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butalik [34]
3 years ago
7

An observer on the earth sees a spaceship approaching at 0.54c. The ship then launches an exploration vehicle that, according to

the earth observer, approaches at 0.82c. What is the velocity of the exploration vehicle relative to the spaceship? Give your answer as a decimal fraction of c, i.e. a β factor.
Physics
1 answer:
AURORKA [14]3 years ago
6 0

Answer:

Explanation:

Expression for relative velocity

= \frac{v_1+v_2}{1+\frac{v_1v_2}{c^2} }

= (.54 + .82 )c/ 1+ \frac{.54 \times.82}{1}

= 1.36 c / 1.4428

= .94 c

β = .94

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It is dangerous to stand near the railway leak when train passing by,why?​
4vir4ik [10]

Answer:

Because a person may be pulled in the direction of the moving train. Thereby causing accident

Explanation:

According to Daniel Bernoulli's theorem, he was widely known as a Mathematician. He stated that due to the higher velocity of a moving train, there is higher kinetic energy in terms of volume around it, while the air pressure between the person and the train becomes lower.

As a result, a person near a moving train may be pulled in the direction of the moving train. Thereby causing accidents that may lead to death.

3 0
3 years ago
A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.
Margarita [4]

Answer:

h = 13.3 m

Explanation:

Given:-

- The mass of ball, mb = 4.80 kg

- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

                                 M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mb*g*H = 0.5*mb*vb^2

                                  vb = √2*g*H

                                  vb = ( 2*9.81*15 ) ^0.5

                                  vb = 17.15517 m/s

- The angular momentum of system before collision is:

                                  M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

                                  M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

                                  M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

                                 M1 = M2

                                 vo = 247.03448 / (5.10*3)

                                 vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mo*g*h = 0.5*mo*vo^2

                                  h = vo^2 / 2*g

                                  h = 16.14604^2 / 2*(9.81)

                                  h = 13.3 m

3 0
3 years ago
Help please ill give brainly and heart !!!!
Nikitich [7]

Answer:

The answer is A ) High pressure

I hope this helps you :)

please let me know if I am wrong

3 0
3 years ago
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What fraction of all the electrons in a 25 mg water
mihalych1998 [28]

Answer:

9.11\times 10^{-15}.

Explanation:

The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of  -40 nC is removed from the water droplet.

The charge on one electron, \rm e=-1.6\times 10^{-19}\ C.

Let the N number of electrons have charge -40 nC, such that,

\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.  

Now, mass of one electron = \rm 9.11\times 10^{-31}\ kg.

Therefore, mass of N electrons = \rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.

It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.

Let it is m times the total mass of the droplet which is 25\ \rm mg = 25\times 10^{-6}\ kg.

Then,

\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.

It is the required fraction of mass of the droplet.

3 0
3 years ago
Mga halimbawa ng metapora​
faust18 [17]

Answer:

UMm If i understood ide answer

Explanation:

4 0
3 years ago
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