All categories

2 years ago

HELP FAST

Physics

1 answer:

katen-ka-za [31]2 years ago

3 0

Answer:

C. 2 and 4

Explanation:

my teacher went over it and the answer was that

You might be interested in

Near the center of our galaxy, hydrogen gas is moving directly away from us in its orbit about a black hole. The electromagnetic

klemol [59]

To solve this problem it is necessary to apply the concept related to wavelength, specifically when the wavelength is observed from a source that is in motion to the observer.

By definition the wavelength is given defined by,

$\lambda_{obs} = \lambda_s \sqrt{\frac{1+u/c}{1-u/c}}$

Where

$\lambda_{obs}$ = Observed wavelength

$\lambda_s$ = Wavelength of the source

c = Speed of light in vacuum

u = Relative velocity of the source to the observer

According to our data we have that the wavelength emitted from the galaxy is 1875nm which is equal to the wavelength from the source, while the wavelength from the observer is $\lambda_{obs}=1945nm$

Therefore replacing in the previous equation we have,

$1945 = 1875 \sqrt{\frac{1+\frac{u}{c} }{1-\frac{u}{c} }}$

$\sqrt{\frac{1+u/c}{1-u/c}} = 1.03733$

$\frac{1+\frac{u}{c} }{1-\frac{u}{c} }=1.03733^2$

$1+\frac{u}{c} =1.03733^2*(1-\frac{u}{c} )$

Solving for u,

$1+\frac{u}{c} =1.03733^2*(1-\frac{u}{c} )$

$1+\frac{u}{c} =1.03733^2-1.03733^2(\frac{u}{c} )$

$\frac{u}{c} +1.03733^2\frac{u}{c} =1.03733^2-1$

$2.88595\frac{u}{c}=1.03733^2-1$

$\frac{u}{c} = \frac{1.03733^2-1}{2.88595}$

$u = \frac{1.03733^2-1}{2.88595}*c$

$u = 0.02635c$

Therefore the speed of the gas relative to earth is 0.02635 times the speed of light.

6 0

3 years ago

A car accelerates from rest at a constant rate of 1.6

SIZIF [17.4K]

Answer:

8m/s

Explanation:

it says that its at a constant rate of 1.6 that means it doesn't change and the car accelerates for 5 seconds so 1.6 X 5 which is 8

7 0

3 years ago

By what factor must we increase the amplitude of vibration of an object at the end of a spring in order to double its maximum sp

strojnjashka [21]

Answer:

$A'=2A$

Explanation:

According to the law of conservation of energy, the total energy of the system can be expresed as the sum of the potential energy and kinetic energy:

$E=U+K=\frac{kA^2}{2}\\E=\frac{kx^2}{2}+\frac{mv^2}{2}=\frac{kA^2}{2}$

When the spring is in its equilibrium position, that is $x=0$ , the object speed its maximum. So, we have:

$\frac{k(0)^2}{2}+\frac{mv_{max}^2}{2}=\frac{kA^2}{2}\\A^2=\frac{mv_{max}^2}{k}\\A=\sqrt{\frac{mv_{max}^2}{k}}$

In order to double its maximum speed, that is $v'{max}=2v_{max}$ . We have:

$A'=\sqrt{\frac{m(v'_{max})^2}{k}}\\A'=\sqrt{\frac{m(2v_{max})^2}{k}}\\A'=\sqrt{\frac{4mv_{max}^2}{k}}\\A'=2\sqrt{\frac{mv_{max}^2}{k}}\\A'=2A$

6 0

3 years ago

How does mass affect weight?

devlian [24]

Answer: Mass affects the weight of an object with the effects of gravity.

Weight is the measure of the force of gravity on an object's mass, while mass is the measure of how much matter there is in an object.

5 0

3 years ago

#86 can an object be moving when its acceleration is zero? can an object be accelerating when its speed is zero?

liraira [26]

Yes, an object that was set in motion in the past by some force, but that is no longer being acted on by a net force, is moving but with zero acceleration, i.e. it is moving at constant velocity.

7 0

3 years ago