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2 years ago

What is the safest method to determine electrical current

Physics

1 answer:

Kisachek [45]2 years ago

4 0

Break the circuit and apace a meter actually within the circuit.

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A student performs a lab measuring the velocities of toy cars of different masses

stiv31 [10]

The answer is Car 1 and Car 2.

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4 years ago

Read 2 more answers

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.40 m/s

tekilochka [14]

Answer:

a) 378Ns

b) 477.27N

Explanation:

Impulse is the defined as the product of the applied force and time taken. This is expressed according to the formula

I = Ft = m(v-u)

m is the mass = 70kg

v is the final velocity = 5.4m/s

u is the initial velocity = 0m/s

Get the impulse

I = m(v-u)

I = 70(5.4-0)

I = 70(5.4)

I = 378Ns

b) Average total force is expressed as

F = ma (Newton's second law)

F = m(v-u)/t

F = 378/0.792

F = 477.27N

Hence the average total force experienced by a 70.0-kg passenger in the car during the time the car accelerates is 477.27N

3 0

3 years ago

label the three planes and axis of movement on the figures. define each plane and axis of movement and axis of movement and give

TiliK225 [7]

Answer: it = kobe

Explanation

5 0

3 years ago

A small object carrying a charge of -2.50 nc is acted upon by a downward force of 18.0 nn when placed at a certain point in an e

Vesna [10]

Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?

B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>

<span>Solution:

A) A charge q </span>under an electric field of intensity E will experience a force F equal to:

$F=qE$

In our problem we have $q=-2.5 nC=-2.5\cdot 10^{-9} C$ and $F=18 nN = 18 \cdot 10^{-9} N$ , so we can find the magnitude of the electric field:

$E= \frac{F}{q}= \frac{18\cdot 10^{-9}N}{2.5\cdot 10^{-9}C}=7.2 V/m$

The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.

B) The proton charge is equal to

$e=1.6\cdot 10^{-19} C$

Therefore, the magnitude of the force acting on the proton will be

$F=eE=1.6\cdot 10^{-19} C \cdot 7.2 V/m=1.15 \cdot 10^{-18} N$

And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.

7 0

3 years ago

What is the final step in the fourth stage of technological design

Ierofanga [76]

Answer:

after a product has been improved and approved? reporting the results finding ways to lower costs selling a prototype determining criteria.

Explanation:

5 0

3 years ago