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1 year ago

An a.c. supply is connected to a wire stretched between the poles of a magnet. Which way does the wire move?

1 answer:

7 0

The wire vibrates back and forth between the poles of the magnet.
The frequency of the vibration is the frequency of the AC supply.

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How would the attractive force between two spheres change if the mass of one sphere was doubled?

Dominik [7]

The force of attraction between two objects can be illustrated using Newton's Law of Universal Gravitation.
The relation between the different parameters is shown in the attached image.

Now, from the relation, we can deduce that the force between the two objects is directly proportional to the masses of the two objects.

This means that, if the mass of one object is doubled, then the force between the two objects will also be doubled.

8 0

2 years ago

An object moving with a speed of 35 m/a and has a kinetic energy of 1500j, what is the mass of the object

EleoNora [17]

Explanation:

Speed or velocity (V) = 35 m/s

Kinetic energy (K. E) = 1500 Joule

mass (m) = ?

We know

K.E = 1/2 * m * v²

1500 = 1/2 * m * 35²

1500 * 2 = 1225m

m = 3000 / 1225

m = 2.45 kg

The mass of the object is 2.45 kg

Hope it will help :)

5 0

2 years ago

Please help I literally don’t understand

Veseljchak [2.6K]

Answer:

A= 2

B=3

C=4

D=5

E=7

F=8

H=12

7 0

2 years ago

A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the

zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

$E= \frac{kqr}{R^3}$

The potential at a distance can be represented as:

V(r) - V(0) = $-\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

V(r) = $-[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

= 0.56 × 10⁻² m

R = 1.29 cm

= 1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

$V(r)$ $= -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}$

$V(r)$ = -2.15 × 10⁻⁴ V

The difference between the radial distance and center can be expressed as:

V(r) - V(0) = $-\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]^R$

V(r) = $-\frac{qR^2}{8 \pi E_0R^3 }$

V(r) = $-\frac{q}{8 \pi E_0R }$

V(r) $= -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}$

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0

2 years ago

A force of 19 newtons is applied on a cart of 2 kilograms, and it experiences a frictional force of 1.7 newtons. What is the acc

Umnica [9.8K]

Frictional force always opposes applied force, so the net force on the cart would have to be 19N - 1.7N. The acceleration can then be solved by using the relation: F = ma. This is shown below:

Net force = 19 - 1.7 = 17.3 N

Acceleration = Force / mass
Acceleration = 17.3 / 2
Acceleration = 8.65 N/m

4 0

2 years ago

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An a.c. supply is connected to a wire stretched between the poles of a magnet. Which way does the wire move?​

An a.c. supply is connected to a wire stretched between the poles of a magnet. Which way does the wire move?