All categories

2 years ago

Please help me

Physics

1 answer:

katrin2010 [14]2 years ago

8 0

The pitch of a sound is determined by its wavelength and speed together, or its SPEED alone.

Helped?

Brainliest?

You might be interested in

What are organelles?<br>Which part of a cell holds organelles?

DanielleElmas [232]

Answer:

An organelle is a structure in a cell that has specific jobs to do. Organelles are held in the cytoplasm

3 0

3 years ago

Read 2 more answers

Explain the process of refraction of light

brilliants [131]

The process of refraction of light occurs when light rays bends when travelling between media of different densities.

What is refraction of light?

Refraction of light is the bending of light rays or the change in the direction of light rays as it travels between media of different densities.

Light waves travel faster in media of less density than media of more density.

The change in density of the media makes light waves to be bend towards or away from the normal.

For example, when light travels from less dense air to more dense water, the light rays are bent towards the normal. However, when light rays travel from water to air, the light rays are refracted away from the normal.

In conclusion, refraction of light waves occur when light crosses the boundary of media of different densities.

Learn more about refraction of light at: brainly.com/question/27932095

#SPJ1

4 0

1 year ago

If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object paral

leonid [27]

Answer:

$\theta = tan^{-1}\mu$

Explanation:

As we know that if the object is placed on the inclined plane then the force of friction on the object is counterbalanced by the component of the weight of the object along the inclined plane.

So we can say

$F_f = mgsin\theta$

now if we increase the inclination of the plane then the component of the weight weight along the inclined plane will increase and hence the friction force will also increase.

As we know that the limiting value or the maximum value of friction force at the static condition is given by

$F_f = \mu N$

$N = mg cos\theta$

so we have

$F_f = \mu (mg cos\theta)$

so we will have

$mg sin\theta = \mu (mg cos\theta)$

so now we have

$tan\theta = \mu$

so maximum possible angle of the inclined plane is

$\theta = tan^{-1}\mu$

3 0

2 years ago

______ had occurred when your body can no longer regulate its temperature

sergejj [24]

Heat stroke had occurred when your body can no longer regulate its temperature. hope it helps :)

4 0

2 years ago

magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,

Mama L [17]

Answer:

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

Explanation:

Given:

height above which the rock is thrown up, $\Delta h=50\ m$

initial velocity of projection, $u=20\ m.s^{-1}$

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

$v=u+g'.t'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0=20-g'.t'$ (-ve sign to indicate that acceleration acts opposite to the velocity)

$t'=\frac{20}{g'}\ s$

The time taken by the rock to reach the top height on the earth:

$v=u+g.t$

$0=20-g.t$

$t=\frac{20}{g} \ s$

Height reached by the rock above the point of throwing on the exoplanet:

$v^2=u^2+2g'.h'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0^2=20^2-2\times g'.h'$

$h'=\frac{200}{g'}\ m$

Height reached by the rock above the point of throwing on the earth:

$v^2=u^2+2g.h$

$0^2=20^2-2g.h$

$h=\frac{200}{g}\ m$

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

$(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2$ (during falling it falls below the cliff)

here:

$u=$ initial velocity= 0 $m.s^{-1}$

$\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2$

$t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}$

$t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'} }$

Similarly on earth:

$t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g} }$

Now the required time difference:

$\Delta t=(t'+t_f')-(t+t_f)$

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

3 0

2 years ago