Ask question

Ask question

All categories

3 years ago

11

A 2.0 kg mass is lifted 4.0 m above the ground. find the change in gravitational energy

1 answer:

Makovka662 [10]3 years ago

6 0

Answer:

78.4 J

Explanation:

Given:

Mass of the object (m) = 2.0 kg

Height or displacement of the mass (h) = 4.0 m

Now, change in gravitational potential energy (ΔU) = ?

Change in gravitational potential energy of an object is the energy stored by the object when it is raised to some height above the reference ground level.

Change in gravitational potential energy of an object of mass 'm' raised to a height 'h' above the ground is given as:

$\Delta U=mgh$

Where, 'g' is the acceleration due to gravity and has a value of 9.8 m/s².

Now, plug in all the given values and solve for ΔU. This gives,

$\Delta U=(2.0\ kg)(9.8\ m/s^2)(4.0\ m)\\\\\Delta U=78.4\ J$

Therefore, the change in gravitational potential energy is 78.4 J.

You might be interested in

Assume light is traveling through an optical medium of n1 and is incident on a boundary with a different optical medium with n2.

Elza [17]

That n2 = 2*n1. That is, the index of refraction is twice as big in medium 2 since v=c/n

5 0

3 years ago

1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (H

JulijaS [17]

Answer:

The angle between the electric field lines and the equipotential surface is 90 degree.

Explanation:

The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.

The electric field lines are always perpendicular to the equipotential surface.

As

$dV = \overrightarrow{E} . d\overrightarrow{r}\\\\$

For equipotential surface, dV = 0 so

$0 = \overrightarrow{E} . d\overrightarrow{r}\\\\$

The dot product of two non zero vectors is zero, if they are perpendicular to each other.

5 0

2 years ago

Which is the 2nd closest star to earth??

IgorLugansk [536]

Answer:

Proxima Centauri

Explanation:

U 2 can help me by marking as brainliest........

8 0

2 years ago

Read 2 more answers

Determine the amount of work done by the engine of a car with a mass of 2500 kg when it accelerates from 45 mph to 65 mph. (Use

Y_Kistochka [10]

Answer:

amount of work done, W = 549.36 kJ

Given:

mass of a car engine, m = 2500 kg

initial velocity, u = 45 mph

final velocity, v = 65 mph

1 mile = 1609

Solution:

We know that 1 hour = 3600 s

Now, velocities in m/s are given as:

u = 45 mph = $\frac{45\times 1609}{3600}$ = 20.11 m/s

v = 65 mph = $\frac{65\times 1609}{3600}$ = 29.05 m/s

Now, the amount of work done, W is given by the change in kinetic energy of the car and is given by:

W = $\frac{1}{2}m\Delta v^{2}$

W = $\frac{1}{2}m\times (v^{2} - u^{2})$

W = $\frac{1}{2}2500\times (29.05^{2} - 20.11^{2})$

W = 549.36 kJ

3 0

3 years ago

An electron is traveling at 0. 85c. what is its mass? (the rest mass is 9. 11 × 10-31 kg. )

egoroff_w [7]

The mass of the electron is $1.72\times 10^{-30}$ Kg.

<h3>How can we calculate the mass of the electron?</h3>

To calculate the mass of the electron we are using the formula,

$m=\frac{m_{0} }{\sqrt{1-\frac{v^{2} }{c^{2} } } }$

Here we are given,

$m_0$ =The rest mass of an electron = $9. 11 \times 10^{-31}$ kg.

v= The velocity of electron = 0. 85c.

c= The velocity of light = c. (c= $3\times 10^8$ m/s)

We have to found the mass of an electron = m Kg.

Now we substitute the known values, we found that,

$m=\frac{9. 11 \times 10^{-31} }{\sqrt{1-\frac{(0.85c)^{2} }{c^{2} } } }$

Or, $m=\frac{9. 11 \times 10^{-31} }{\sqrt{1-(\frac{0.85c}{c})^2 } }$

Or, $m=\frac{9. 11 \times 10^{-31} }{\sqrt{1-({0.85})^2 } }$

Or, $m=1.72\times 10^{-30}$ Kg.

From the above calculation we can conclude that, The mass of the electron is $1.72\times 10^{-30}$ Kg.

Learn more about electron:

brainly.com/question/860094

#SPJ4

4 0

1 year ago