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3 years ago

10

An object to charge 2.00 c is a pretty from a second object with the same charge by a distance of 1.50 m what is the electric fo

rce acting between the charges

1 answer:

Orlov [11]3 years ago

5 0

F = q₁q₂C / r²

F force
q charge
C Coulomb constant
r separation between charges

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Applying the Law of Conservation of Momentum

IgorC [24]

answer

the final velocity is half of train car B's initial velocity

explanation

the conservation of momentum states that the initial and final total momentum are equal

m1v1 + m2v2 for initial momentum

(m1 + m2)v3 for final momentum because both cars stick together so their masses are combined

m1v1 + m2v2 = (m1 + m2)v3

since the mass of both are the same, m1 = m2

m1v1 + m1v2 = (m1 + m1)v3

m1(v1 + v2) = 2m1v3

divide both sides by m1

v1 + v2 = 2v3

since train car A is initially at rest, v1 = 0

v2 = 2v3

v3 = v2 * 1/2

the final velocity is half of train car B's initial velocity

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3 years ago

A car accelerates from rest at 2 m/s. what is the speed after 8 sec?

beks73 [17]

Answer:

16m/s

Explanation:

$v_{f}=v_{i}+at$

$v_{f}=0+2\cdot8$

$v_{f}=16\ \frac{m}{s}$

Therefore, the speed after 8 seconds is 16m/s

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2 years ago

What happens to a species if the death rate is higher then the birthrate?

AURORKA [14]

Answer:

The population size decreases.

Explanation:

If more of a species are dying than being born, the population size will decrease.

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2 years ago

The crisis in Erikson’s identity versus role confusion stage of psychosocial development is that the individual must __________.

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C. realize who or what one wants to be

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3 years ago

Read 2 more answers

A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k

Luba_88 [7]

Answer:

The frictional force $F_{fri} =$ 6.446 N

The acceleration of the block a = 6.04 $\frac{m}{s^{2} }$

Explanation:

Mass of the block = 3.9 kg

$\theta = 40$ °

$\mu$ = 0.22

(a). The frictional force is given by

$F_{fri} = \mu R_{N}$

$R_{N} = mg \cos \theta$

$R_{N} =$ 3.9 × 9.81 × $\cos 40$

$R_{N} =$ 29.3 N

Therefore the frictional force

$F_{fri} =$ 0.22 × 29.3

$F_{fri} =$ 6.446 N

(b). Block acceleration is given by

$F_{net} = F - F_{fri}$

F = 30 N

$F_{fri}$ = 6.446 N

$F_{net}$ = 30 - 6.446

$F_{net}$ = 23.554 N

The net force acting on the block is given by

$F_{net} = ma$

23.554 = 3.9 × a

a = 6.04 $\frac{m}{s^{2} }$

This is the acceleration of the block.

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2 years ago