HELP ME PLEASEE!!!THIS IS SCIENCE

a rock is dropped from a height of 80 m and is in free fall what is the velocity of as it reaches the ground 4.0 seconds later

Marta_Voda [28]

Velocity = displacement (distance)/time

v=80m/4s

v=20m/s

velocity = 20 meters per second

8 0

3 years ago

A periodic wave travels from one medium to another. Which pair of variables are likely to change in the process? A. velocity and

Lesechka [4]

Answer:

A. velocity and wavelength

Explanation:

When a wave travels from one medium to another it undergoes refraction which results to the change in direction.

Refraction of a wave is one of the property of waves that occurs when a wave changes direction when it passes from one medium to another. This occurs as a result of bending of the wave which occurs since the mediums involved are of different density and refractive index.

Apart from the change in direction, refraction is accompanied by the change in wavelength of a wave and thus a change in speed.

5 0

3 years ago

A particle moving along a straight line is subjected to a deceleration ???? = (−2???? 3 ) m/???? 2 , where v is in m/s. If it ha

dlinn [17]

Answer:

(a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

Explanation:

Given that,

The deceleration is

$a=(-2v^3)\ m/s^2$

We need to calculate the velocity at t = 25 s

The acceleration is the first derivative of velocity of the particle.

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=-2v^3$

$\dfrac{dv}{-v^3}=2dt$

On integrating

$int{\dfrac{dv}{-v^3}}=\int{2dt}$

$\dfrac{1}{2v^2}=2t+C$

$v^2=\dfrac{1}{4t+2C}$ ....(I)

At t = 0, v = 10 m/s

$10^2=\dfrac{1}{4\times0+2C}$

$C=\dfrac{1}{200}$

Put the value of C in equation (I)

$v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}$

$v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}$

$v=0.099\ m/s$

The velocity is 0.099 m/s.

(b). We need to calculate the position at t = 25 sec

The velocity is the first derivative of position of the particle.

$\dfrac{ds}{dt}=v$

On integrating

$\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt$

$s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'$

At t = 0, s = 15 m

$15=\dfrac{200}{800}+C'$

$C'=15-\dfrac{200}{800}$

$C'=14.75$

Put the value in the equation

$s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75$

$s=19.75\ m$

The position is 19.75 m.

Hence, (a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

3 0

3 years ago

Cara is swimming in a river. She can swim downstream up to a maximum speed of 66 meters per-minute. If she is swimming downstrea

makvit [3.9K]

It would be 429 because in the question it said “ if she was swimming downstream of HALF of her maximum speed, how far will she swim for 13 minutes” so first you would divide 66 by 2 and get 33 then you would multiply 33 by 13 to get you’re answer.

8 0

3 years ago

Read 2 more answers

What is the biological impact of minimum catch sizes on a population of fish?

ivann1987 [24]

The population comes to be dominated by smaller, slower-growing individuals

7 0

3 years ago