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Slav-nsk [51]
3 years ago
9

To prevent accidental electric shock, most electrical equipment is A. painted. B. isolated. C. coated. D. grounded.

Physics
2 answers:
Nesterboy [21]3 years ago
4 0
D. It is Grounded below ground 
Aleksandr [31]3 years ago
4 0

Answer:

Grounded

Explanation:

Three wires are use while wiring and even in appliances. They are live, neutral and earthing wire.

Grounding or earthing is the process through which we safeguard us from electric shock.

The earthing wire is directly connected to a box buried in ground at certain level So any kind of leakage current that can cause shock will directly pass to the ground making equipment and humans safe.

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A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him
nataly862011 [7]

Answer:

a) u_c=0\ m.s^{-1}       &        m_c.v_c=m_b.v_b\times \cos\theta

b) v_c=0.0566\ m.s^{-1}

c) p_e=2.9218\ kg.m.s^{-1}

Explanation:

Given:

mass of the book, m_b=1.35\ kg

combined mass of the student and the skateboard, m_c=97\ kg

initial velocity of the book, v_b=4.61\ m.s^{-1}

angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

where:

v_c= final velcotiy of the student after throwing the book

b)

m_c.v_c=m_b.v_b\times \cos\theta

97\times v_c=1.35\times 4.61\cos28

v_c=0.0566\ m.s^{-1}

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

p_e=m_b.v_b\sin\theta

p_e=1.35\times 4.61\times \sin28^{\circ}

p_e=2.9218\ kg.m.s^{-1}

6 0
2 years ago
2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The t
serious [3.7K]

Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v (1)

Where:

m_{S}, m_{C} - Masses of the semi truck and the car, measured in kilograms.

v_{S}, v_{C} - Initial velocities of the semi truck and the car, measured in meters per second.

v - Final speed of the system after collision, measured in meters per second.

If we know that m_{S} = 2200\,kg, m_{C} = 2000\,kg, v_{C} = 45\,\frac{m}{s} and v = -15\,\frac{m}{s}, then the initial velocity of the semi truck is:

m_{S}\cdot v_{S} = (m_{S}+m_{C})\cdot v -m_{C}\cdot v_{C}

v_{S} = \frac{(m_{S}+m_{C})\cdot v - m_{C}\cdot v_{C}}{m_{S}}

v_{S} = \left(1+\frac{m_{C}}{m_{S}} \right)\cdot v - \frac{m_{C}}{m_{S}} \cdot v_{C}

v_{S} = v +\frac{m_{C}}{m_{S}}\cdot (v-v_{C})

v_{S} = -15\,\frac{m}{s}+\left(\frac{2000\,kg}{2200\,kg} \right) \cdot \left(-15\,\frac{m}{s}-45\,\frac{m}{s}  \right)

v_{S} = -69.545\,\frac{m}{s}  

The semi truck travels at an initial speed of 69.545 meters per second downwards.

3 0
2 years ago
A power plant produces 1000 MW to suply a city 40Km away.Current flows from the power plant on a single wire of resistance0.050
Westkost [7]

Answer:

The current in wire resistance 2Ω

a). 8696 A

b). fraction power 15.1% a 115kV

Explanation:

Resistance

R=0.05Ω/Km*40km

R=2Ω

P=1000 MW

a).

P=V*I\\I=\frac{P}{V}=\frac{1000x10^{6}W}{115x10^{3}k }  =8696.65A

Using law ohm

b).

V=I*R\\I=\frac{V}{R}

P=I*I*R\\P=I^{2} *R\\P=8696.65^{2}*2\\P=151.228 x10^{6}  W

e=\frac{151.228x10^{6} }{1000x10^{6} }*100= 15.12%

8 0
3 years ago
What is the net force on this object?
damaskus [11]

Upward and downward forces cancel out. Net force is 8 newtons to the right

5 0
3 years ago
Read 2 more answers
The total mass of an object can be assumed to be focused at one point, which is called its center of _______. A. force B. veloci
QveST [7]
C.) <span>The total mass of an object can be assumed to be focused at one point, which is called its center of "Mass"

Hope this helps!</span>
4 0
2 years ago
Read 2 more answers
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