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3 years ago

Gravitational energy can be negative?

Physics

1 answer:

AlladinOne [14]3 years ago

6 0

answer:

yes

explanation:

At a separation of the surface of Earth (r=6400km) gravity wants pull the test mass closer and closer. ... So the work done by gravity is NEGATIVE. The gravitational potential energy is negative because us trying to do the opposite of what gravity wants needs positive energy.

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The band gab of semi conductor?

Lerok [7]

Answer:

Properties of semiconductors are determined by the energy gap between valence and conduction bands. To understand, what is semiconductor, we have to define these terms. In solid-state physics, the energy gap or the band gap is an energy range between valence band and conduction band where electron states are forbidden.

8 0

3 years ago

A boy is holding a ball 1 m from the ground with a force of 20 N. He holds it still for 60seconds. How much power in watts is be

Sergio [31]

Answer:

Power = 0.33 Watts

Explanation:

Given the following data;

Distance = 1m

Force = 20N

First of all, we would solve for the work done by the boy.

Workdone = force * distance

Substituting into the equation, we have;

Workdone = 20*1 = 20J

Now to find power;

Power = workdone/time

Power = 20/60

Power = 0.33 Watts.

8 0

3 years ago

1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

$\tau=(25 N)(0.5 m)=12.5 Nm$

4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity ( $\omega$ ) is:

$L=I\omega$

In this problem, we have

$I=0.007875 kgm^2$

$\omega=6.28 rad/s$

So, the angular momentum is

$L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s$

6 0

3 years ago

Determine whether the given value is a statistic or a parameter. a homeowner measured the voltage supplied to his home on 20 day

Gelneren [198K]

<span>For this example, the value presented would be considered a statistic. The value is a statistic as it represents a numerical measurement of a sample. If it were a parameter, it would need to represent a numerical measurement of a population.</span>

4 0

3 years ago

A box slides down a 30.0° ramp with an acceleration of 1.20 m/s^2. Determine the coefficient of kinetic friction between the box

Zielflug [23.3K]

m = mass of the box

N = normal force on the box

f = kinetic frictional force on the box

a = acceleration of the box

μ = coefficient of kinetic friction

perpendicular to incline , force equation is given as

N = mg Cos30 eq-1

kinetic frictional force is given as

f = μ N

using eq-1

f = μ mg Cos30

parallel to incline , force equation is given as

mg Sin30 - f = ma

mg Sin30 - μ mg Cos30 = ma

"m" cancel out

a = g Sin30 - μ g Cos30

inserting the values

1.20 = (9.8) Sin30 - (9.8) Cos30 μ

μ = 0.44

4 0

3 years ago