A pot of soup because metal is a better conductor of heat and it stores more heat as well. thermal energy is heat energy.
Answer:
<em>The force of friction acting on the block has a magnitude of 15 N and acts opposite to the applied force.</em>
Explanation:
<u>Net Force
</u>
The Second Newton's law states that an object acquires acceleration when an unbalanced net force is applied to it.
The acceleration is proportional to the net force and inversely proportional to the mass of the object.
If the object has zero net force, it won't get accelerated and its velocity will remain constant.
The m=2 kg block is being pulled across a horizontal surface by a force of F=15 N and we are told the block moves at a constant velocity. This means the acceleration is zero and therefore the net force is also zero.
Since there is an external force applied to the box, it must have been balanced by the force of friction, thus the force of friction has the same magnitude acting opposite to the applied force.
The force of friction acting on the block has a magnitude of 15 N opposite to the applied force.
Answer:
Explanation:
Using the magnification formula.
Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)
M = v/u = H1/H2
v/u = H1/H2...1
3) Given the radius of curvature of the concave lens R = 20cm
Focal length F = R/2
f = 20/2
f = 10cm
Object distance u = 5cm
Object height H2= 5cm
To get the image distance v, we will use the mirror formula
1/f = 1/u+1/v
1/v = 1/10-1/5
1/v = (1-2)/10
1/v =-1/10
v = -10cm
Using the magnification formula
(10)/5 = H1/5
10 = H1
H1 = 10cm
Image height of the peg is 10cm
4) If u = 15cm
1/v = 1/f-1/u
1/v = 1/10-1/15
1/v = 3-2/30
1/v = 1/30
v = 30cm
30/15 = H1/5
15H1 = 150
H1/= 10cm
5) if u = 20cm
1/v = 1/f-1/u
1/v = 1/10-1/20
1/v = 2-1/20
1/v = 1/20
v = 20cm
20/20 = H1/5
20H1 = 100
H1 = 5cm
6) If u = 30cm
1/v = 1/f-1/u
1/v = 1/10-1/30
1/v = 3-1/30
1/v = 2/30
v = 30/2 cm
v =>15cm
15/30 = Hi/5
30H1 = 75
H1 = 75/30
H1 = 2.5cm
Answer:
C) 2.44 × 106 N/C
Explanation:
The electric flux through a circular loop of wire is given by
where
E is the electric field
A is the cross-sectional area
is the angle between the direction of the electric field and the normal to A
The flux is maximum when 
, so we are in this situation and therefore 
, so we can write
Here we have:
is the flux
d = 0.626 m is the diameter of the coil, so the radius is
r = 0.313 m
and so the area is
And so, we can find the magnitude of the electric field:
