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MrRa [10]
3 years ago
7

A penny dropped into a wishing well reaches the bottom in 2.50 seconds. What was the velocity at impact?

Physics
2 answers:
evablogger [386]3 years ago
7 0
Does anyone wanna premarital seggs?
12345 [234]3 years ago
4 0

Answer:

Taking it was dropped with inital velocity of 0m/s.

v = u+at

v = 0 + 9.8*1.5

v = 14.7s

(a=9.8m/s^2 << That's the acceleration due to gravity)

Explanation:

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Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.
Ksivusya [100]

|v| =\sqrt{ G \cdot M / r}, where

  • M the mass of the planet, and
  • G the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

<em>Equation 1</em>  (see below) relates net force the object experiences, \Sigma F to its orbit velocity v and its mass m required for it to stay in orbit :

\Sigma F = m \cdot v^{2} / r <em>(equation 1)</em>

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force \Sigma F acting on the soccer ball shall equal to its weight, W = m \cdot g where g the gravitational acceleration constant. Thus

\Sigma F = W = m \cdot g <em>(equation 2)</em>

Substitute equation 2 to the left hand side of <em>equation 1</em> and solve for v; note how the mass of the soccer ball, m, cancels out:

m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0) <em>(equation 3)</em>

<em>Equation 4 </em> gives the value of gravitational acceleration, g, a point of negligible mass experiences at a distance r from a planet of mass M (assuming no other stellar object were present)

g = G \cdot M/ r^{2} <em>(equation 4)</em>

where the universal gravitation <em>constant</em> G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}

Thus

\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}

3 0
3 years ago
a car accelerates from rest to a velocity of 5meters/second in 4 seconds. what is its average acceleration over this period of t
Pani-rosa [81]

Answer:

1.25 m/s²

Explanation:

Average acceleration is the change in velocity over change in time.

a = Δv / Δt

a = (5 m/s − 0 m/s) / 4 s

a = 1.25 m/s²

6 0
3 years ago
If a 6 kilogram box is moved from the floor to a storage compartment 2 meters above the floor by how many joules does it's gravi
kifflom [539]
Joules is work so you wood use the equation w=f(d)
5 0
3 years ago
(a) What is the acceleration of gravity on the surface of the Moon? The mass of the moon is 7.35 x 1022 kg and its radius is 1.7
Liono4ka [1.6K]

Answer:

Part a)

a = 1.62 m/s/s

Part b)

a = 3.70 m/s/s

Explanation:

Part A)

Acceleration due to gravity on the surface of moon is given as

a = \frac{GM}{R^2}

here we know that

M = 7.35 \times 10^{22} kg

R = 1.74 \times 10^6 m

now we have

a_g = \frac{(6.67 \times 10^{-11})(7.35 \times 10^{22})}{(1.74 \times 10^6)^2}

a_g = 1.62 m/s^2

Part B)

Acceleration due to gravity on surface of Mercury is given as

a = \frac{GM}{R^2}

here we know that

M = 3.30 \times 10^{23} kg

R = 2.44 \times 10^6 m

now we have

a_g = \frac{(6.67 \times 10^{-11})(3.30 \times 10^{23})}{(2.44 \times 10^6)^2}

a_g = 3.70 m/s^2

8 0
3 years ago
Plz answer the question below
GaryK [48]
Action verb. 100% sure

To soared 
Soar
its an action 
3 0
3 years ago
Read 2 more answers
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