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allochka39001 [22]

3 years ago

15

What is the condition for an object experiencing free fall?

1 answer:

S_A_V [24]3 years ago

3 0

Answer:

The body acts under the influence of gravity.

Explanation:

An object experiencing free fall is acting under the influence of gravitational force and the acceleration due gravity is positive for any falling object. The body is able to fall freely due to the effect of gravity on it. This gravity effect causes the body to get attracted to the earth's gravitational surface due to gravitational pull exerted on the body.

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Uses of hydrometer in points

Arisa [49]

Answer:

A hydrometer is an instrument used to determine specific gravity. It operates based on the Archimedes principle that a solid body displaces its own weight within a liquid in which it floats. Hydrometers can be divided into two general classes: liquids heavier than water and liquids lighter than water

7 0

2 years ago

Two cars are driving at the same constant speed on a straight road, with car 1 in front of car 2. car 1 suddenly starts to brake

Ganezh [65]

<span> Maths delivers! Braking distance ... If the </span>car<span> is initially travelling at u</span>m<span>/s, then the stopping distance d </span>m<span> ... the </span>speed<span> of the </span>car<span> at the </span>instant<span> the </span>brakes<span> are applied. ... An object with </span>constant acceleration<span> travels the </span>same<span> distance as it would ... We </span>start<span> with the second equation of motion:.</span>

4 0

3 years ago

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

8 0

2 years ago

Read 2 more answers

Strongest Element on Earth

Alex777 [14]

Tungsten has the highest tensile strength of any natural metal, but it's brittle and tends to shatter on impact.

Titanium has a tensile strength of 63,000 PSI. ...

Chromium, on the Mohs scale for hardness, is the hardest metal around.

8 0

2 years ago

A charged particle is moving in a uniform magnetic field at a speed of 8.2Ã10^3 m/s in a direction 87Â° from the direction of th

77julia77 [94]

Answer:

Magnetic field, B = 0.042 T

Explanation:

It is given that,

Speed of charged particle, $v=8.2\times 10^3\ m/s$

Angle between velocity and the magnetic field, $\theta=87$

Charge, $q=5.7\ \mu C=5.7\times 10^{-6}\ C$

Magnetic force, F = 0.002 N

The magnetic force is given by :

$F=qvB\ sin\theta$

B is the magnetic field

$B=\dfrac{F}{qv\ sin\theta}$

$B=\dfrac{0.002}{5.7\times 10^{-6}\times 8.2\times 10^3\times sin(87)}$

B = 0.042 T

So, the strength of the magnetic field is 0.042 Tesla. Hence, this is the required solution.

5 0

3 years ago