What is the complete back-and-forth motion of an object called?

If a 6V battery is connected to a light bulb whose resistance is 55,000Ω How much current will flow in the circuit?

notka56 [123]

Answer:

Current, I = 0.000109 Amps

Explanation:

Given the following data;

Voltage = 6V

Resistance = 55,000 Ohms

To find the current flowing through the circuit;

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

$V = IR$

Where;

V represents voltage measured in voltage.

I represents current measured in amperes.

R represents resistance measured in ohms.

Making current the subject of formula, we have;

$I = \frac {V}{R}$

Substituting into the formula, we have;

$I = \frac {6}{55000}$

Current, I = 0.000109 Amps

5 0

3 years ago

Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c

Mariulka [41]

Answer:

A. $v_{3}=12.17m/s$

B. $v_{car}=6.3m/s\\v_{truck}=-6.3m/s$

C. ΔK $=-4.13x10^3J$

Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

$v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg$

A. Since the two vehicles become entangled the final mass is:

$m_{3}=102kg+103kg=205kg$

From linear momentum we got that:

$p_{1}=p_{2}$

$m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}$

$v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}$

$v_{3}=12.17m/s$

B. The change in velocity of both vehicles are:

For the car

$v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s$

For the truck

$v_{truck}=12.17m/s-18.5m/s=-6.3m/s$

C. The change in kinetic energy is:

ΔK= $K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})$

ΔK= $\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J$

ΔK $=-4.13x10^{3}J$

6 0

3 years ago

The particles of a gas are packed tightly together.

finlep [7]

Answer:

False

Explanation:

4 0

3 years ago

Read 2 more answers

The ratio of the output work to input force is the _ of the machine ?

Colt1911 [192]

<h3><u>Answer;</u></h3>

Mechanical advantage

<h3><u>Explanation;</u></h3>

Mechanical advantage is the ratio of force output from a machine divided by the force input into the machine.
Mechanical advantage measures the machine's force-magnifying effect. It is an advantage gained by using simple machines to accomplish work with less effort.
The formula is; M.A = output force/ Input force

3 0

3 years ago

Read 2 more answers

An athlete whirls a 7.00 kg hammer 1.8 m from the axis of rotation in a horizontal circle, as

iogann1982 [59]

Answer:

A-500 N

Explanation:

The computation of the tension in the chain is shown below

As we know that

F = ma

where

F denotes force

m denotes mass = 7

And, a denotes acceleration

Now for the acceleration we have to do the following calculations

The speed (v) of the hammer is

v = Angular speed × radius

where,

Angular seed = 2 × π ÷ Time Period

So, v = 2 × π × r ÷ P

v = 2 × 3.14 × 1.8 ÷ 1

= 11.304 m/s

Now

a = v^2 ÷ r

= 70.98912 m/s^2

Now the tension is

T = F = m × a

= 7 × 70.98912

= 496.92384 N

= 500 N

5 0

3 years ago