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irakobra [83]
2 years ago
6

You and your friend are both 14 years old. Your friend however moves to a planet where just 30 minutes on that planet is the sam

e as 5 years on Earth. It has been 10 years on Earth since your friend moved.
1) How much time has passed for your friend?
2) How old is your friend?
3) How old are you?
4) Who is older, you or your friend?
Physics
1 answer:
vaieri [72.5K]2 years ago
5 0

Answer:

ok so

Explanation:

1.) 1 hour

2.) 14 years old

3.)

24 years old

4.) you are

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Speed of 26.7 m/s in 3.06 s. How far had the car traveled by the time the final speed was achieved?
Alik [6]

Answer:

Below

Explanation:

You can use this equation to find the distance :

     distance = velocity x time

     distance = (26.7)(3.06)

                    = 81.702 m

Rounding to 3 sig figs

     = 81.7 m

Hope this helps

6 0
2 years ago
Which of the following describes an accelerated motion
Naily [24]
Accelerated motion such as a vehicle (car) or a moving item such as a (football thrown in the air)
3 0
3 years ago
One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories
ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = 9.11\times 10^{-31}\ kg

B = Magnetic field = 0.044 T

q = Charge of electron = 1.6\times 10^{-19}\ C

The centripetal force and the magnetic forces are conserved

m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}

Velocity of first electron

v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s

Velocity of second electron

v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s

Total kinetic energy is given by

K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J

Converting to eV

1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV

The energy of incident electron is 114.92749 keV

5 0
3 years ago
A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is movin
gogolik [260]

Answer:

a) 23 m/s

Explanation:

  • Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

       p_{o} = p_{f}  (1)

  • The initial momentum p₀, can be written as follows:

       p_{o} =  m_{1}  * v_{1o} + m_{2}* v_{2o} =   6.0 kg * 5.0 m/s + 2.0 kg * v_{2o}  (2)

  • The final momentum pf, can be written as follows:

        p_{f} = (m_{1} + m_{2} )* v_{f}  = 8.0 kg* (-2.0 m/s)  (3)

  • Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

       v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg}  = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s  (4)

  • This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.
6 0
3 years ago
Help me pls
Zina [86]

Answer:

Explanation:

the first one is D

8 0
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