Collisions between galaxies are thought to:

Mark creates a graphic organizer to review his notes about electrical force. Which labels belong in the regions marked X and Y?

sveta [45]

Answer:

The correct answer is A

Explanation:

The question requires as well the attached image, so please see that below.

Coulomb's Law.

The electrical force can be understood by remembering Coulomb's Law, that describes the electrostatic force between two charged particles. If the particles have charges $q_1$ and $q_2$ , are separated by a distance r and are at rest relative to each other, then its electrostatic force magnitude on particle 1 due particle 2 is given by:

$|F|=k \cfrac{q_1 q_2}{r^2}$

Thus if we decrease the distance by half we have

$r_1 =\cfrac r2$

So we get

$|F|=k \cfrac{q_1 q_2}{r_1^2}$

Replacing we get

$|F|=k \cfrac{q_1 q_2}{(r/2)^2}\\|F|=k \cfrac{q_1 q_2}{r^2/4}$

We can then multiply both numerator and denominator by 4 to get

$|F|=k \cfrac{4q_1 q_2}{r^2}$

So we have

$|F|=4 \left(k \cfrac{q_1 q_2}{r^2}\right)$

Thus if we decrease the distance by half we get four times the force.

Then we can replace the second condition

$q_{2new} =2q_2$

So we get

$|F|=k \cfrac{q_1 q_{2new}}{r_1^2}$

which give us

$|F|=k \cfrac{q_1 2q_2}{r_1^2}\\|F|=2\left(k \cfrac{q_1 q_2}{r_1^2}\right)$

Thus doubling one of the charges doubles the force.

So the answer is A.

8 0

3 years ago

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A heat pump receives heat from a lake that has an average winter time temperature of 6o C and supplies heat into a house having

Natalija [7]

Answer:

Explanation:

60000 kJ / h = 60000 x 1000 / (60 x 60 )

= 16667 J /s

a)

$\frac{Q_H}{W} =\frac{T_H}{T_H-T_C}$

where $Q_H$ and $W$ are heat supplied and heat given from outside source .

Here $Q_H$ = 16667 J/s

$\frac{T_H}{T_H-T_C}=\frac{298}{298-279}$

= 15.684

$Q_H = 16667 J/s$

16667 / W = 15.684

W= minimum power supplied = 1062.7 W.

b )

If $T_C=283K$

$\frac{T_H}{T_H-T_C}=\frac{298}{298-283}$

=19.87

$\frac{Q_H}{W} = 19.87$

$W=\frac{Q_H}{19.87}$

$W=\frac{16667}{19.87}$

= 838.8 W .

4 0

3 years ago

A child is holding a wagon from rolling straight back down in a driveway that inclined at 20 degree horizontal. if the wagon wei

Digiron [165]

Answer:

F = 51.3°

Explanation:

The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

$F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\$

where,

W = Weight of Wagon = 150 N

θ = Angle of Inclinition = 20°

Therefore,

$F = (150\ N)Sin\ 20^o$

5 0

3 years ago

A locomotive is pulling 15 freight cars, each of which is loaded with the same amount of weight. The mass of each freight car (w

VMariaS [17]

Answer:

298,220 N

Explanation:

Let the force on car three is T_23-T_34

Since net force= ma

from newton's second law we have

T_23-T_34 = ma

therefore,

T_23-T_34 = 37000×0.62

T_23= 22940+T_34

now, we need to calculate

T_34

Notice that T_34 is accelerating all 12 cars behind 3rd car by at a rate of 0.62 m/s^2

F= ma

So, F= 12×37000×0.62= 22940×12= 275280 N

T_23 =22940+T_34= 22940+ 275280= 298,220 N

therefore, the tension in the coupling between the second and third cars

= 298,220 N

3 0

3 years ago

2. Before 0.5-kg ball is dropped from a third story window 20 m above the sidewalk. What is the potential

Flauer [41]

Answer:

C

Explanation:

C

20*9.8*0.5

98

4 0

3 years ago

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