Answer:
hydrogen bridge
Explanation:
Joule's relationship to heat and temperature is true for all materials where we assume that interatomic forces are linear, when atoms separate these forces decrease. There is a point where the separation between atoms is enough that thermal agitation can separate the molecules and there is a change of state, generally from solid to liquid and from liquid to vapor. When these changes of state are occurring all the energy supplied is used to break the links, so the temperature does not change.
In the specific case of water, there is a bond called a hydrogen bridge that breaks around 4ºC, therefore, at this temperature there is a deviation from the curve since this link is being broken, this does not lead to a change of macroscopic state.
For the other temperatures the water behaves like the other bodies.
Answer:
size get smaller maybe there is no object remaining
ANSWER:
F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.
STEP-BY-STEP EXPLANATION:
F(h) is Horizontal Force = 200 N
V is Speed = 2.4 m/s
The total weight increase by 42%
coefficient of rolling friction decrease by 19%
Since the velocity is constant so acceleration is zero; a=0
Now the horizontal force required to move the pickup is equal to the frictional force.
F(h) = F(f)
F(h) = mg* u
m is mass
g is gravitational acceleration = 9.8 m/s^2
200 = mg*u
Since weight increases by 42% and friction coefficient decreases by 19%
New weight = 1+0.42 = 1.42 = (1.42*m*g)
New friction coefficient = μ = 1 - 0.19 = 0.81 = 0.81 u
F(h) = (0.81μ) (1.42 m g)
= (0.81) (1.42) (μ m g)
= (0.81) (1.42) (200)
= 230 N
If you're moving, then you have kinetic energy.
If you're not at the bottom yet, then you still have
some potential energy left.
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
