Climate is best described as the

what would the mass be of an object that was moving at a velocity of 35 m/s and has a kinetic energy of 500j be?

bearhunter [10]

Answer:

0.816 kg

Explanation:

$E_k=\frac{1}{2}mv^{2}\\$

so $m=\frac{2E_k}{v^2}=\frac{2\times500}{35^2}=0.816 kg$

5 0

2 years ago

The bending of light as it passes into a transparent material of different optical intensity is known as A. conversion. B. aberr

stiks02 [169]

Answer:

D. refraction

Explanation:

Refraction: Refraction is change in direction of light rays. Refraction occurs whenever light rays travels from a transparent medium to another transparent medium of different density. The abrupt change in direction at the surface of the surface of the two media is referred to as refraction.

Refraction occurs when light travels from air to glass or from air to liquid.

Laws Of Refraction:

(i) The incident ray, the refracted ray and the normal, all at the point of incident lies in the same plane.

(ii) The ratio of the sine of the angle of incident to the sine of the angle of refraction is a constant for a given pair of media.

Thus the right option is D. refraction

6 0

3 years ago

What occurs when light changes direction after colliding with particles of matter

Slav-nsk [51]

Scattering occurs when light changes direction after colliding with particles of matter.

7 0

3 years ago

Worth 50 point!!!

stira [4]

Answer:

fact one

Explanation:

fact two is going slower and a longer distance than fact one so fact one will get there first.

hope this helps

4 0

3 years ago

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

5 0

3 years ago