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kifflom [539]
3 years ago
9

A light bulb is connected to a 2V supply and experiences a current of 6.4A. What is the power rating of the bulb?

Physics
1 answer:
Olenka [21]3 years ago
8 0

Answer:

12.8 Watts

Explanation:

P = VI

P = (2 V) (6.4 A)

P = 12.8 W

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You pick up a 10-newton book off the floor and put it on a shelf 2 meters high. How much work did you do?​
Murljashka [212]

Answer:

20 J

Explanation:

Given:

Weight of the book is, W=10\ N

Height or displacement of the book is, d=2\ m

The work done on the book to raise it to a height of 2 m on a shelf is against gravity. The gravitational force acting on the book is equal to its weight. Now, in order to raise it, an equal amount of force must be applied in the opposite direction.

So, the force applied by me should be equal to weight of the body and in the upward direction. The displacement is also in the upward direction.

Now, work done by the applied force is equal to the product of force applied and displacement of book in the direction of the applied force.

Therefore, work done is given as:

Work=W\times d\\Work=10\times 2=20\ J

Therefore, the work done to raise a book to a height 2 m from the floor is 20 J.

3 0
3 years ago
Read 2 more answers
Calculate the work done by an applied force of 76.0 N on a crate for the following. (Include the sign of the value in your answe
telo118 [61]

Answer:

a) 400.4Joules

b) 262.69Joules

Explanation:

Work is said to be done if the force applied to an object cause the object to move through a distance

Workdone = Force × Distance

Given

Force = 76N

Distance= 5.2m

Work done = 77 × 5.2

Work done = 400.4Joules

b) If the force is exerted at an angle of 41°

Work done = Fdsin theta

Work done = 77(5.2)sin41

Work done = 400.4sin41

Work done = 262.69Joules

6 0
3 years ago
A 1369.4 kg car is traveling at 28.9 m/s when the driver takes his foot off the gas pedal. It takes 5.1 s for the car to slow do
Darya [45]

Answer:

F = 2389.603 N

Explanation:

Given:

Mass m = 1,369.4 kg

Initial velocity u = 28.9 m/s

Final velocity v = 20 m/s

Time t = 5.1 s

Find:

Net force

Computation:

a = (v - u)/t

a = (20 - 28.9)/5.1

a = -1.745 m/s²

F = ma

F = (1369.4)(1.745)

F = 2389.603 N

7 0
3 years ago
Show how the alternative definition of power, found in your book, can be derived by substituting the definitions of work and spe
Harman [31]

Let us consider body moves a distance S due to the force F.

Hence the work by the body W = FS

If the force is not along the direction of displacement,then the work by a body for travelling a distance S will be -

                                       W=[ Fcos\theta]*S  where    Fcos\theta is the component of the force along the direction of displacement.

                                  Hence\ W= FScos\theta

                                                        = F.S

As per the question the power P is given as -

                                                  P=\frac{W}{\delta t}

                                                         =\frac{F.S}{\delta t}

                                                         = F.\frac{S}{\delta t}

                                                         = \ F.V

Hence alternative definition of power P = F.V


8 0
3 years ago
Read 2 more answers
For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

4 0
3 years ago
Read 2 more answers
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