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Maslowich
3 years ago
14

A small child gives a plastic frog a big push at the bottom of a slippery 2.0 meter long, 1.0 meter high ramp, starting it with

a speed of 5.0 m/s. What is the frog's speed as it flies off the top of the ramp?

Physics
1 answer:
valentinak56 [21]3 years ago
7 0
Refer to the diagram shown below.

Because the ramp is slippery, ignore dynamic friction.
Let m =  the mass of the frog.
g = 9.8 m/s²

The KE (kinetic energy) at the bottom of the ramp is
KE₁ = (1/2)*(m kg)*(5 m/s)² = 12.5 m J

Let v =  the velocity at the top of the ramp.
The KE at the top of the ramp is
KE₂ = (1/2)*m*v²= 0.5 mv² J
The PE (potential energy) at the top of the ramp relative to the bottom is
PE₂ = (m kg)*(9.8 m/s²)*(1 m) = 9.8m J

Conservation of energy requires that
KE₁ = KE₂ + PE₂
12.5m = 0.5mv² + 9.8m
0.5v² = 2.7
v = 2.324 m/s

Answer: 2.324 m/s

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Answer:

The objects were 1.8m apart.

Explanation:

We will start stating the Coulomb's Law. It says that:

F_e=\frac{Kq_1q_2}{r^{2}}

Where F_e is the electric force between the objects, q_1 and q_2 are the magnitude of the charge of the objects, r is the distance between them and K is the Coulomb's constant (K=8.9*10^{9} \frac{Nm^{2} }{C^{2} } in vacuum). Solving for the distance r we have:

r=\sqrt{\frac{Kq_1q_2}{F_e} }

Plugging the given values into this equation, we obtain:

r=\sqrt{\frac{(8.9*10^{9}\frac{Nm^{2} }{C^{2} })(2.56*10^{-6}C)(3.34*10^{-7}C)}{2.26*10^{-3}N}}=1.8m

In words, the two charged objects were 1.8m apart.

6 0
3 years ago
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1. A 65 kg student, starting from rest, slides down an 16.2 m high water slide. On the way down, friction does 5700 J of work on
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Answer:

11.94

Explanation:

Remark

Find the Potential Energy at the top.

Givens

m = 65 kg

h = 16.2 m

g = 9.81

PE = 65 * 9.81 * 16.2

PE = 10329.93

The tricky part is what do you do about Friction?

Formula

PE = Friction + KE

Solution

PE = 10329.93 Joules

Friction = 5700 Joules

Find the KE

10329.93 = 5700 + KE

KE = 10329.93 - 5700

KE = 4629.93

Find V from the KE formula

KE = 4629.93

m = 65

KE = 1/2 m v^2

KE = 1/2 65 v^2

4629.93 = 1/2 65 v^2

v^2 = 142.46

v = √142.46

v = 11.94

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