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baherus [9]
3 years ago
15

How could you tell whether or not you are in an electrical field

Physics
1 answer:
dalvyx [7]3 years ago
3 0
This may help you out a little bit: http://www.physicsclassroom.com/class/estatics/Lesson-4/Electric-Field-Intensity
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Two carts undergo an inelastic collision where they stick together. Cart A has an initial velocity v0, and the second cart B is
dimulka [17.4K]

Answer:

Explanation:

Initial kinetic energy of the system = 1/2 mA v0²

If Vf be the final velocity of both the carts

applying conservation of momentum

final velocity

Vf = mAvo / ( mA +mB)

kinetic energy ( final ) =  1/2 (mA +mB)mA²vo² /  ( mA +mB)²

= mA²vo²  / 2( mA +mB)

Given 1/2 mA v0²  / mA²vo²  / 2( mA +mB) = 6

mA v0² x ( mA +mB) / mA²vo² = 6

( mA +mB) / mA = 6

mA + mB = 6 mA

5 mA = mB

mB / mA = 5 .

3 0
3 years ago
During sexual reproduction, each parent contributes
Snowcat [4.5K]

Answer:

I don't understand what exactly you're asking, but this statement is true.

Explanation:

4 0
3 years ago
10. A diver is collecting shells under water. They see a very large and beautiful shell so they reach out to pick it up. After s
dexar [7]

Answer:

It is the depth of the water that allowed it to be like day

Explanation:

what we see abovewater is not the actual thing or ssize we would see when insife the water

4 0
1 year ago
a 10-N force is exerted on a box, moving it 20 m in the same direction. What is the magnitude of work done on the box?
-Dominant- [34]
F = 10 N

d = 20 m

θ = 0°

W = F (dot product) D = F * D * cos(angle between them)

W = FDcosθ

W = 10 * 20 * cos0 = 200 J
8 0
2 years ago
A car accelerates uniformly from rest to 24.8 m/s in 7.88 s along a level stretch of road. Ignoring friction, determine the aver
nevsk [136]

Answer:

When he weight of the car is 8.55 x 10^{3} N then power = 314.012 KW

When he weight of the car is 1.10 x 10^{4}  N then power =  43.76 KW

Explanation:

Given that

Initial velocity V_{1} = 0

Final velocity V_{2} = 24.8 \frac{m}{s}

Time = 7.88 sec

We know that power required to accelerate the car is given by

P = \frac{change \ in \ kinetic \ energy}{time}

Change in kinetic energy Δ K.E = \frac{1}{2} m (V_{2}^{2} - V_{1}^{2}   )

Since Initial velocity V_{1} = 0

⇒ Δ K.E = \frac{1}{2} m V_{2}  ^{2}

⇒ Power P = \frac{1}{2} \frac{m}{t}  V_{2} ^{2}

⇒ Power P = \frac{1}{2} \frac{W}{g\ t}  V_{2} ^{2}   -------- (1)

(a). The weight of the car is 8.55 x 10^{3} N = 8550 N

Put all the values in above formula

So power P = \frac{1}{2} \frac{8550}{(9.81)\ (7.88)} (24.8) ^{2}

P = 314.012 KW

(b). The weight of the car is 1.10 x 10^{4} N = 11000 N

Put all the values in equation (1) we get

P = \frac{1}{2} \frac{11000}{(9.81)\ (7.88)} (24.8) ^{2}

P = 43.76 KW

5 0
3 years ago
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