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Hatshy [7]
2 years ago
14

"How did your current and voltage measurements differ between the series and parallel circuits you created

Physics
1 answer:
irakobra [83]2 years ago
4 0

Answer:

Series circuit:

The voltage that is measured across the circuit is different.

The current measured in a series circuit remains the same at all points in the circuit.

Parallel circuit:

The current measured across each resistor varies

The voltage measured across a parallel circuit will remain the same

Explanation:

Series and parallel circuits behave differently when it comes to the circulation of current and the interaction with a potential difference.

In a series circuit, the resistances are connected end to end. As a result, the voltage that is measured across the circuit is different once resistance is encountered. However, the current measured in a series circuit remains the same at all points in the circuit.

A parallel circuit behaves in an exactly opposite manner to the series circuit. In a parallel circuit, the resistances are connected side by side. As a result of this, the current measured across each resistor varies as there are circuit branches through which electric current can flow into. On the other hand, the voltage measured across a parallel circuit will remain the same

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Atoms of two different elements must have different
Tresset [83]

Answer:

Number of neautrons

Explanation:

Atoms from two different elements may have the same number of neutrons, but never the same number of protons. The number of protons is unique to the element and represents the atomic number. (sorcastic.org)

4 0
3 years ago
The alarm at a fire station rings and a 87.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance o
blsea [12.9K]

Answer:

F_f=840N

Explanation:

From the question we are told that

Weight of fireman W_f= 87.5kg

Pole distance D=4.10m

Final speed is V_f 1.75m/s

Generally the equation for velocity is mathematically represented as

v^2 = v_0^2 + 2 a d

Therefore Acceleration a

a'= v^2 / 2 d

a'= 0.21m/s^2

Generally the equation for Frictional force F_f is mathematically given as

F_f=m*a

F_f=m*(g-0.21)

F_f=87.5*(9.81-0.21)

Therefore

F_f=840N

6 0
3 years ago
A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.
LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

W = \frac{1}{2}(50)(5.61^2) = 786.8025 J

Now, we can define work:

\large\boxed{W = Fdcos\theta}}

Now, plug in the values:

786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta

Solve for theta:

cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}

4 0
2 years ago
Nicole throws a ball at 25 m/s at an angle of 60 degrees abound the horizontal. What is the range of the ball?
just olya [345]

Answer:

the range or the ball is 48.81 m

Explanation:

given;

Nicole throws a ball at 25 m/s at an angle of 60 degrees abound the horizontal.

find:

What is the range of the ball?

solution:

let Ф = 25°

Vo = 25 m/s

<u>consider x-motion using time of fight: x = Vox * t</u>

where x = R = range

t =<u> 2 Voy </u>

      g

R =<u> Vo² sin (2Ф)</u>

           g

plugin values into the formula:

R = <u>(25)² sin (2*25) </u>

               9.81

R = 48.81 m

therefore, the range or the ball is 48.81 m

4 0
3 years ago
Read 2 more answers
If 3600 j of work is done in 3.0 s what is the power<br>0.00083W<br>1200W<br>3600W<br>11000W
Viktor [21]

Answer:

1200 W

Explanation:

Power is given by the ratio between work done and time taken:

P=\frac{W}{t}

where W is the work done and t the time taken.

In this problem, W = 3600 J and t = 3.0 s. Therefore, the power in this exercise is

P=\frac{3600 J}{3.0 s}=1200 W

5 0
3 years ago
Read 2 more answers
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