1) Focal length
We can find the focal length of the mirror by using the mirror equation:

(1)
where
f is the focal length

is the distance of the object from the mirror

is the distance of the image from the mirror
In this case,

, while

(the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:

from which we find

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.
3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.
4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
Answer:
B.The force of friction between the block and surface will decrease.
Explanation:
The force of friction is given by

where
is the coefficient of friction and
is the normal force.
When the student pulls on the block with force
at an angle
, the normal force on the block becomes

and hence the frictional force becomes
.
Now, as we increase
,
increases which as a result decreases the normal force
, which also means the frictional force decreases; Hence choice B stands true.
<em>P.S: Choice D is tempting but incorrect since the weight </em>
<em> is independent of the external forces on the block. </em>
Answer:
0.0613°C
Explanation:
the given parameters are m=15gm=15×10⁻³ V₁=865m/s V₂=534m/s
the bullet moves with different kinetic energies before and after the penetration, therefore
Kinetic energy before - kinetic energy after = 1/2 × m × ( V₁² - V₂²)
=
× 15×10⁻³ × (865² - 534²)
= 3.47 × 10⁻³J
this loss in energy is transferred to the water, therefore
change in temperature = 
where c = heat capacity of water = 4.19 x 10^3
m = mass of water = 13.5 kg
= {3.47 × 10⁻³} / {13.5 x 4.19 x 10^3 }
=0.0613°C