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3 years ago

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

2 answers:

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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A less than youthful 82.6 kg physics professor decides to run the 26.2 mile (42.195 km) Los Angeles Marathon. During his months

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2 years ago

What is a positive impact of artificial fertilizers

dybincka [34]

<span>The top benefit of pesticides is their effectiveness against pests that would otherwise decimate crops large and small. </span>

5 0

3 years ago

Ceres is a dwarf planet located in the main asteroid belt.

Pie

Yes that is correct, it <span>lies between the orbits of Mars and Jupiter and is the largest in the belt.

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3 years ago

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