If the mass of Earth increased, with no change in radius, your weight would

Physics

1 answer:

amm18122 years ago

6 0

Answer:

Increase

Explanation:

If the mass of the earth increase, with no change in radius, one's weight would also increase considerably.

The reason for this is that, the gravitational force of attraction between the two bodies will increase.

The gravitational force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
So, increasing the mass of the earth will increase the gravitational force on one's body and hence, an increase in weight.

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According to Newton’s second law of motion what is force equals to

Maksim231197 [3]

Answer:force equals to rate of change of momentum

Explanation:

F=force

t=time

m=mass

v=final velocity

u=initial velocity

(mv-mu)/t=rate of change of momentum

Force=rate of change of momentum

F=(mv-mu)/t

8 0

3 years ago

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Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s

Ne4ueva [31]

The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
$F=qvB \sin \theta$
where $\theta$ is the angle between the directions of v and B.

1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is $q=2.7 \mu C=2.7 \cdot 10^{-6}C$ . In this case, v and B are perpendicular, so $\theta=90^{\circ}$ , therefore we have:
$B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T$

2) In this second case, the angle between v and B is $\theta=55^{\circ}$ . The charge is now $q=42.0 \mu C=42.0 \cdot 10^{-6}C$ , and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
$F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N$