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3 years ago

Define unit of electric work (JOULE) in relation to quantity of charge and potential difference.

1 answer:

6 0

1 volt = 1 joule per coulomb

It takes 1 joule of work to force a coulomb of charge enough closer to a charge
with the same sign to raise its potential 1 volt.

If you allow 1 coulomb of charge to fall to where its potential is 1 volt less,
it gives up 1 joule of energy.

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1. The Moon's mass is 7.34 x 1022 kg, and it is 3.8 x 105 km away from Earth. Earth's

sineoko [7]

Answer:

2.03 x 10²⁴N

Explanation:

Given parameters:

Mass of moon = 7.34 x 10²²kg

Mass of the earth = 5.97 x 10²⁴kg

Distance = 3.8 x 10⁵km

Unknown:

Gravitational force of attraction = ?

Solution:

To find the gravitational force of attraction between the masses, we use the expression below;

F = $\frac{Gm_{1} m_{2} }{r^{2} }$

G is the universal gravitation constant

m is the mass

1 and 2 represents moon and earth

r is the distance

F = $\frac{6.67 x 10^{-11} x 7.34 x 10^{22} x 5.97 x 10^{24} }{(3.8 x 10^{5})^{2} }$

F = $\frac{2.92 x 10^{35} }{1.44 x 10^{11} }$ = 2.03 x 10²⁴N

8 0

2 years ago

Which events may occur when ocean salinity increases? Check all that apply.

Pavlova-9 [17]

The answer is number two, number four, and number one

3 0

2 years ago

Read 2 more answers

A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

7 0

3 years ago

As the distance between two charged objects increases, the strength of the electrical force between the objects

GuDViN [60]

Answer:

I believe the answer is It increases

4 0

2 years ago

A waitperson carrying a tray with a platter on it tips the tray at an angle of 12 degrees below the horizontal. If the gravitati

Tasya [4]

Answer:1.04 N

Explanation:

Given

Gravitational Force on the Platter is $5 N$

Tray makes an angle of $\theta =12^{\circ}$

This gravitational Force has components along and Perpendicular to Platter

Perpendicular Force $W_p=W\cos \theta$

$W_p=5\times \cos 12=4.89 N$

Along the Tray

$W_{along}=W\sin \theta$

$W_{along}=5\times \sin 12=1.04 N$

Thus 1.04 N is the magnitude of force that will cause Platter to slide down

7 0

3 years ago