volume}=\frac{0.072moles}{3.9L}=0.018M  

The given balanced equilibrium reaction is,

                 I_2(g)\rightleftharpoons 2I(g)

Initial conc.         0.018 M            0

At eqm. conc.    (0.018-x) M      (2x) M  

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[I]^2}{[I_2]}

K_c=\frac{(2x)^2}{0.2-x}

we are given :  K_c=1.60\times 10^{-3}

Now put all the given values in this expression, we get :

1.60\times 10^{-3}=\frac{(2x)^2}{(0.018-x)}

x=0.0025

So, the concentrations for the components at equilibrium are:

[I]=2\times x=2\times 0.0025=0.0050

[I_2]=0.018-x=0.018-0.0025=0.0155

Hence, concentrations of I_2 and I are 0.0050 M ad 0.0155 M respectively.

4 0
4 years ago
When Dr. Hewitt immerses an object in water the second time and catches the water that is displaced by the object, how does the
Dimas [21]

Answer:

Explanation:

- The volume of water displaced by immersing the object is equal the amount of water spilled and caught by Dr. Hewitt.

- The amount of water is proportional to the volume of object of fraction of object immersed in water will lead to the same fraction of water displaced and caught by Dr. Hewitt.  

- When the object is immersed the force of Buoyancy acts against the weight and reducing the scale weight.

- The amount of Buoyancy Force is proportional to the fraction of Volume of object immersed in water; hence, the same amount is spilled/lost.

5 0
3 years ago
The image below The image below The image below The image below The image below The image below The image below The image below
larisa [96]

Answer:

wow i have know idea

Explanation:

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Thepotemich [5.8K]

Answer:

cant understand anything

Explanation:

3 0
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