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Korvikt [17]
3 years ago
12

According to the Doppler effect, objects moving away from Earth would have a cooling appearance. blue shift. red shift. warming

appearance.
Physics
2 answers:
jeka943 years ago
7 0

When you take the Doppler effect into account, it turns out that
the visible light emitted from a source moving away from Earth
will appear to have a wavelength longer than it really has, that is,
shifted toward the red end of the visible spectrum.

This is the only way we have of measuring whether a source of
light is moving away from us, and how fast.  We measure the
wavelengths of its light, and measure how much longer they are
than they should be. 

The main trick in this whole process is: When you see the light
from a star or a galaxy, how do you know what its wavelength
should be ?

Cloud [144]3 years ago
5 0

The correct answer is:

redshift.

Explanation:

Redshift and blueshift describe how light shifts approaching shorter or longer wavelengths as objects in space (such as stars or galaxies) move closer or removed away from us. The knowledge is key to charting the universe's expansion. Visible light is a spectrum of colors, which is clear to anyone who has observed at a rainbow.

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Sounds travel faster in Question 1 options: warmer air. cooler air. Temperature does not influence the speed of sound. a vacuum.
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Answer:

warmer air

Explanation:

the particles are more excited which increases the probability that the particles will bump into each other

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2 years ago
A softball with a mass of 0.11 kg goes moves at a speed of 12 m/s, then the ball is hit by a bat and rebounds in the opposite di
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3 years ago
A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra
lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

m_{1}=5.8 kg is the mass of the bowling ball

V_{1}=4.34 m/s is the velocity of the bowling ball

m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg is the mass of the ping-pong ball

V_{2} is the velocity of the ping-pong ball

Now, the momentum p_{1} of the bowling ball is:

p_{1}=m_{1}V_{1} (1)

p_{1}=(5.8 kg)(4.34 m/s)  

p_{1}=25.172 kg m/s (2)

And the momentum p_{2} of the ping-pong ball is:

p_{2}=m_{2}V_{2} (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

p_{1}=p_{2} (4)

m_{1}V_{1}=m_{2}V_{2} (5)

Isolating V_{2}:

V_{2}=\frac{m_{1}V_{1}}{m_{2}} (6)

V_{2}=\frac{25.172 kg m/s}{0.002214 kg} (7)

Finally:

V_{2}=11369.46 m/s

6 0
3 years ago
6)An electric field of 6 N/C points in the positive X direction. What is the electric flux through a surface that is 4 m2, if it
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Answer:

Flux is 21 Nm^2/C.

Explanation:

Electric field, E = 6 N/C along X axis

Electric filed vector, E = 6 i N/C

Area, A = 4 square meter

Area vector

\overrightarrow{A} = 4 (cos30 \widehat{i} + sin 30 \widehat{j})\\\\\overrightarrow{A} = 3.5 \widehat{i} + 2 \widehat{j}\\

The flux is given by

\phi= \overrightarrow{E}.\overrightarrow{A}\\\\\phi = 6 \widehat{i} .  \left (3.5 \widehat{i} + 2 \widehat{j}  \right )\\\\\phi = 21 Nm^2/C

3 0
3 years ago
A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
jeyben [28]

Answer:

V_0

Explanation:

Given that, the range covered by the sphere, M, when released by the robot from the height, H, with the horizontal speed V_0 is D as shown in the figure.

The initial velocity in the vertical direction is 0.

Let g be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. V_0 will remain constant throughout the projectile motion.

So, if the time of flight is t, then

D=V_0t\; \cdots (i)

Now, from the equation of motion

s=ut+\frac 1 2 at^2\;\cdots (ii)

Where s is the displacement is the direction of force, u is the initial velocity, a is the constant acceleration and t is time.

Here, s= -H, u=0, and a=-g (negative sign is for taking the sigh convention positive in +y direction as shown in the figure.)

So, from equation (ii),

-H=0\times t +\frac 1 2 (-g)t^2

\Rightarrow H=\frac 1 2 gt^2

\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)

Similarly, for the launched height 2H, the new time of flight, t', is

t'=\sqrt {\frac {4H}{g}}

From equation (iii), we have

\Rightarrow t'=\sqrt 2 t\;\cdots (iv)

Now, the spheres may be launched at speed V_0 or 2V_0.

Let, the distance covered in the x-direction be D_1 for V_0 and D_2 for 2V_0, we have

D_1=V_0t'

D_1=V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_1=\sqrt 2 D [from equation (i)]

\Rightarrow D_1=1.41 D (approximately)

This is in the 3 points range as given in the figure.

Similarly, D_2=2V_0t'

D_2=2V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_2=2\sqrt 2 D [from equation (i)]

\Rightarrow D_2=2.82 D (approximately)

This is out of range, so there is no point for 2V_0.

Hence, students must choose the speed V_0 to launch the sphere to get the maximum number of points.

7 0
3 years ago
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