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Zanzabum
3 years ago
6

Which part of a circuit creates an electric force field that makes it possible for the circuit to work? A. bulb B. buzzer C. mot

or D. voltage source
Physics
2 answers:
Grace [21]3 years ago
8 0

D. voltage source. ttyl

castortr0y [4]3 years ago
6 0

Answer:

D. voltage source

Explanation:

A voltage source can maintain a potential difference between two points in a circuit, which allows current to be supplied if it is connected to a resistor. The potential difference creates an electric field that exerts force on the charges, causing current.

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A camper walks away from camp. He walks 23.72m north, then 39.25m south and
Arte-miy333 [17]

Answer:

89.53    North

Explanation:

Total N-S displacement

   23.72  -  39.25 + 105.06   =  + 89.53

5 0
1 year ago
What are 3 ways acceleration can change?
Karolina [17]

Velocity, direction, or both at the same time.

3 0
3 years ago
At baseball practice, Mason and Alfredo both picked up the same bat and neither would let go until one of them had it for himsel
Illusion [34]

Answer:

Option B. O because the net force was 5 N in Alfredo's direction

Explanation:

To know the the correct answer to the question given above, we shall determine the net force acting on the bat. This can be obtained as follow:

Force of pull by Mason (Fₘ) = 15 N

Force of pull by Alfredo (Fₐ) = 20 N

Net force (Fₙ) =?

Fₙ = 20 – 15

Fₙ = 5 N in Alfredo's direction

From the calculation made above, we can see that the net force is 5N in Alfredo's direction. This is the reason why Alfredo end up having the bat.

7 0
3 years ago
The length of the side of a cube is 10.0 ±0.3 cm. What is the absolute uncertainty in the perimeter of one side of the cube?
Nimfa-mama [501]

Answer:Stuff

Explanation: Nun

4 0
3 years ago
You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th
PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is 6.378\times 10^{10}N/m^2

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight  = m\times g

m = mass of the ball = 10 kg

g = acceleration due to gravity = 9.81m/s^2

l = length of wire  = 2.6 m

A = area of cross section  = \pi r^2

r = radius of the wire = \frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m      (Conversion factor:  1 m = 1000 mm)

\Delta l = change in length  = 1.99 mm = 1.99\times 10^{-3}m

Putting values in above equation, we get:

Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2

Hence, the Young's modulus for the wire is 6.378\times 10^{10}N/m^2

3 0
3 years ago
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