Determine the speed of sound on a rainy day with the temperature of 18 degrees celsius.

The power in an electrical circuit is given by the equation P= RR, where /is

Rainbow [258]

Answer:D

Explanation:Electric power=I*I*R

=12*12*100

=14400watts

6 0

3 years ago

In which situation will the lowest resistance occur?

MrMuchimi

C. Thick wire and cold temperature.

Explanation:

The resistance of a wire is given by: R = (ρL)/A

where ρ is the resistivity of the material, L is the length of the wire, A is the cross-sectional area of the wire.

From the formula, we see that the thicker the wire, the larger A, therefore the smaller the resistivity. so, a thick wire will have lower resistivity.

Moreover, the resistance of a wire increases with the temperature. In fact, high temperatures mean more motion of the atoms/electrons inside the wire, so more resistance to the flow of current through it. Therefore, colder temperature means lower resistance.

So, the correct option is thick wire and cold temperature.

7 0

3 years ago

A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro

grigory [225]

Answer:

The correct answer is B

Explanation:

Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity

Φ $._{E}$ = ∫ E. dA = $q_{int}$ / ε₀

For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

∫ E dA = $q_{int}$ / ε₀

The area of a sphere is

A = 4π r²

E 4π r² = $q_{int}$ / ε₀

E = (1 /4πε₀ ) q / r²

Having the solution of the problem let's analyze the points:

A ) r = 3R / 4 = 0.75 R.

In this case there is no charge inside the Gaussian surface therefore the electric field is zero

E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

E = (1 /4πε₀ ) Q / (1.25R)²

E = (1 /4πε₀ ) Q / R2 1 / 1.56²

E₀ = (1 /4π ε₀ ) Q / R²

$E_{B}$ = Eo /1.56 ²

$E_{B}$ = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

$E_{B}$ =(1 /4π ε₀ ) Q 1/(2R)²

$E_{B}$ = (1 /4π ε₀ ) q/R² 1/4

$E_{B}$ = Eo 1/4

$E_{B}$ = 0.25 Eo

D) False the field changes with distance

The correct answer is B

4 0

3 years ago

A car travels 82 meters do North and 14 seconds the car turns around and travels 44 m due south in four seconds what is the magn

Yakvenalex [24]

Answer:

2.11 m/s

Explanation:

Take north to be positive and south to be negative.

Average velocity = displacement / time

v = (82 m + -44 m) / (14 s + 4 s)

v = 2.11 m/s

The velocity is positive, so it is 2.11 m/s north. The magnitude of the velocity is 2.11 m/s.

8 0

3 years ago

A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).

Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c. The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω² + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh = 1/2Iω² + 1/2mv² = 1/2Iω² + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω² + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω² + 1/2mr²ω²

mgh = mr²ω²/5 + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0

3 years ago