PLEASE ASAP ILL GIVE BRAINLIEST.

What properties are different between blue light and red light?

Oksi-84 [34.3K]

Blue light has a shorter wavelength than does red light, and since it travels at the same speed (in vacuum) as red light, has a higher frequency. This gives it a higher energy per photon.

3 0

3 years ago

HELP ASAP due in 5m

Papessa [141]

Answer:

I belive it is the one at the bottom of the table visable in the picture with 27.4 radioactive age

Explanation:

8 0

3 years ago

On planet X, an object weighs 7.04 N. Onplanet B where the magnitude of the free-fallacceleration is 1.91g(whereg= 9.8 m/s2isthe

timurjin [86]

<h2>Mass of object in Earth is 1.37 kg</h2>

Explanation:

On planet B where the magnitude of the free-fall acceleration is 1.91g , the object weighs 25.74 N.

We have

Weight = Mass x Acceleration due to gravity

On planet B

25.74 = Mass x 1.91 g

25.74 = Mass x 1.91 x 9.81

Mass = 1.37 kg

Mass is constant for an object. It will not change with location.

Mass of object in Earth = Mass of object in Planet B

Mass of object in Earth = 1.37 kg

4 0

3 years ago

An dog is transported from the earth to the moon. which measurement is the same in both places

lys-0071 [83]

The dog's length, width, height, mass, volume, density, and color are the same in both places.

4 0

3 years ago

A person jumps from the roof of a house 3.5-m high. When he strikes the ground below, he bends his knees so that his torso decel

Korvikt [17]

The force exerted on his torso by his legs during the deceleration is 4365 N.

<u>Explanation:</u>

Mass of the torso m=45kg

Height of the building s=3.5 m

Decelerating distance=0.71 m

when he jumps to the ground, the only acceleration is acceleration due to gravity g

<u>motion1 from top to ground </u>

initial velocity u=0

we have to calculate final velocity v using the following equation of motion.

$v^2-u^2=2gs\\v^2-0^2=2\times 9.8\times3.5=68.6\\v=\sqrt{68.6} \\=8.3$

use height of the building as the distance s as the jump from top to the ground is only described here.

<u>Motion 2 on the ground</u>

v=0

u=8.3(final velocity of motion 1)

The deceleration after striking the ground can be calculated from the equation of motion

$v^2-u^2=2as\\\\a=v^2-u^2/2\times 0.71\\=0^2-8.3^2/0.71=97 m/s^2$

The decelerating distance is used in the place of s since since the motion after hitting the ground is described in this case.

The equation of force is

$F=ma\\=45\times 97=4365 N$

6 0

3 years ago