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3 years ago

Does gravity affect your weight???

2 answers:

7 0

Yes. beacuse the gravity pulls you down. Example:If you are in the space you will weight less, beacuse there's no gravity.
hope this help. :)

Degger [83]3 years ago

5 0

Weight is a force caused by gravity. The weight of an object is the gravitational force between the object and the Earth. The more mass the object has the greater its weight will be.

You might be interested in

Explain, using the kinetic theory of matter, why liquids and solids are more denser than gases

AlekseyPX

Solid and liquids are much more denser than gas because their molecules are close to each other and with that the molecules of them can't move that freely unlike the gas molecules. Also, because of being near to each other the molecules of solid and liquids became heavy making them dense.

5 0

3 years ago

A wave of infrared light has a speed of 6 m/s and a wavelength of 12 m. What is the frequency of this wave?

givi [52]

I'll be happy to solve the problem using the information that
you gave in the question, but I have to tell you that this wave
is not infrared light.

If it was a wave of infrared, then its speed would be close
to 300,000,000 m/s, not 6 m/s, and its wavelength would be
less than 0.001 meter, not 12 meters.

For the wave you described . . .

   Frequency = (speed) / (wavelength)

   = (6 m/s) / (12 m)

   = 0.5 / sec

   =    0.5 Hz .

(If it were an infrared wave, then its frequency would be
greater than 300,000,000,000 Hz.)

5 0

3 years ago

Read 2 more answers

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

3 years ago

Objects change motion due to forces. When they become unbalanced, their motion changes, while balanced forces cause constant mot

Neko [114]

Answer:

Balanced forces: When a number of forces acting on a body do not cause any change in its state of rest or of uniform motion along a straight line then the forces are said to be balanced forces. In other words, a body is said to be underbalanced forced when the resulting force acting on the body is zero.

The balanced forces:

⋅ Cannot set any stationary body into motion.

⋅ May change the shape and size of soft objects.

⋅ Cannot change the speed/velocity of a moving body.

Unbalanced forces:

When the resultant of all the forces acting on a body is not zero, then forces are called unbalanced forces.

Example:

⋅ Game of tug of war: When the forces exerted by both the teams are equal, then the rope does not move. But, if the force applied by team A is greater than team B, then the rope, as well as members of the weaker team, i.e., B, will be pulled towards A. The unbalanced force can (a) Set a stationary body in motion.

⋅ Set a moving body at rest.

⋅ Change the direction of motion.

Explanation:

give me an one thanks please

5 0

3 years ago

A force of 25N acts on a mass of 5.0kg, initially at rest. Calculate the distance travelled before achieving a velocity of 20m/s

spin [16.1K]

Answer:

40m

Explanation:

let's calculate the acceleration first

force = mass × acceleration

rearranging to find acceleration:

acceleration = force ÷ mass

force = 25N, mass = 5.0kg

acceleration = 25 ÷ 5 = 5ms^-2

we can now use the formula v^2 = u^2 + 2as where v = final velocity, u = initial velocity, a = acceleration and s = distance

rearranging v^2 = u^2 + 2as the distance is

s = (v^2 - u^2) ÷ 2a

v = 20, u = 0, a = 5

s = (20^2 - 0^2) ÷ (2 × 5) = 40m

the distance is 40m

6 0

2 years ago