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Lunna [17]
3 years ago
10

A test charge of +2 μC is placed halfway between a charge of +6 μC and another of +4 μC separated by 10 cm. (a) What is the magn

itude of the force on the test charge? (b) What is the direction of this force (away from or toward the +6 μC charge)?
Physics
1 answer:
sashaice [31]3 years ago
8 0

Answer:

ᵃ. Fₙ= 1.424 x 10¹³ μN

ᵇ. away from the +6 charge

Explanation:

PART A

Using Coulomb's law FOR CHARGES +2 AND +4

F= (Q . q)÷4(3.142)ε(5/100)²

F= 2.848 x 10¹³ μN

for charges of 2 and 6

F₂=(Q . q)÷4(3.142)ε(5/100)

F₂= 4.272 x 10¹³ μN

<em>NET FORCE IS EQUAL TO THE SUM OF ALL THE FORCES ACTING AT A POINT.</em>

<em>FORCE IS A VECTOR QUANTITY, THEREFORE, WE ASSIGN A NEGATIVE SIGN TO ONE OF THE FORCES, as both of them are acting in opposite directions while repelling the +2. </em>

Fₙ= F-F₂

Fₙ= 4.272 x 10¹³ - 2.848 x 10¹³

Fₙ= 1.424 x 10¹³ μN

PART B

The force caused by the +6 charge is twice as much as the one caused by the +4 charge on the same +2 charge, therefore the repulsion caused by the greater charge is greater and has an effect of pushing the +2 charge away from it. However there is also a force of repulsion by the +4 charge which is NOT AS STRONG as compared to the +6 charge, therefore it is less likely to push away the +2 charge

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How much force is needed to accelerate a 1,800kg car at rate of 1.5 m/s2?
topjm [15]

<u>Given data:</u>

acceleration (a) = 1.5 m/s² ,

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        Determine F = ?

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                          F = m.a   N

                              = 1800× 1.5

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3 years ago
Water waves approach an underwater "shelf" where the velocity changes from 2.8 m/s to 2.1 m/s. If the incident wave crests make
otez555 [7]

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24°

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7 0
3 years ago
A ski lift has a one-way length of 1 km and a vertical rise of200m. The chairs are spaced 20 m apart, and each chair can seatthr
Strike441 [17]

Answer:

a) 68.125 KW

b) 43.04 KW

Explanation:

Distance =d= 1 km

Height = h = 200 m

Spacing between chairs =D= 20 m

No. of people per chair = 3

Speed = V= 10 km/h= 10000m\3600 s=2.8 m/s

mass of chair = 250 kg

Work to operate sky lift

W= mgh

Number of chairs any moment operational = N= 1 km/20=1000m/20=50

So, total mass of chairs = 50 × 250 =12500kg

so, w= mgh=12500×9.8×200=2452500 j

Power is rate of work - we need time for operation time of lift

Time= t = distance/speedf= 1km/(10km/h)=1km/(10 km/3600s)=360s

So, Power P= Work/time=w/t=12500/360=68125 j/s=68.125 KW

Now calculating power for operating speed in 5 sec

We calculate accelleration=a for 5 sec

a= speed / time= V/52.8/5=0.28 m/sec2

for vertical acceleration we calculate θ angle first;

tanθ= height /distance= 200/1000= 0.2

==> θ=11.34°

Vertical acceleration = a₁=a sinθ= 0.28× sin 11.34=0.10835 m/sec2

to calculate height gained during startup use;

S=vit+1/2at2

here s=H

vi=0m/s

t=5 sec

==> H = 0+1/2a₁×t=0.5×0.10835 ×5²=0.5×0.10835×5×5=1.362 m

Total Work =mgH+0.5×m×V²=12500(9.8×1.362+0.5×2.8×2.8)=215240.56 j

Again power = work / time=215240.56/5=43048.112 W=43.04 KW

6 0
2 years ago
radar wave is transmitted and later reflected off an aircraft and recieved 1.4×10^3 sec after being sent out. how far is the air
lutik1710 [3]

I think there's a typo because the answer I'm getting is very large.

This is what I'm getting

--------------------------------------

c = speed of light

c = 3.0 x 10^8 m/sec approximately

This is roughly 300 million meters per second

The time it takes the signal to reach the aircraft and come back is 1.4 x 10^3 seconds. Half of this time period is going one direction (say from the radar station to the aircraft), so (1.4 x 10^3)/2 = 7.0 x 10^2 seconds is spent going in this one direction.

distance = rate*time

d = r*t

d = (3.0 x 10^8) * (7.0 x 10^2)

d = (3.0*7.0) x (10^8*10^2)

d = 21.0 x 10^(8+2)

d = 21.0 x 10^10

d = (2.1 x 10^1) * 10^10

d = 2.1 x (10^1*10^10)

d = 2.1 x 10^11 meters

d = 210,000,000,000 meters (this is 210 billion meters; equivalent to roughly 130,487,950 miles)

3 0
3 years ago
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