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Volgvan
3 years ago
6

The negatively charged student touches a metal tap and receives am electric shock. Explain why.

Physics
2 answers:
svetlana [45]3 years ago
8 0

Answer:

The potential difference between the student and the tap causes electrons to transfer to the tap which is earthed.

Hope this helps my sis and others

-Amelia

rosijanka [135]3 years ago
3 0

Answer:

An electric shock occurs when a person comes into contact with an electrical energy source. Electrical energy flows through a portion of the body causing a shock. If you scuff your feet on your living room rug, you pick up extra electrons and have a negative charge. Electrons move more easily through certain materials like metal, which scientists call conductors. The shock he received was a result of the quick movement of these electrons.

- Benny

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To find the answer follow the following steps.
1. 6524.96 x .25 = X

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5 0
2 years ago
What must the diver's minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, wh
Dafna1 [17]

Answer:

1.08 m/s

Explanation:

This can be solved with two steps, first we need to find the time taken to fall 9.5 m, then we can divide the horizontal distance covered with time taken to calculate the velocity.

Time taken to fall 9.5 m

vertical acceleration = a = 9.8 m/s^2.

vertical velocity = 0, (since there is only horizontal component for velocity, )

distance traveled  s = 9.5 m.

Substituting these values in the equation

s= u \timest+0.5at^{2}

t= \sqrt{\frac{2s}{g} }

t=\sqrt{\frac{2\times9.5}{9.8} }

⇒ t= 1.392 sec

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4 0
3 years ago
A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m
Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k

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Initial position d=(2.75)i-(2.00)j+(5.00)k

Final position d=-(5.00)i+(4.50)j+(7.00)k

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We need to calculate the work done on the object by the force in the 5.60 s interval.

Using formula of work done

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W=F\cdot(d_{f}-d_{i})

Put the value into the formula

W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)

W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)

W=2.75\times-7.75+7.50\times6.50+6.75\times2.00

W=40.93\ J

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.

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