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AysviL [449]
3 years ago
14

A wheel of mass 48 kg is lifted to a height of 0.8 m. How much gravitational potential energy is added to the wheel? Acceleratio

n Due to gravity is g = 9.8 m/s2.
A. 30.4 J
B. 3.1 J
C. 297.9 J
D. 11,321 J
Physics
1 answer:
ANTONII [103]3 years ago
5 0

Answer: 376.32 J

Explanation: Gravitational potential energy is the energy possessed by a body by virtue of its position.

P.E=mgh

m= mass of the body = 48 kg

g= acceleration due to gravity= ]9.8m/s^2

h= height of body= 0.8 m

P.E=48kg\times 9.8m/s^2\times 0.8m=376.3J (1kgm^2s^{-2}=1 J)

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6 months with no sunrise and the other 6 without a sunset, at some places on Earth, are the result of the orientation of Earth's axis.
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PLEASE HELP, NO ONE IS HELPING MEEEE :(
Blizzard [7]

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4 0
3 years ago
Read 2 more answers
A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th
Mila [183]

Answer:

x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0

\frac{3}{(r + a)^2} = \frac{2}{r^2}

\frac{r+a}{r} = \sqrt{\frac{3}{2}}

1 + \frac{a}{r} =\sqrt{\frac{3}{2}}

\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}

so we will have

r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}

so the x coordinate of this position is given as

x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}

6 0
4 years ago
A crate is pushed horizontally by a horizontal force 527.018 N . Sliding friction resists the motion, and the kinetic coefficien
gulaghasi [49]

Answer:

m= 10 kg a = 52 m / s²

Explanation:

For this problem we must use Newton's second law, let's apply it to each axis

X axis

      F - fr = ma

The equation for the force of friction is

    -fr = miu N

Axis y

     N- W = 0

     N = mg

Let's replace and calculate laceration

     F - miu (mg) = ma

    a = F / m - mi g

    a = 527.018 / m - 0.17 9.8

We must know the mass of the body suppose m = 10 kg

    a = 527.018 / 10 - 1,666

    a = 52 m / s²

5 0
4 years ago
Which of the following would decrease in size during the contraction of a sarcomere? The width of the I-bands The width of the A
ANEK [815]

Hi!


The correct answer would be: the width of I-bands


The sacromere is the smallest contractile unit of striated muscles. These units comprise of filaments (fibrous proteins) that, upon muscle contraction or relaxation, slide past each other. The sacromere consists of thick filaments (myosin) and thin filaments (actin).


<em>Refer to the attached picture to clearly see the structure of a sacromere.</em>


<u>When a sacromere contracts, a series of changes take place which include:</u>

<em>- Shortening of I band, and consequently the H zone</em>

<em>- The A line remains unchanged</em>

<em>- Z lines come closer to each other (and this is due to the shortening of the I bands) </em>

The only changes that take place occur in the zones/areas in the sacromere (as mentioned), not in the filaments (actin and myosin) that make the up the sacromere; hence all other options are wrong.


Hope this helps!

8 0
3 years ago
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