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KengaRu [80]
4 years ago
7

The force required to make an object travel along a circular path is:

Physics
2 answers:
AVprozaik [17]4 years ago
6 0

it the centripetal force

trust me

garik1379 [7]4 years ago
6 0

Answer:

centripetal force

Explanation:

just did this on a question

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A 75 kg bungee jumper leaps from a bridge. When he is 30 meters above the water, and moving at a speed of 20 m/s, the bungee cor
jasenka [17]

Answer:

k = 52.2 N / m

Explanation:

For this exercise we are going to use the conservation of mechanical energy.

Starting point. When it is 30 m high

        Em₀ = K + U = ½ m v² + m g h

Final point. Right when you hit the water

        Em_{f} = K_{e} = ½ k x²

in this case the distance the bungee is stretched is 30 m

        x = h

as they indicate that there are no losses, energy is conserved

        Em₀ = Em_{f}

       ½ m v² + m g h = ½ k h²

       k = \frac{m (v^{2} + 2 g h)}{h^{2} }

let's calculate

       k = \frac{75 \ ( 20^{2}  + 2 \ 9.8 \ 30)}{30^{2} }

       k = 52.2 N / m

3 0
3 years ago
A proton of mass is released from rest just above the lower plate and reaches the top plate with speed . An electron of mass is
xenn [34]

Answer:

  v = √ 2e (V₂-V₁) / m

Explanation:

For this exercise we can use the conservation of the energy of the electron

At the highest point. Resting on the top plate

         Em₀ = U = -e V₁

At the lowest point. Just before touching the bottom plate

        Emf = K + U = ½ m v² - e V₂

Energy is conserved

         Em₀ = Emf

          -eV₁ = ½ m v² - e V₂

           v = √ 2e (V₂-V₁) / m

Where e is the charge of the electron, V₂-V₁ is the potential difference applied to the capacitor and m is the mass of the electron

3 0
4 years ago
The spring constant in a given oscillating mass spring may be changed by
tia_tia [17]
(D) None of the above.
5 0
3 years ago
Three carts of masses 4.0 kg, 10kg, and 3.0 kg move on a frictionless track with speeds of v1 = 5.0m/s, v2=3.0m/s, and v3=-3.6 m
gogolik [260]

2.24 m/s is the calculated velocity.

Initial velocity (u) squared plus two times the acceleration (a) times the displacement equals final velocity (v) squared (s). Final velocity (v) is equal to the square root of initial velocity (u) squared plus two times the acceleration (a) times displacement when v is the variable being solved for (s).

The cart's masses and speeds are known.

M1 = 4.00 kg, M2 = 10.0 kg, M3 = 3.00 kg, etc.

v1 = 5.00 m/s = 5.00 m/s, v2 = 3.00 m/s = 3.00 m/s, v3 = -4.00 m/s = 4.00 m/s, and m1v1+m2v2+m3v3 = (m1+m2+m3) v=d frac m 1v 1+m 2v 2+m 3v 3, where (m1+m2 + m3) is the product of (v1 v 1+m2v2+m3v3).

"m 1+m 2+m 3" is equivalent to "m 1+m2+m3/m1v 1+m2v2 +m3v3"

the three carts' final velocities are calculated as follows: v=d frac

{4.00kg\sdot5.00m/s+10.0kg\sdot3.00m/s-3.00kg\sdot4.00m/s} 4.24m/s = 4.50kg+10.0kg+3.00kg vs. 4.50kg+10.0kg+3.00kg

5.00m/s/4.00kg/5.00m/s+10.0m/s/3.00m/s/4.00m/s =2.24m/s.

2.24 m/s is the calculated final velocity.

Learn more about velocity here-

brainly.com/question/18084516

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7 0
2 years ago
Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a sp
Alex787 [66]

Answer:

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.

Explanation:

i want the answer i don't know

7 0
3 years ago
Read 2 more answers
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