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2 years ago

in a circuit, where do electric charges go once they have left the voltage source and traveled along the wires?

1 answer:

7 0

In an electric circuit, the free electrons are moving around. Since equal charges repel and opposite charges attract, the electrons move from the negatively charged to the positively charged pole of the voltage source.

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What os the konectic energy of 620.0kg coaster moving with a velocity of 9.00m/s

luda_lava [24]

Hope this helps you.

4 0

3 years ago

Can someone please help

ki77a [65]

Answer:

Acceleration of that planet is 30 $\frac{m}{s^{2} }$ .

Given:

initial speed of hammer = 0 $\frac{m}{s}$

time = 1 s

distance = 15 m

To find:

Acceleration due to gravity = ?

Formula used:

Distance covered by hammer is given by,

s = ut + $\frac{1}{2} a t^{2}$

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

Solution:

Distance covered by hammer is given by,

s = ut + $\frac{1}{2} a t^{2}$

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

u = 0

t = 1 s

s = 15 m

a = g

Thus substituting these value in above equation.

15 = 0 + $\frac{1}{2} g 1^{2}$

g = 15 × 2

g = 30 $\frac{m}{s^{2} }$

Thus, acceleration of that planet is 30 $\frac{m}{s^{2} }$ .

8 0

3 years ago

Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at

PSYCHO15rus [73]

Answer:

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

$\mu= \frac{m_1m_2}{m_1+m_2}$

now, since both the masses have mass m

therefore,

$\mu= \frac{m^2}{2m}$

= m/2

The final K.E. of the particles is

$KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2$

Distance between two particles is d and the gravitational potential energy between them is given by

$PE_{Gravitational}= \frac{Gmm}{d}$

By law of conservation of energy we have

$KE_{initial}+KE_{final}= PE_{gravitaional}$

Now plugging the values we get

$0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}$

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

$=\sqrt{\frac{Gm}{d} }$

This the required relation between G,m and d

5 0

2 years ago

Maya made this picture to represent a chemical reaction:

Natalka [10]

The statement that best explains the type of chemical reaction represented by Maya's picture is that it is neither a synthesis reaction nor a decomposition reaction because two reactants form two products. That is option B.

<h3>What is a chemical reaction?</h3>

A chemical reaction is the combination of two elements to yield a new product through the formation of bonds.

A chemical reaction is said to be a synthesis reaction when when two different atoms or molecules interact to form a different molecule or compound.

A chemical reaction is said to be a decomposition reaction when one reactant breaks down into two or more products.

Therefore, from the picture, the chemical reaction is neither a synthesis reaction nor a decomposition reaction because two reactants form two products.

Learn more about chemical reaction here:

brainly.com/question/16416932

#SPJ1

5 0

1 year ago

Holding onto a tow rope moving parallel to a frictionless ski slope, a 68.7 kg skier is pulled up the slope, which is at an angl

Furkat [3]

Answer:

a) $F = 78.606\,N$ , b) $F = 88.911\,N$