Ask question

Ask question

All categories

4 years ago

12

What are three properties that magnets have?

1 answer:

hichkok12 [17]4 years ago

3 0

1. Magnets attract objects of iron, cobalt and nickel.
2. The force of attraction of a magnet is greater at its poles than in the middle.
3. Like poles of two magnets repel each other.

You might be interested in

You do 25 J of work to move a 4N object 5 meters. Find your efficiency and write down the formula.

daser333 [38]

4*5=20; ration of 20:25,
20/25= %80

3 0

3 years ago

A solid circular plate has a mass of 0.25 kg and a radius of 0.30 m. It starts rolling from rest at the top of a hill 10 m long

KATRIN_1 [288]

To solve the problem it is necessary to apply the equations related to the conservation of both <em>kinetic of rolling objects</em> and potential energy and the moment of inertia.

The net height from the point where it begins to roll with an inclination of 30 degrees would be

$h=Lsin30$

$h=10sin30$

$h=5m$

In the case of Inertia would be given by

$I = \frac{mR^2}{2}$

In general, given an object of mass m, an effective radius k can be defined for an axis through its center of mass, with such a value that its moment of inertia is

$I = mk^2$

$\frac{mR^2}{2}= mk^2$

$\frac{k^2}{R^2}=\frac{1}{2}$

Replacing in Energy conservation Equation we have that

Potential Energy = Kinetic Energy of Rolling Object

$mgh = \frac{1}{2}mv^2(1+\frac{k^2}{r^2})$

$9.8*5=\frac{1}{2}v^2(1+\frac{1}{2})$

$v^2 (1.5) = 98$

$v=8.0829m/s$

Therefore the correct answer is C.

3 0

3 years ago

If it required 575 cal to convert a sample of liquid water to a vapor, how many calories would be released when the vapor conden

Maru [420]

B because when a liquid converts to a vapor is it going to be less

5 0

3 years ago

How much energy (in joules) is required to evaporate 0.0005 kg of liquid ammonia to vapor at the same temperature?

Lana71 [14]

Answer:

685.6 J

Explanation:

The latent heat of vaporization of ammonia is

L = 1371.2 kJ/kg

mass of ammonia, m = 0.0005 Kg

Heat = mass x latent heat of vaporization

H = 0.0005 x 1371.2

H = 0.686 kJ

H = 685.6 J

Thus, the amount of heat required to vaporize the ammonia is 685.6 J.

6 0

3 years ago

(From Tipler, problem 8-40 on p. 282). A wedge of mass M is placed on a frictionless horizontal surface, and a block of mass m i

dybincka [34]

Answer:

Explanation:

The loss of potential energy of the block is converted into kinetic energy of both the block and wedge.

loss of potential energy

= mgh

If v₁ and v₂ be the velocity of block and wedge at the time of separation

applying conservation of energy in horizontal direction ( no force acting in horizontal direction )

m v₁ = M v₂

Applying conservation of mechanical energy

mgh = 1/2 mv₁² + 1/2 M v₂²

1/2 mv₁² + 1/2 M ( mv₁ / M )² = mgh

1/2 mv₁² ( 1 + m/M ) = mgh

v₁² = 2gh / ( 1 + m/M )

v₁ = $\sqrt{\frac{2Mgh}{m+M} }$

v₂ =( m / M) v₁

v₂= m / M $\sqrt{\frac{2Mgh}{m+M} }$

If M is much greater

M+m = M

v₁ = $\sqrt{\ 2gh }$

v₂ = $\frac{m}{M} \times\sqrt{\ 2gh }$

5 0

3 years ago