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jok3333 [9.3K]
3 years ago
10

An egg suspended above the ground has 93.1 J of gravitational potential energy. The egg is then dropped and falls to the ground.

What is the kinetic energy of the egg just as it reaches the ground?
Physics
1 answer:
Marina86 [1]3 years ago
3 0

93.1 J of kinetic energy

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What is the magnitude and direction (right or left) of the
Sunny_sXe [5.5K]

Answer: 12 N to the right

Explanation:

If we calculate the net force acting on the box, we will have:

<u>In y-component:</u>

Fy_{net}=F_{n}+F_{g} (1)

Where F_{n}=12 N is the Normal force, directed upwards and F_{g}=-12 N is the weight of the box (gravity force), directed downwards.

Fy_{net}=12 N-12 N (2)

Fy_{net}=0 N (3) Hence the net force in the vertical component is zero

<u>In x-component:</u>

Fx_{net}=F_{left}+F_{right} (4)

Where F_{left}=-3 N and F_{right}= 15 N

Fx_{net}=-3 N + 15 N (5)

Fx_{net}=12 N (6) This is the net force in the horizontal component

Therefore, the total net force acting on the box is 12 N directed to the right

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3 years ago
The electric force between two charged objects of charge +Q0 that are separated by a distance R0 is F0. The charge of one of the
vovikov84 [41]

Answer:

F=3F_o(\frac{R_o}{R_{o2}})^2

Explanation:

This problem is approached using Coulomb's law of electrostatic attraction which states that the force F of attraction or repulsion between two point charges, Q_1 and Q_2 is directly proportional to the product of the charges and inversely proportional to the square of their distance of separation R.

F=\frac{kQ_1Q_2}{R^2}..................(1)

where k is the electrostatic constant.

We can make k the subject of formula  as follows;

k=\frac{FR^2}{Q_1Q_2}...........(2)

Since k is a constant, equation (2) implies that the ratio of the product of the of the force and the distance between two charges to the product of charges is a constant. Hence if we alter the charges or their distance of separation and take the same ratio as stated in equation(2) we will get the same result, which is k.

According to the problem, one of the two identical charges was altered from Q_o to 3Q_o and their distance of separation from R_o to R_{o2}, this also made the force between them to change from F_o to F_{o2}. Therefore as stated by equation (2), we can write the following;

\frac{F_oR_o^2}{Q_o*Q_o}=\frac{F_{o2}R_{o2}^2}{3Q_o*Q_o}.............(3)

Therefore;

\frac{F_oR_o^2}{Q_o^2}=\frac{F_{o2}R_{o2}^2}{3Q_o^2}.............(4)

From equation (4) we now make the new force F_{o2} the subject of formula as follows;

{F_oR_o^2}*{3Q_o^2}=F_{o2}R_{o2}^2*{Q_o^2}

Q_o then cancels out from both side of the equation, hence we obtain the following;

3{F_oR_o^2}=F_{o2}R_{o2}^2.............(4)

From equation (4) we can now write the following;

F_{o2}=\frac{3F_oR_o^2}{R_{o2}^2}

This could also be expressed as follows;

F_{o2}=3F_o(\frac{R_o}{R_{o2}})^2

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