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2 years ago

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An object of mass 5 m is lifted to a height of 5 h above the floor. What is the potential energy the object gained in this proce

ss (in the unit of mgh)? The potential change is

1 answer:

Gekata [30.6K]2 years ago

7 0

Answer:

Potential energy, E = 25 mgh

Explanation:

Given that,

Mass of the object, m = 5 m

It is lifted to a height of, h = 5 h

It is required to find the potential energy gained by the object. It is given by the product of object's mass, acceleration due to gravity and heigh above ground. It is given by :

$E=m\times g\times h$

$E=5\ m\times g\times 5\ h$

E = 25 mgh

So, the potential energy gained by the object is (25 mgh). Hence, this is the required solution.

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An open pipe is 1.42 m long.

Varvara68 [4.7K]

Answer:

f(3) = 362.32 Hz

Explanation:

the formula of frequency of open pipe is

f(n) = (n+1)v/2L

n = 0,1,2,3,... (the order)

0 for the first

1 for the second

v = speed of sound

L = the lenght of pipe

f(3) = (2+1) * 343 / 2 * 1,42

f(3) = 3 * 343 / 2.84

f(3) = 362.324 Hz (or you can write 362.32 Hz

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2 years ago

In an experiment, one of the forces exerted on a proton is F⃗ =−αx2i^, where α=12N/m2. What is the potential-energy function for

Blizzard [7]

Answer:

The potential energy is $-4x^3$

Explanation:

Given that,

Force $F=-\alpha x^2 i$

We need to calculate the potential energy

Using formula of work done

$\Delta U=F(x) dx$

Put the value of F

$\Delta U=-\alpha x^2\ dx$

On integration

$U=-\alpha \dfrac{x^3}{3}+C$

$U=-\dfrac{\alpha x^3}{3}+C$ ...(I)

U = 0, x = 0 so C = 0

Put the value of c and α in equation (I)

$U=-\dfrac{12x^3}{3}+0$

$U=-4x^3$

Hence, The potential energy is $-4x^3$

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2 years ago

Explain the process of why the balloon is attracted to the wall, and why electrons are not transferred in this process. Is the w

strojnjashka [21]

Answer:

The process by which the balloon is attracted and possibly sticks to the wall is known as static electricity which is the attraction or repulsion between electric charges which are not free to move.

The wall is an insulator.

Explanation:

When a balloon is blown and tied off, and then the balloon is rubbed on the woolly object once in one direction, and the side that was rubbed against the wool is brought near a wall and then released, it is observed that the balloon is attracted to and sticks to the wall. The above observation is due to static electricity.

Static electricity refers to electric charges that are not free to move or that are static. One of the means of generating such charges is by friction. When the balloon is rubbed on the woollen material, electrons are given away to the balloon's surface. Since the balloon is an insulator (materials which do not allow electricity to pass through them easily), the electrons are not free to move. When the balloon is brought near to a wall, there is a rearrangement of the charges present on the wall. Negative charges on the wall move farther away while the positive charges on the wall are attracted to the electrons on the balloon's surface. Because the wall is also an insulator, the charges are not discharged immediately. Therefore, this attraction between opposite charges as well as the static nature of the charges results in the balloon sticking to the wall.

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2 years ago

Please help me, I’m confused on where to start.

zimovet [89]

Answer:

D

Explanation:

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2 years ago

Read 2 more answers

The current theory of the structure of the

Mariana [72]

Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

$K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2}$ (6)

$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):

$v=\sqrt{\frac{2 K}{m}}$ (6)

$v=\sqrt{\frac{2 (1413 J)}{77 kg}}$

Finally:

$v=36.62 m/s$ This is the speed of the jogger

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2 years ago