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Nataly [62]
3 years ago
9

How energy is obtained due to flow of charges?

Physics
1 answer:
dmitriy555 [2]3 years ago
7 0
Electrons in atoms can act as our charge carrier, because every electron carries a negative charge. If we can free an electron from an atom and force it to move, we can create electricity. ... Electrons orbit at varying distances from the nucleus of the atom
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This photo shows sugar dissolved into a solution with a _______ at the bottom of the jar.
prohojiy [21]

Answer:

D

Explanation:

It dose not dissolve

6 0
3 years ago
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A rock falls off a cliff that is 82 m high. How fast is the rock moving when it hits the ground?
drek231 [11]

Answer:

40.1m/s

Explanation:

This is a kinematics problem

x = 82m

a = -9.8m/s^2

v(initial) = 0

Find: v(final)

Using kinematics equation:

vf^2 = vi^2 + 2a(delta)x

vf^2 = 0 + 2*(-9.8)*(82)

vf = sqrt(1607.2) = 40.09m/s

The is traveling with a velocity of 40.1m/s when it hits the ground.

7 0
3 years ago
Hewo pweese help
OlgaM077 [116]

Answer: 18 m/s/s

Explanation:

3 times 6 equals 18 and 3 seconds every second makes this equation

8 0
3 years ago
if the distance of separation begin two objects is doubled, is the gravitational force between the objects increased or decrease
SashulF [63]

Answer:

force is decreased by a factor of 4.

Explanation:

According to the Newton's law of gravitation, the force of gravitation between the two object is inversely proportional to the square of distance between them. Now the distance is doubled, so the force between the two objects becomes one forth.

Force is decreased by a factor or 4.

7 0
3 years ago
A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta
Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, \theta = 54^{\circ}

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, R = \frac{v^{2}sin2\theta}{g}

R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m

Time of flight, t' = \frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m

Distance traveled vertically:

s_{y} = vcos\theta - \frac{1}{2}gt^{2}

s_{y} = 12.4sin54^{\circ}\times 1.42 -  \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m

Now,

s_{x} = \frac{mx + m'x'}{M}

10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}

x' = 3.535 m

Similarly,

s_{y} = \frac{my + m'y'}{M}

4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m

The location of the other fragment is at (3.535, 1.162)

5 0
4 years ago
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