Answer:
I think its radiation
Explanation:
Conduction is the transfer of heat through solids (A)
Convection is the transfer of heat through liquids or gasses (B)
Radiation is the transfer of heat through em waves (C)
Answer:
a) 
b) 
c) 
d)
or 18.3 cm
Explanation:
For this case we have the following system with the forces on the figure attached.
We know that the spring compresses a total distance of x=0.10 m
Part a
The gravitational force is defined as mg so on this case the work donde by the gravity is:

Part b
For this case first we can convert the spring constant to N/m like this:

And the work donde by the spring on this case is given by:

Part c
We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

And if we solve for the initial velocity we got:

Part d
Let d1 represent the new maximum distance, in order to find it we know that :

And replacing we got:

And we can put the terms like this:

If we multiply all the equation by 2 we got:

Now we can replace the values and we got:


And solving the quadratic equation we got that the solution for
or 18.3 cm because the negative solution not make sense.
The answer would be (A) Protons
Answer:
20.4m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Distance = 53m
Time = 5.2s
Unknown:
Acceleration = ?
Solution:
This is a linear motion and we use the right motion equation;
S = ut +
at²
S is the distance
u is the initial velocity
a is the acceleration
t is the time
Insert the parameters and solve;
53 = (0x 5.2) +
x a x 5.2
53 = 2.6a
a =
= 20.4m/s²