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lilavasa [31]
3 years ago
12

A ball drops some distance and gains 30 J of kinetic energy. How much gravitational potential energy did the ball start with? Do

not ignore the effects of air resistance.
Physics
1 answer:
Dovator [93]3 years ago
4 0

Answer:

The gravitational potential energy of the ball is more than 30 J.

Explanation:

Given that,

Gains kinetic energy = 30 J

We need to calculate the gravitational potential energy

According to law of conservation of energy

If there is no air resistance then the potential energy of the ball fully converted into kinetic energy.

If the air resistance is considered then the potential energy used to do work against air resistance.

\Delta U\geq \Delta K.E

According to question,

A ball drops some distance and gains 30 J of kinetic energy and considers air resistance then 30 J of potential energy can be loosed by the ball.

Hence, The gravitational potential energy of the ball is more than 30 J.

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It is interesting to speculate on the properties of a universe with different values for the fundamental constants.
qwelly [4]

Answer:

Part a)

\lambda = 0.345 m

Part b)

\Delta x = 0.274 m

Part c)

r = 2.8 \times 10^{11} m

Explanation:

Part a)

De broglie wavelength is given as

\lambda = \frac{h}{mv}

\lambda = \frac{1}{(0.145)(20)}

\lambda = 0.345 m

Part b)

By principle of uncertainty we know that

\Delta x \times \Delta P = \frac{h}{4\pi}

\Delta x \times (0.145)(21 - 19) = \frac{1}{4\pi}

\Delta x = 0.274 m

Part c)

As we know that

\frac{kq_1q_2}{r^2} = \frac{mv^2}{r}

also we know

mvr = \frac{nh}{2\pi}

v = \frac{h}{2\pi mr}

now we have

\frac{ke^2}{r} = \frac{mh^2}{4\pi^2m^2 r^2}

r = \frac{h^2}{4\pi^2mke^2}

r = 2.8 \times 10^{11} m

5 0
3 years ago
What is the energy of a baby who weighs 20 N sitting on a 1.5 m high chair
guajiro [1.7K]
The energy of the baby is gravitational potential energy, and it is equal to the weight of the baby times its height from the ground:
U=(mg) h
where
mg=20 N is the weight (the mass times the gravitational acceleration)
h=1.5 m is the height from the ground
If we plug the numbers into the equation, we find
U=(mg)h=(20 N)(1.5 m)=30 J
3 0
4 years ago
Why machines are never 100 percent efficient
dem82 [27]
Machines are never 100% efficient because they are made by humans and we make errors and nothing we make is perfect, that's why the machines we build are not 100% accurate or efficient.
8 0
3 years ago
Read 2 more answers
1. When asteroids collided some of the broken materials fall into Earth's orbit. What do
Kazeer [188]
I believe it is meteoroids
6 0
3 years ago
Read 2 more answers
A toy rocket launcher can project a toy rocket at a speed as high as 35.0 m/s.
Anestetic [448]

Answer:

(a) 62.5 m

(b) 7.14 s

Explanation:

initial speed, u = 35 m/s

g = 9.8 m/s^2

(a) Let the rocket raises upto height h and at maximum height the speed is zero.

Use third equation of motion

v^{2}=u^{2}+2as

0^{2}=35^{2}- 2 \times 9.8 \times h

h = 62.5 m

Thus, the rocket goes upto a height of 62.5 m.

(b) Let the rocket takes time t to reach to maximum height.

By use of first equation of motion

v = u + at

0 = 35 - 9.8 t

t = 3.57 s

The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.

8 0
3 years ago
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