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Licemer1 [7]
3 years ago
6

Which law explains why a book flying in the air hitting a person would hurt more than a feather hitting a person?

Physics
2 answers:
OlgaM077 [116]3 years ago
7 0
Because of force, the force would keep growing from the air, while the feather glides with the wind. So force is the reason why.
RoseWind [281]3 years ago
4 0
Newton's laws of motion
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In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a 200 WW electric im
ivolga24 [154]

Answer:

The heat is 115478.4 J.

Explanation:

Given that,

Mass of water = 0.400 kg

Power = 200 W

Suppose, we determine how much heat must be added to the water to raise its temperature from 20.0°C to 89.0°C?

We need to calculate the heat

Using formula of heat

Q=mc\Delta T

Where, m = mass of water

c = specific heat

Put the value into the formula

Q=400\times4.184\times(89-20)

Q=115478.4\ J

Hence, The heat is 115478.4 J.

7 0
3 years ago
A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor
STatiana [176]

Answer:

The magnitude of the second charge is \rm 1.062\times 10^{-7}\ C or \rm 0.1062\ \mu C.

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge q_o at a point in the electric field of another charge q is given by the product of the amount of charge q_o and electric potential at that point due to the charge q.

U = q_o\ V.

The electric potential at that point is given by

V = \dfrac{kq}{r}.

where k is the Coulomb's constant.

Therefore,

U=q_o\ \dfrac{kq}{r}.

Now, We have given two charges q_1 = +44\ \mu C = +44\times 10^{-6}\ C and q_2, whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

U_i = \dfrac{kq_1q_2}{\infty}=0.

When the coordinates of position of the two charges are

(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).

The distance between the two charges is given by

r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.

The electric potential energy of the charges in this configuration is given by

U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.

6 0
3 years ago
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