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kondaur [170]
3 years ago
5

The SI unit of average spped and velocity.​

Physics
1 answer:
Verdich [7]3 years ago
3 0

<em>meter per</em><em> </em><em>second</em><em> </em><em>is </em><em>the </em><em>main </em><em>answer </em><em>of</em><em> </em><em>both</em>

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A skydiver jumps out of a hovering helicopter. A few seconds later, another diver jumps out, so they both fall along the same ve
Sergio039 [100]

Answer:

distance difference would a) increase

speed difference would f) stay the same

Explanation:

Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.

Their equations of motion for distance and velocities are

s_2 = gt^2/2

s_1 = g(t + \Delta t)^2/2

v_2 = gt

v_1 = g(t + \Delta t)

Their difference in distance are therefore:

\Delta s = s_1 - s_2 = g(t + \Delta t)^2/2 - gt^2/2

\Delta s = g/2((t + \Delta t)^2 - t^2)

\Delta s = g/2(t + \Delta t - t)(t + \Delta t + t) (AsA^2 - B^2 = (A-B)(A+B)

\Delta s = g\Delta t/2(2t + \Delta t)

So as time progress t increases, Δs would also increases, their distance becomes wider with time.

Similarly for their velocity difference

\Delta v = v_1 - v_2 = g(t + \Delta t) - gt

\Delta v = gt + g\Delta t - gt = g\Delta t

Since g and Δt both are constant, Δv would also remain constant, their difference in velocity remain the same.

This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.

8 0
3 years ago
Read 2 more answers
A 350-N child is in a swing that is attached to a pair of ropes 2.10 m long. Find the gravitational potential energy of the chil
o-na [289]

Answer:

a)  U = 735 J , b) U = 125.7 J , c)   U = 0 J

Explanation:

The gravitational power energy is

      U = mg y - mg y₀

The last value is a constant, for simplicity we can make it zero, if the lowest point is at the origin of the coordinate system, which in this case we will place in the lowest part

a) Rope is horizontal

The height in this case is the same length of the rope

     y = 2.10 m

    w = mg = 350 N

    U = 350 2.10

    U = 735 J

b) when the angle is 34º

     y = L - L cos 34

    y = L (1- cos34)

    y = 2.10 (1- cos 34)

    y = 0.359 m

    U = 350 0.359

    U = 125.7 J

c) in this case this point coincides with the reference system

     y = 0

     U = 0 J

4 0
4 years ago
The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc
kumpel [21]
The Answer is 11,121 Kg
6 0
3 years ago
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An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver
Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, E=2.3\times 10^{-10}\ N/C

Vertical distance covered, s=1\ \mu m=10^{-6}\ m

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

v^2-u^2=2as

v^2=2as.............(1)

Electric force is F_e and force of gravity is F_g. As both forces are acting in downward direction. So, total force is:

F=mg+qE

F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}

F=4.57\times 10^{-29}\ N

Acceleration of the electron, a=\dfrac{F}{m}

a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}

a=50.21\ m/s^2

Put the value of a in equation (1) as :

v=\sqrt{2as}

v=\sqrt{2\times 50.21\times 10^{-6}}

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

6 0
4 years ago
The opposite poles of two magnets will do which of the following
madam [21]

To what i see, the answer is....

C.

6 0
3 years ago
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