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vovikov84 [41]
3 years ago
7

Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect a

ny change in its brightness as it moves toward or away from the earth. Instead we can use the Doppler effect to determine its relative speed. For this problem we are going to look at the spectral lines from hydrogen, specifically the one with a wavelength of 656.46 nm.
Required:
What is V(star) , the speed at which the star is moving relative to the earth?
Physics
1 answer:
rjkz [21]3 years ago
8 0

Answer:

The speed will be "18km/s". A further explanation is given below.

Explanation:

According to the question, the values are:

Wavelength,

\lambda = 656.46 \ nm

\Delta \lambda = 0.04

c=3\times 10^8

As we know,

⇒  \frac{\Delta \lambda}{\lambda} =\frac{v}{c}

On substituting the values, we get

⇒  \frac{656.46}{0.04} =\frac{v}{3\times 10^8}

⇒         v=\frac{656.46}{0.04} (3\times 10^8)

⇒            =16411.5\times 3\times 10^8

⇒            =18280 \ m/s

or,

⇒            =18 \ km/s

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d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}

the elastic potencial energy is given by:

U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J

The work is defined as:

W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J

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The gravitational potencial energy at 2.5 m aboves is given by:

U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s

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3 years ago
What is a conservative force?
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Explanation:

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Answer:

13.6 N

Explanation:

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Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

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So, Mg - mgcos57 = F  

F = Mg - mgcos57

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We find F, thus

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F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0
3 years ago
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