Objects in free fall are weightless. true or false???

An electric lamp consumes 60W at 220 volts. How many dry cells of 1.5 V and internal resistance 1 Ohm are required to glow the l

GenaCL600 [577]

Answer:

1. Number of dry cells of 1.5 V required is 40.

2. Number of internal resistance of 1 ohm required is 807

Explanation:

We'll begin by calculating the resistance. This can be obtained as follow:

Power (P) = 60 W

Voltage (V) = 220 V

Resistance (R) =?

P = V²/R

60 = 220² / R

Cross multiply

60 × R = 220²

60 × R = 48400

Divide both side by 60

R = 48400 / 60

R ≈ 807 Ohm

1. Determination of the number of dry cells of 1.5 V required.

Voltage (V) = 220

Dry Cells = 1.5 V

Number of dry cells (n) =?

n = Voltage / Dry cells

n = 60 / 1.5

n = 40

2. Determination of the number of internal resistance of 1 ohm required.

Resistance (R) = 807 Ohm

Internal resistance (r) = 1 ohm

Number of internal resistance (n) =?

n = R/r

n = 807 / 1

n = 807

SUMMARY:

1. Number of dry cells of 1.5 V required is 40.

2. Number of internal resistance of 1 ohm required is 807

3 0

3 years ago

A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

Umnica [9.8K]

Answer:

Explanation:

Given

Charge Q is uniformly spread over large non-conducting Elastic sheet

Electric field due to non-conducting Elastic sheet

$E=\frac{\sigma }{2\epsilon }$

where $\sigma =$ surface charge density $=\frac{q}{d^2}$

$E=\frac{\frac{q}{d^2}}{2\epsilon }$

for side 2d Electric Field is given by

$E'=\frac{\frac{q}{2d^2}}{2\epsilon }$

$E'=\frac{1}{4}\times \frac{\frac{q}{d^2}}{2\epsilon }$

$E'=\frac{E}{4}$

8 0

3 years ago

20 points please answer

Anna [14]

Another way to test your question is to build your own miniature buildings. Depending on how in-depth you go, building could get a little pricey, but if you keep it basic there shouldn't be a problem. Decide on a certain number of foundations to test [maybe 3 or so] and try simulating an earthquake.
<span>Hope this helps! </span>

6 0

3 years ago

If a wire lies withina magnetic field what must be true for the magnetic field to produce an electric current in the wire

BigorU [14]

Answer:

The magnetic field through the wire must be changing

Explanation:

According to Faraday's law, the induced emf, ε in a metallic conductor is directly proportional to the rate of change of magnetic flux,Φ through it. This is stated mathematically as ε = dΦ/dt.

Now for the wire, the magnetic flux through it is given by Φ = ABcosθ where A = cross-sectional area of wire, B = magnetic field and θ = angle between A and B.

So, dΦ/dt = dABcosθ/dt

Since A and B are constant,

dΦ/dt = ABdcosθ/dt = -(dθ/dt)ABsinθ

Since dθ/dt implies a change in the angle between A and B, since A is constant, it implies that B must be rotating.

So, <u>for an electric current (or voltage) to be produced in the wire, the magnetic field must be rotating or changing</u>.

5 0

3 years ago

If Earth were 10.0 times farther away from the Sun than it is now, how many times weaker would the gravitational force between t

choli [55]

If Earth were 10.0 times farther away from the Sun than it is now, 100 times weaker would the gravitational force between the Sun and Earth.

What is Gravitational Force?

According to Newton's universal law of gravitation, The force of attraction between any two bodies is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them.

What causes gravitational force?

Earth's gravity comes from all its mass. All its mass makes a combined gravitational pull on all the mass in your body. That's what gives you weight. And if you were on a planet with less mass than Earth, you would weigh less than you do here.

Learn more about gravitational force:

brainly.com/question/862529

#SPJ4

5 0

2 years ago