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Artemon [7]
3 years ago
9

Star #1 is approaching the Earth with speed v. Star #2 is receding from the Earth with the same speed v. Measurements of the sam

e spectral line from each of the stars show Doppler shifts in frequency. The light from which star will have the larger magnitude shift in frequency? a. star #1 b. Both stars will have the same shift. c. The value of the speed must be known before an answer can be found. d. star #2
Physics
1 answer:
leva [86]3 years ago
4 0

Answer:

b. Both stars will have the same shift.

Explanation:

It's a very simple problem to solve. Star 1 is approaching toward Earth with a speed v, so let's assume that the change in Doppler Shift is +F and Star 2 is moving away so the change in Doppler shift is -F. But it's time to notice the speed of both stars and that is same but only directions are different. speed is the main factor here. The magnitude of both shifts is F as we can see and + and - are showing there direction of motion. So, because of same amount of speed, both stars will have same shift magnitude. (Just the directions are different)

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frez [133]

Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

a(t)=(5-10t)\ m/s^2

(a) Since, a=\dfrac{dv}{dt}

v = velocity

dv=a.dt

v=\int(a.dt)

v=\int(5-10t)(dt)

v=5t-\dfrac{10t^2}{2}=5t-5t^2

(b) v=\dfrac{dx}{dt}

x = position

x=\int v.dt

x=\int (5t-5t^2)dt

x=\dfrac{5}{2}t^2-\dfrac{5}{3}t^3

(c) Velocity function is given by :

v=5t-5t^2

5t-5t^2=0

t = 1 seconds

So, at t = 1 second the velocity of the particle is zero.

7 0
2 years ago
A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball
adell [148]

Answer:

Part a)

T = 2\sqrt{\frac{R}{3g}}

Part b)

v_x = \frac{\sqrt{3Rg}}{2}

Part c)

v_y = \sqrt{Rg/3}

Part d)

v = \frac{1}{2}\sqrt{13Rg}

Part e)

\theta_i = 33.7 degree

Part f)

H = \frac{13R}{8}

Part g)

X = \frac{13R}{4}

Explanation:

Initial speed of the launch is given as

initial speed = v_i

angle = \theta_i degree

Now the two components of the velocity

v_x = v_i cos\theta_i

similarly we have

v_y = v_i sin\theta_i

Part a)

Now we know that horizontal range is given as

R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}

maximum height is given as

H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}

so we have

v_i sin\theta = \sqrt{Rg/3}

time of flight is given as

T = \frac{2v_isin\theta_i}{g}

T = \frac{2\sqrt{Rg/3}}{g}

T = 2\sqrt{\frac{R}{3g}}

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

R = v_x T

R = v_x 2\sqrt{\frac{R}{3g}}

v_x = \frac{\sqrt{3Rg}}{2}

Part c)

Initial vertical velocity is given as

v_y = v_i sin\theta_i

v_i sin\theta = \sqrt{Rg/3}

Part d)

Initial speed is given as

v = \sqrt{v_x^2 + v_y^2}

so we will have

v = \sqrt{Rg/3 + 3Rg/4}

v = \frac{1}{2}\sqrt{13Rg}

Part e)

Angle of projection is given as

tan\theta_i = \frac{v_y}{v_x}

tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}

\theta_i = 33.7 degree

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

H = \frac{v^2}{2g}

H = \frac{13R}{8}

Part g)

For maximum range the angle should be 45 degree

so maximum range is

X = \frac{v^2}{g}

X = \frac{13R}{4}

3 0
3 years ago
In a "worst-case" design scenario, a 2000 kg elevator with broken cables is falling at 4.00 m/s when it first contacts a cushion
Leno4ka [110]

Answer:

4 m/s²

Explanation:

 When the elevator is 1 m below point of contact , compression will be 1 m.

Restoring force in the spring will be 10600 N. Friction force of 17000N  will also act in upward direction . The weight of 2000 x 9.8 N will act downwards

Force in down ward direction = 2000 x 9.8

= 19600 N

Force in upward direction

= 10600 + 17000

= 27600 N

Net force in upward direction

= 27600 - 19600

= 8000 N

Acceleration in upward direction

= 8000 / 2000

= 4 m/s²

5 0
3 years ago
The tension in the rope securing the upper pulley and the force that must be applied to keep the system in equilibrium ​
chubhunter [2.5K]

Answer:

Explanation:

In a frictionless system with no acceleration, the tension in the rope must be F along its entire length

FBD analysis of the lower pulley has two upward acting tension vectors F and one downward acting weight vector W

2F = W

F = W/2

FBD analysis of the upper pulley has one upward acting support vector T and three downward acting tension vectors F

T = 3F

T = 3(W/2)

T = 1.5W

8 0
2 years ago
Land plants are adapted to survive on land through stomata to regulate water regulation, specialized cells, structures that coll
Law Incorporation [45]

Answer:

stomata to regulate water

specialized cells

Explanation:

Land plants are able to successfully adapt to staying on land through the use of stomata to regulate water loss and through the development of specialized cells that give support to grow erect.

Water regulation is important for the survival of plants on land, otherwise, they may face desiccation as a result of the heat from the sun. <u>The stomata is a structure found on the leaves of land plants that open and close to regulate the amount of water lost to the environment through evapotranspiration.</u> When there is adequate water and the plant can afford to lose some water, the stomata open, otherwise, the stomata close.

Support is also important for land plant's survival and the development of specialized cells in the form of <em>sclerenchyma cells</em> take care of this. The <em>fib</em>ers and <em>sclerids</em> are two types of sclerenchyma cells that primarily function to support the land plants.

8 0
2 years ago
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