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luda_lava [24]
3 years ago
15

A ball is thrown straight upward on the Moon. Is the maximum height it reaches less than, equal to, or greater than the maximum

height reached by a ball thrown upward on the Earth with the same initial speed (no air resistance in both cases)?
Physics
2 answers:
coldgirl [10]3 years ago
7 0
They will have the same kinetic energy as the release for both. (1/2)mV^2=mgh
so h=5V^2/g
if g is 1/6 that on Earth the h will be 6 times higher on the moon. Also look up the question on google because someone posted it on o p e n s t u d y and had a better and more clear explanation then me.
IgorC [24]3 years ago
6 0
The maximum height on moon is HIGHER than the maximum height on earth
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Which statements describe the image produced by a concave lens?check all that apply.
kotegsom [21]

<em>Answer:</em>

<em>The answers are: </em>

  • <em>A-which is the image is always right side up.</em>
  • <em>E-the image is virtual</em>

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<em>Explanation: MY EXPLANATION IS YOU ARE WELCOME BIG DOG 100..</em>

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7 0
3 years ago
Read 2 more answers
A swinging pendulum has a total energy of <img src="https://tex.z-dn.net/?f=E_i" id="TexFormula1" title="E_i" alt="E_i" align="a
Zolol [24]

Answer:

\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta} (for small oscillations)

Explanation:

The total energy of the pendulum is equal to:

E_{1} = m\cdot g \cdot (1-\cos \theta)\cdot L

For small oscillations, the equation can be re-arranged into the following form:

E_{1} \approx m\cdot g \cdot (1-\theta) \cdot L

Where:

\theta = \frac{A}{L^{2}}, measured in radians.

If the amplitude of pendulum oscillations is increase by a factor of 4, the angle of oscillation is 4\theta and the total energy of the pendulum is:

E_{2} \approx m\cdot g \cdot (1-4\theta)\cdot L

The factor of change is:

\frac{E_{2}}{E_{1}} \approx \frac{1 - 4\theta}{1-\theta}

\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}

3 0
3 years ago
What constant acceleration, in SI units, must a car have to go from zero to 60 mph in 10 s? How far has the car traveled when it
nalin [4]

Answer:

Explanation:

initial velocity, u = 0

final velocity, v = 60 mph = 26.8 m/s

time t = 10 s

Let a be the acceleration and s be he distance traveled.

Use first equation of motion

v = u + a t

26.8 = 0 +  a x 10

a = 2.68 m/s

Use second equation of motion

s = ut + 1/2 at²

s = 0 + 0.5 x 2.68 x 10 x 10

s = 134 m

As, 1 m = 3.28 ft

So, s = 134 x 3.28 ft

s = 439.6 ft

7 0
3 years ago
PLEASE HELP
Mashutka [201]

Answer:

It cannot be constant because if it does not change and each time it increases its strength and speed.

Explanation:

4 0
3 years ago
Why is the motion of an athlete moving along the circular path with Constant speed considered to be an accelerated motion?
Strike441 [17]

The speed is changing its direction all the time. There is an acceleration which changes the direction of the speed – that is called centripetal acceleration. Only uniform linear motions are considered to have no acceleration.

This is the general formula for acceleration

a = dv/dt

When calculating dv, you should keep in mind the change in the velocity vector’s direction. You can easily see in a graph that with dt tending to 0 (so the length of the arc covered is also tending to 0), the difference between vectors Vf and V0 has a direction which is perpendicular to velocity (the shorter the arc, the closest the angle is to 90 degrees).

There is a formula (which can be deducted from the previous formula) which allows you to calculate the acceleration:

a = v^2/r

Let’s talk about the units:

v is in m/s

r is in m

so v^2/r

is in (m/s)^2/m = (m^2/s^2)/m = m/s^2

which is the same unit as dv/dt:

dv/dt = (m/s)/s= m/s^2

5 0
3 years ago
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