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2 years ago

What is the only force that can act on an object in free fall? gravity friction air resistance speed

Physics

2 answers:

sdas [7]2 years ago

8 0

Answer: It is gravity!

Explanation:

motikmotik2 years ago

4 0

Answer: Gravity

Explanation:When the only force acting on an object is gravity, so no air resistance. the constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling prevents further acceleration.

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Planet B has a tilt of 45 degrees. What seasonal changes would be expected on this planet?

12345 [234]

... Extreme temperature changes between seasons.

... Extreme changes in length of daylight and darkness.

... Larger areas around both poles where daylight/darkness
can last more than a whole day.
Altogether, these areas cover half of the planet.

5 0

3 years ago

What is the net force required to accelerate a 884 kg car at 2 m/sec2?

EastWind [94]

Answer:

1768 N

Explanation:

We can solve the problem by using Newton's second law:

$F=ma$

where

F is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, we have a car of mass

m = 884 kg

And its acceleration is

$a=2 m/s^2$

Substituting into the equation, we find the net force on the car:

$F=(884)(2)=1768 N$

3 0

2 years ago

True or False: When looking at kinetic vs. thermodynamic products the kinetic product predominates at low temperature.

zheka24 [161]

Answer:

true answer this question

3 0

2 years ago

A ball is projected at an angle of elevation of 60 ° with an initial velocity of 120m/s.calculate

Mrac [35]

Explanation:

It is given that,

The angle of projection is 60 degrees

Initial velocity of the ball is 120 m/s

We need to find the time taken to get to the maximum height and the time of flight.

Time taken to reach the maximum height is given by :

$T=\dfrac{u^2\sin^2\theta}{2g}$

g is acceleration due to gravity

$T=\dfrac{(120)^2\times \sin^2(60)}{2\times 10}\\\\T=540\ s$

(ii) Time of flight,

$t=\dfrac{2u\sin\theta}{g}$

So,

$t=\dfrac{2\times 120\times \sin(60)}{10}\\\\t=20.78\ s$

Hence, this is the required solution.

3 0

3 years ago

A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,

Rus_ich [418]

Answer:

$\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\$

Explanation:

Given that

$Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m\\$

As both charges are negative so there exist force of repulsion in direction as shown in figure.

$F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N$

Angle at which force F12 is acting is

$\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o$

$F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\$

$\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=$

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

$F_{21}=-5.63\times 10^{-11}N$

$\vec{F}_{21}=$

7 0

3 years ago