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valkas [14]
2 years ago
6

A roller coaster is released from the top of a track that is 57 m high. Find the roller coaster speed when it reaches ground lev

el.
Physics
1 answer:
Art [367]2 years ago
7 0

Hi there!

Assuming the track is frictionless:

E_i = E_f\\\\PE = KE\\\\mgh = \frac{1}{2}mv^2

Cancel out the masses and rearrange to solve for velocity:

gh = \frac{1}{2}v^{2}\\\\v = \sqrt{2gh}\\\\

Plug in the given height and let g = 9.8 m/s²:

v = \sqrt{2(9.8)(57)} = \boxed{33.42 m/s}

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A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i
erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

\mu_k = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem,  we have

Work done by the friction force = change in kinetic energy of the mass 2

or

0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring =  Kinetic energy of the mass 2

or

\frac{1}{2}kx^2=\frac{1}{2}mv_3^2

on substituting the values, we get

\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2

or

x = 2.86 m

8 0
3 years ago
A concave mirror always forms a virtual image of a real object. O True O False
kifflom [539]

Answer:

The given statement is false.

Explanation:

The spherical mirrors are the mirror that are a part of a sphere. Concave and convex mirrors are two types of spherical mirrors.

A concave mirror always forms real and inverted image. A convex mirror forms real and virtual images.

For concave mirror, the value of magnification is less that 1. Also, the focal length is negative for concave mirrors.

So, the given statement is false as a concave mirror always forms a real and inverted image. Hence, this is the required solution.

4 0
3 years ago
(EARTH SPACE SCIENCE QUESTION)
Sergeeva-Olga [200]

*may be egg shaped

*has no new stars being formed

*has almost no gas or dust between stars

I literally just took notes on elliptical galaxies a week ago wow

8 0
2 years ago
Read 2 more answers
At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl
Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

s=ut+0.5at^2 \\

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\

Solving both equations

600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\

Passing of B occurs at 4108.31 height.

6 0
3 years ago
A golfer hits a golf ball upwards at an angle. We can ignore air resistance on the ball.
Natalka [10]

Answer:

Check the diagram from the photo

Explanation:

5 0
2 years ago
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