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3 years ago

5

What will happen to the absorption and/or emission spectral lines of an object moving away from Earth at high speed?

1 answer:

Rufina [12.5K]3 years ago

4 0

Answer:

c. They will be shifted toward the red end of the spectrum.

Explanation:

When the absorption or emission lines of a moving object away from from the earth at very high speed then the spectral lines will be shifted toward the red end of the spectrum.This effect can be easily understand by Doppler effect..

So the our option c is correct.

c. They will be shifted toward the red end of the spectrum.

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A piece of rope is pulled by two people in a tug-of-war. each exerts a 400-n force. what is the tension in the rope?

Leokris [45]

Newtons 3.law: Action = Reaction

If a body exerts a force on a rope of 400 N the rope exerts a force on the body of 400N also. So the tension in the rope is 400N. See pictures below.

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4 years ago

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Ms. Sparkle bought 12 cans of diet soda. Each can

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3 years ago

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the <em>x</em>-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its <em>y</em>-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

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3 years ago

A ranger in a national park is driving at 56 km/h when a deer jumps onto the road 65 m ahead of the vehicle. After a reaction ti

vagabundo [1.1K]

Answer:

t = 1.58 s

Explanation:

given,

Speed of ranger, v = 56 km/h

v = 56 x 0.278 = 15.57 m/s

distance, d = 65 m

deceleration,a = 3 m/s²

reaction time = ?

using stopping distance formula

$d = v. t + \dfrac{v^2}{2a}$

$t = \dfrac{d}{v} -\dfrac{v}{2a}$

t is the reaction time

$t = \dfrac{65}{15.57} -\dfrac{15.57}{2\times 3}$

t = 1.58 s

hence, the reaction time of the ranger is equal to 1.58 s.

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3 years ago

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kenny6666 [7]

Answer:

to prevent cuz now for my mind when you prevent it it will not happen to you or nothing happened to you

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3 years ago