Ask question

Ask question

All categories

3 years ago

6

a DJ uses a 110-volt outlet to plug in a strobe light. if the current flowing through the light is 0.050 amps, how much resistan

ce is within the circuit?

1 answer:

tigry1 [53]3 years ago

5 0

Ohms Law: V = IR

V is the voltage in volts

I is the current in amps

R is the resistance in Ohms

Rearrange: R = V/I

R = (110)/(0.050)

R = 2200

There are 2200 Ohms of resistance in the circuit.

You might be interested in

Which of these is equivalent to the law of conservation of energy?

Korolek [52]

The Law of the Conservation of Energy is stating that the total mechanical energy is always conserved or in simpler terms, not used or saved.

7 0

2 years ago

Read 2 more answers

Can all rocks be weathered and eroded?how do you know?

skelet666 [1.2K]

Answer:yes

Explanation: Rock can be weathered by ice, acids, animals, temp and etc because the breaking down or dissolving of rock and minerals on the earths surface.

rocks can be eroded by wind, water, ice and, gravity be cause of an action, and friction.

(I think going off of what i learned.)

8 0

2 years ago

Read 2 more answers

a car was traveling at 30.0 m/s when a small child darted onto the road 60.0 m away. Reacting instantly the driver slams the bra

Galina-37 [17]

2 seconds,,,,,,,,,,,,,,,,,,,,,,,

5 0

2 years ago

Answer and explanation please!!

expeople1 [14]

Answer:

Option 3

Explanation:

O Option C is NEGATIVELY CHARGED, meaning it has GAINED ELECTRONS resulting in a GREATER number of ELECTRONS than PROTONS.

8 0

2 years ago

Read 2 more answers

If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

$U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m$

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

$W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J$

So, the required work done is 16 J.

7 0

2 years ago