Answer:
An object decreases in size due to the collision of materials. An object increases in size due to the addition of materials. Gas particles are formed from solar nebula materials.
Answer:
A
Explanation:
When friction slows a sliding block, <u>the kinetic energy of the block is transformed into internal energy
.</u>
<em>The frictional movement of two surfaces over one another leads to the conversion of some of their kinetic energies to another energy - heat or thermal energy. Hence, the temperatures of the objects are raised in the process. </em>
<u>Therefore, when a sliding block is slowed down due to friction, some of the kinetic energy of the block would be transformed into internal energy in the form of heat.</u>
The correct option is A.
Answer:
The angle between the magnetic field and the wire’s velocity is 19.08 degrees.
Explanation:
Given that,
Potential difference, V = 53 mV
Length of the wire, l = 12 cm = 0.12 m
Magnetic field, B = 0.27 T
Speed of the wire, v = 5 m/s
Due to its motion, an emf is induced in the wire. It is given by :

Here,
is the angle between magnetic field and the wire’s velocity

So, the angle between the magnetic field and the wire’s velocity is 19.08 degrees.
A) No, the equations presented above are the product of the derivation of position and velocity when the acceleration is constant.
The equations change to polynomial function of the second degree for the description of the acceleration when described as a function of time.
B) Yes, when the acceleration is zero it is concluded that the velocity is constant, therefore they could be used to describe the position as a function of the change in velocity.
Answer:
The minimum wall thickness Tmin required for the spherical tank is 65.90mm
Explanation:
Solution
Recall that,
Tmin = The minimum wall thickness =PD/2бp
where D = diameter of 8.0 m
Internal pressure = 1.62 MPa
Then
The yield strength = 295MPa/3.0 = 98.33
thus,
PD/2бp = 1.62 * 8000/ 2 *98.33
= 12960/196.66 = 65.90
Therefore the wall thickness Tmin required for the spherical tank is 65.90mm