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3 years ago

What Doppler shift would an astronomer notice for the spectral light from a galaxy moving away from Earth? Choose

Physics

1 answer:

Misha Larkins [42]3 years ago

4 0

Answer:

In contrast, the light from a star moving away from us seems to shift towards longer wavelengths. As this is towards the red end of the spectrum, astronomers call it redshift

Explanation:

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1. Which object is farthest from the origin at t=2sec

Stolb23 [73]

Answer:

that one i know only pe not that sorry again

6 0

3 years ago

According to Newton’s first law of motion, when will an object at rest begin to move?

Natasha_Volkova [10]

Newton's first law of motion says something like "An object remains
in constant, uniform motion until acted on by an external force".

Constant uniform motion means no change in speed or direction.
If an object changes from rest to motion, that's definitely a change
of speed. So it doesn't remain in the state of constant uniform
motion (none) that it had when it was at rest, and that tells us
that an external force must have acted on it.

8 0

4 years ago

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At a particular temperature, the speed of sound is 330 m/s and its frequency is 990 Hz. Choose the correct wavelength of this so

koban [17]

Answer:

1/3 M

I hope this helps :) sorry its late

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3 years ago

Read 2 more answers

Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father

Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity $\omega_o = 0$

angular acceleration $\alpha$ = 4.44 rad/s²

Using the formula:

$\omega = \omega_o+ \alpha t$

Making t the subject of the formula:

$t= \dfrac{\omega- \omega_o}{ \alpha }$

where;

$\omega = 1.53 \ rad/s^2$

∴

$t= \dfrac{1.53-0}{4.44 }$

t = 0.345 s

Using the formula:

$\omega ^2 = \omega _o^2 + 2 \alpha \theta$

here;

$\theta$ = angular displacement

∴

$\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }$

$\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }$

$\theta =0.264 \ rad$

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

$x = \dfrac{0.264 \times 1}{2 \pi}$

x = 0.042 revolutions

Here; force = 270 N

radius = 1.20 m

The torque = F * r

$\tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm$

However;

From the moment of inertia;

$Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}$

given that;

I = 84.4 kg.m²

$\alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2$

For re-tardation; $\alpha=-3.84 \ rad/s^2$

Using the equation

$t= \dfrac{\omega- \omega_o}{ \alpha }$

$t= \dfrac{0-1.53}{ -3.84 }$

$t= \dfrac{1.53}{ 3.84 }$

t = 0.398s

The required time it takes= 0.398s

5 0

3 years ago

two circular loops of wire, each containing a single turn, have the same radius of 4.0 cm and a common center. the planes of the

REY [17]

The magnetic field at center of circular loops of wire is 3.78 x 10¯⁵ T.

We need to know about the magnetic field at the center of circular loops of wire to solve this problem. The magnetic field at the center can be determined as

B = μ₀ . I / 2r

where B is magnetic field, μ₀ is vacuum permeability (4π×10¯⁷ H/m), I is the current and r is radius.

From the question above, we know that:

r = 4 cm = 0.04 m

I = 1.7 A

By substituting the parameter, we get

B = μ₀ . I / 2r

B = 4π×10¯⁷ . 1.7 / (2.0.04)

B = 2.67 x 10¯⁵ T

Due to the perpendicular plane of loops, the total magnetic field at center will be

Btotal = √(2(B²))

Btotal = √(2(2.67 x 10¯⁵²))

Btotal = 3.78 x 10¯⁵ T

Find more on magnetic field at: brainly.com/question/7802337

#SPJ4

3 0

2 years ago