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PSYCHO15rus [73]
3 years ago
15

Calculate the electrical energy expended in a device across which the circuit voltage drops by 50 volts in moving a charge of 6

coulombs.
Physics
2 answers:
solmaris [256]3 years ago
5 0
1 volt = 1 joule per coulomb 50 volts = 50 joules per coulomb 50 joules/coulomb times 6 coulombs = 300 joules
olasank [31]3 years ago
4 0
I=\dfrac{Q}{t} \\  \\  \\ W=VIt=V\dfrac{Q}{t}t=VQ=50\times 6=300
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It's located in the nucleus
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A 26-g steel-jacketed bullet is fired with a velocity of 630 m/s toward a steel plate and ricochets along path CD with a velocit
Alecsey [184]

Answer:

  F = - 3.53 10⁵ N

Explanation:

This problem must be solved using the relationship between momentum and the amount of movement.

          I = F t = Δp

To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio

        v = d / t

        t = d / v

Reduce SI system

          m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg

          d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m

Let's calculate

         t = 50 10⁻³ / 600

         t = 8.33 10⁻⁵ s

With this value we use the momentum and momentum relationship

        F t = m v - m v₀

As the bullet bounces the speed sign after the crash is negative

       F = m (v-vo) / t

       F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵

       F = - 3.53 10⁵ N

The negative sign indicates that the force is exerted against the bullet

5 0
3 years ago
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LuckyWell [14K]

Answer:

To the right

Weight

Explanation:

8 0
2 years ago
An inattentive driver is traveling 18.0 m/s when he notices a red light ahead. his car is capable of decelerating at a rate of 3
ryzh [129]

Speed of the car given initially

v = 18 m/s

deceleration of the car after applying brakes will be

a = 3.35 m/s^2

Reaction time of the driver = 0.200 s

Now when he see the red light distance covered by the till he start pressing the brakes

d_1 = v* t

d_1 = 18* 0.200 = 3.6 m

Now after applying brakes the distance covered by the car before it stops is given by kinematics equation

v_f^2 - v_i^2 = 2 a s

here

vi = 18 m/s

vf = 0

a = - 3.35

so now we will have

0^2 - 18^2 = 2*(-3.35)(s)

s = 48.35 m

So total distance after which car will stop is

d = d_1 + d_2

d = 48.35 + 3.6 = 51.95 m

So car will not stop before the intersection as it is at distance 20 m

3 0
3 years ago
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