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wariber [46]
3 years ago
11

Please help!! (Picture attached)

Physics
1 answer:
KonstantinChe [14]3 years ago
3 0
Well it is definitely answer B because when light is on for a long time it heats up a lot.the thermometer obviously went up which means the light bulb had more energy and was hotter than the start of it
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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 ball is thrown 20.0 m/s at an angle of 40.0° with the horizontal. Assume the ball is thrown at ground level.
TEA [102]
The ball's horizontal component of velocity (ie it's horizontal speed) is 20 cos 40degrees. Without knowing the distance of the ball to the wall it's difficult to go further ...
8 0
3 years ago
A girl stand 5m away from a large plane mirror.How far must she walk ti be 2m away from her image?​
KonstantinChe [14]

Answer:

3m

Explanation:

If a girl is 5m away from a mirror, she needs to walk a further 3m to be 2m away from her image.

5m-3m = 2m

4 0
1 year ago
A force of 100 newtons Is applled to a box at an angle of 36° with the horizontal. If the mass of the box Is 25 kilograms, what
faust18 [17]
Horizontal component of force = 100cos(36)= 80.9 N
5 0
3 years ago
In Challenge Example 11.9 (p. 280), after the explosion, suppose that the m1 fragment shot directly north at 12 m/s and the m3 f
Inga [223]

The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.

Solution:

It is given that mass of object before explosion is,m = 10 g

Speed of object before explosion, v = 2 m/s

Let $m_1, m_2 \text{ and}\ m_3$ be the masses of the three fragments.

Let $v_1, v_2 \text{ and}\ v_3$ be the velocities of the three fragments.

Therefore, according to the law of conservation of momentum,

$mv=m_1v_1 +m_2v_2+m_3v_3$

$10 \times 2  \hat i=3 \times 12 \hat{j} + 3(v_{2x} \hat{i}+v_{2y} \hat{j})-4 \times 9 \hat{j}$

So the x- component of the velocity of the m2 fragment after the explosion is,

$3v_{2x} = 20$

∴ $v_{2x} = 6.67 \ m/s$

6 0
3 years ago
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