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4 years ago

What is the temperature at which a solid melts?

1 answer:

3 0

Generally speaking, solid turns to a liquid at it's melting point. Ice turns to water at 0 degrees Celcius. Chocolate melts at 25 degrees Celcius-Yum! Ice (solid) thaws when the temperature rises above 32 degrees Fahrenheit, becoming water (liquid). Other solids (oddly) vary. your welcome

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Each tyre of a car has an area of 100 cm2 in contact with the ground. The car has a mass of 1600 kg. The weight of the car is eq

vlabodo [156]

Answer:

b is the answer

Explanation:

tq friend b is the answer

3 0

3 years ago

Read 2 more answers

A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul

zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

$\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}$

$\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}$

$\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 + \omega^2}$

$\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2} }$

$\mathbf{\omega^2=\dfrac{39.24 }{2}}$

$\mathbf{\omega=\sqrt{19.62 } \ rad/sec}$

$\mathbf{\omega=4.429 \ rad/sec}$

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

4 0

2 years ago

Read 2 more answers

When atoms _____, they break down into atoms of different elements and release energy.

SCORPION-xisa [38]

Answer:

It may be combine?

Do you have multiple choice i can see?

Explanation:

5 0

3 years ago

C) An identical spring is pulled with a force or 75 N The elastie limit of the spring is 72N

tiny-mole [99]

Answer:

Spring cannot return to its original, since a part of its deformation is plastic, not elastic.

Explanation:

Physically speaking, stress is equal to the axial force divided by effective transversal area of spring. In addition, springs have usually a linear relationship between stress and strain in elastic region, since they are made of ductile materials. Axial force is directly proportional to axial stress, which is also directly proportional to axial strain.

Then, if force is greater than force associated with elastic limit of the spring, then spring cannot return to its original, since a part of its deformation is plastic, not elastic.

8 0

3 years ago

Two disks of polaroid are aligned so that they polarize light in the same plane. Calculate the angle through which one sheet nee

Olegator [25]

Answer: The unpolarized light's intensity is reduced by the factor of two when it passes through the polaroid and becomes linearly polarized in the plane of the Polaroid. When the polarized light passes through the polaroid with the plane of polarization at an angle $\theta$ with respect to the polarization plane of the incoming light, the light's intensity is reduced by the factor of $\cos^2\theta$ (this is the Law of Malus).

Explanation: Let us say we have a beam of unpolarized light of intensity $I_0$ that passes through two parallel Polaroid discs with the angle of $\theta$ between their planes of polarization. We are asked to find $\theta$ such that the intensity of the outgoing beam is $I_2$ . To solve this we follow the steps below:

Step 1. It is known that when the unpolarized light passes through a polaroid its intensity is reduced by the factor of two, meaning that the intensity of the beam passing through the first polaroid is

$I_1=\frac{I_0}{2}.$

This beam also becomes polarized in the plane of the first polaroid.

Step 2. Now the polarized beam hits the surface of the second polaroid whose polarization plane is at an angle $\theta$ with respect to the plane of the polarization of the beam. After passing through the polaroid, the beam remains polarized but in the plane of the second polaroid and its intensity is reduced, according to the Law of Malus, by the factor of $\cos^2\theta.$ This yields $I_2=I_1\cos^2\theta$ . Substituting from the previous step we get

$I_2=\frac{I_0}{2}\cos^2\theta$

yielding

$\frac{2I_2}{I_0}=\cos^2\theta$

and finally,

$\theta=\arccos\sqrt{\frac{2I_2}{I_0}}$

3 0

3 years ago