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Setler79 [48]
3 years ago
11

The electrical system that provides power to your school has two major parts: The transmission line which sends power to your lo

cal area and the distribution lines which transmit the power locally. For safety and efficiency, the voltage must be ______________ in the transmission lines and ____________ in the distribution lines.
Physics
1 answer:
7nadin3 [17]3 years ago
8 0

Answer:

The answer is below

Explanation:

Because of the resistance of transmission lines, high current along the line leads to large voltage drop. This voltage drop causes low efficiency because the voltage at the receiving end is far less than that at the sending end.

Therefore there is need to transmit at a higher voltage (smaller current) than that required by distribution because of efficiency.

Hence, For safety and efficiency, the voltage must be large in the transmission lines and small in the distribution lines.

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Two spheres, 1.00 kg each, whose centers are 2.00 m apart, would have what gravitational force between them? A. 3.14 X 10-17 N
Angelina_Jolie [31]

Answer: B

Explanation: the teacher just told us the answer

4 0
3 years ago
A crate is sliding on the floor. If there is a total force acting on the crate in the same direction as it is sliding, the crate
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But we do not know whether the force is pushing or pulling (the same direction (both forces are parallel) but: .........[ ]<-F-- or .......[ ]--F-->). I suppose the correct answer is B

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A toy car having mass m = 1.50 kg collides inelastically with a toy train of mass M = 3.60 kg. Before the collision, the toy tra
baherus [9]

Answer:

  v_{f} = 3,126 m / s

Explanation:

In a crash exercise the moment is conserved, for this a system formed by all the bodies before and after the crash is defined, so that the forces involved have been internalized.

the car has a mass of m = 1.50 kg a speed of v1 = 4.758 m / s and the mass of the train is M = 3.60 kg and its speed v2 = 2.45 m / s

Before the crash

    p₀ = m v₁₀ + M v₂₀

After the inelastic shock

    p_{f}= m v_{1f} + M v_{2f}

    p₀ =  p_{f}

    m v₀ + M v₂₀ = m v_{1f}  + M v_{2f}

We cleared the end of the train

     M v_{2f} = m (v₁₀ - v1f) + M v₂₀

Let's calculate

     3.60 v2f = 1.50 (2.15-4.75) + 3.60 2.45

     v_{2f}  = (-3.9 + 8.82) /3.60

      v_{2f}  = 1.36 m / s

As we can see, this speed is lower than the speed of the car, so the two bodies are joined

set speed must be

      m v₁₀ + M v₂₀ = (m + M) v_{f}

      v_{f}  = (m v₁₀ + M v₂₀) / (m + M)

      v_{f}  = (1.50 4.75 + 3.60 2.45) /(1.50 + 3.60)

      v_{f} = 3,126 m / s

4 0
4 years ago
A 0.40 kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.0 m on a frictionless horizontal su
DerKrebs [107]
<span>The centripetal force for such an arrangement can be found through the equation Fc = mv^2/r where m is the mass of the rotating object, v is that object's velocity, and r is the radius of rotation. In this case, we know that the maximum Fc that can be tolerated by the cord is 64N. Thus we set the equation up and solve for the value of v for which Fc = 64.
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3 0
3 years ago
The mass of the Sun is 2x1030 kg, and the mass of Mars is 6.4x1023 kg. The distance from the Sun to Mars is 2.3X1011 m. Calculat
user100 [1]

Answer: force from sun to mars = 1.62*10^{19} N, force from mars to sun = -1.62*10^{19} N

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F =\frac{Gm_{s} m_{m} }{r^{2} }

where G = 6.67 * 10^{-11} , m_{s} = 2*10^{30}kg      m_{m}  = 6.4 *10^{23} kg\\r = 2.3 *10^{11} m

G = gravitational constant, m_{m}= mass of mars, m_{s} = mass of sun

F = \frac{6.64 * 10 ^{-11}* 2 * 10^{30} *6.4*10^{23}} {2.3*10^{11} * 2.3* 10^{11} } =\frac{85.376 *10^{42} }{5.25 *10^{22} }  \\\\F = 1.62 * 10 ^{19} N\\

the mars also exerts the same magnitude of force above on the sun but in the opposite direction, thus the force mars exerts on the sun is -1.62*10^{19} N

8 0
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