Abody is moving according to the relation:vf=21,then its displacement after 5s

Coulomb's law states that the force F of attraction between two oppositely charged particles varies jointly as the magnitude of

Mars2501 [29]

Answer: Force F will be one-sixteenth of the new force when the charges are doubled and distance halved

Explanation:

Let the charges be q1 and q2 and the distance between the charges be 'd'

Mathematical representation of coulombs law will be;

F1=kq1q2/d²...(1)

Where k is the electrostatic constant.

If q1 and q2 is doubled and the distance halved, we will have;

F2 = k(2q1)(2q2)/(d/2)²

F2 = 4kq1q2/(d²/4)

F2 = 16kq1q2/d²...(2)

Dividing equation 1 by 2

F1/F2 = kq1q2/d² ÷ 16kq1q2/d²

F1/F2 = kq1q2/d² × d²/16kq1q2

F1/F2 = 1/16

F1 = 1/16F2

This shows that the force F will be one-sixteenth of the new force when the charges are doubled and distance halved

4 0

3 years ago

A balloon is filled at sea level on a spring day (temperature 20°C) to a volume of 0.500 m3 and then brought at op of Mount Ever

blagie [28]

Answer:

Volume increases

Explanation:

The balloon when filled at sea level being comparatively close to the center of the earth will have higher pressure due to the influence of gravity and when this balloon is taken to the top of the mountain being away from the center of earth, it will experience a lesser pressure due to low gravity where the amount of force exerted by the air on the object is lesser as compared to to that at the sea level.

Therefore, there will be an increase in volume of the balloon as there is expansion of air on the inside of the balloon as a result of low pressure.

There is not much effect of temperature at both the sea level and the mountain top as the temperature does not impart any energy to the air molecules so as to decrease the volume.

Therefore,there is an increase in the volume of the balloon at the top of the mountain.

Percentage change in volume is given by:

$% change = \farc{V - 0.5}{0.5}times 100$

8 0

3 years ago

A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the m

konstantin123 [22]

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, $\gamma _0$ = 0.875

mass of the object in oil, $M_o$ = 0.013 kg

mass of the object in water, $M_w$ = 0.012 kg

let the mass of the object in air = $M_a$

weight of the oil, $W_0 = M_a - 0.013$

weight of the water, $W_w = M_a - 0.012$

The relative density of the oil is given as;

$\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg$

Therefore, the mass of the body is 0.02 kg.

6 0

3 years ago

A patient has an ongoing history of cancer. She has a tumor in the abdominal region, and has been undergoing treatment for it. T

Vadim26 [7]

Answer:

the answer is at the BOTTOM OF THEIR QUESTION

Explanation:

IT IS CORRECT BTW

6 0

3 years ago

"4.What determines how many figures are significant in reported measurement values? What

MAXImum [283]

As the number of significant figures increases, the more accurate or precise the measurement is.

<h3>What is significant figure?</h3>

The term significant figures refers to the number of important single digits in the coefficient of an expression in scientific notation.

Significant figures are the digits in a value that are known with some degree of confidence.

The effect of reporting more or fewer figures or digits than are significant;

As the number of significant figures increases, the more accurate or precise the measurement is.

As precision of a measurement increases, so does the number of significant figures.

Learn more about significant figures here: brainly.com/question/24491627

#SPJ1

8 0

1 year ago