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3 years ago

What is the potential energy of a 3kg ball that is on the ground?

1 answer:

5 0

This is where we have to admit that gravitational potential energy is
one of those things that depends on the "frame of reference", or
'relative to what?'.

Potential energy = (mass) x (gravity) x (height).

So you have to specify height above what .

-- With respect to the ground, the ball has zero potential energy.
(If you let go of it, it will gain zero kinetic energy as it falls to
the ground.)

-- With respect to the floor in your basement, the potential energy is

(3) x (9.8) x (3 meters) = 88.2 joules.

(If you let go of it, it will gain 88.2 joules of kinetic energy as it falls
to the floor of your basement.)

-- With respect to the top of that 10-meter hill over there, the potential
energy is
(3) x (9.8) x (-10) = -294 joules

(Its potential energy is negative. After you let go of it, you have to give it
294 joules of energy that it doesn't have now, in order to lift it to the top of
the hill where it will have zero potential energy.)

You might be interested in

What are the primary colors? What are the secondary colors?

Jet001 [13]

Color Basics

Three Primary Colors (Ps): Red, Yellow, Blue.

Three Secondary Colors (S'): Orange, Green, Violet

5 0

2 years ago

Read 2 more answers

A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

$y(x, t) = Acos(kx -wt)$

The relationship between velocity and propagation constant is

$v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\$

v = 15.87m/s

Time taken, $t = \frac{\lambda}{v}$

$= \frac{1.3}{15.87}\\\\=0.0819 sec$

t = 0.0819s

The velocity of transverse wave is given by

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{W}{\frac{m}{\lambda}}}$

mass of string is calculated thus

mg = 0.0125N

$m = \frac{0.0125N}{9.8N/s}$

m = 0.00128kg

$\omega = \frac{v^2m}{\lambda}$

$\omega = \frac{(15.87^2)(0.00128)}{1.30}$

$\omega =$ 0.25N

The propagation constant k is

$k=\frac{2\pi}{\lambda}$

hence

$\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}$

$\lambda =$ 0.036 m

No of wavelengths, n is

$n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\$

n = 36

The equation of wave travelling down the string is

$y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]$

$without, unit\\\\y(x , t)= Acos[172x + 2730t]$

7 0

2 years ago

A 10.0 g bullet moving at 300m/s is fired into a 1.00 kg block at rest. The bullet emerges (the bullet does not get embedded in

chubhunter [2.5K]

Answer:

v' = 1.5 m/s

Explanation:

given,

mass of the bullet, m = 10 g

initial speed of the bullet, v = 300 m/s

final speed of the bullet after collision, v' = 300/2 = 150 m/s

Mass of the block, M = 1 Kg

initial speed of the block, u = 0 m/s

velocity of the block after collision, u' = ?

using conservation of momentum

m v + Mu = m v' + M u'

0.01 x 300 + 0 = 0.01 x 150 + 1 x v'

v' = 0.01 x 150

v' = 1.5 m/s

Speed of the block after collision is equal to v' = 1.5 m/s

5 0

2 years ago

A car with a mass of 1500 kg accelerates when the traffic light turns green. If the net force on the car is 3750 newtons, what i

timurjin [86]

Answer:

2.5 ms⁻²

Explanation:

By Newton's 2nd law,

The rate of change of momentum is directly proportional to the unbalance force applied on the object,

By that you can get the equation,

F = ma ⇒ a = F/m

where terms are in usual meaning

a = 3750/1500 = 2.5 ms⁻²

8 0

2 years ago

A cruise ship made a trip to Guam and back. The trip there took 12 hours and the trip

vekshin1

15 because i did the math, had this assignment, and also watch a video

6 0

2 years ago