Answer:
= 0.417 m/s
Explanation:
Momentum before throwing the rock: m*V = 95.0 kg * 0.460 m/s
= 44.27 N*s
A) man throws the rock forward
mass of rock m1 = 0.310 kg
V1 = 15.5 m/s, in the same direction of the sled with the man
sled and man:
m2 = 95 kg - 0.310 kg = 94.69 kg
v2 = ?
Conservation of momentum:
momentum before throw = momentum after throw
44.27N*s = 0.310kg * 15.5m/s + 94.69kg*v2
⇒ v2 = [44.27 N*s - 0.310 * 15.5N*s ] / 94.69 kg
= 0.417 m/s
In order for the object not to slip, the component of the weight parallel to the surface must be equal to the frictional force (which acts in the opposite direction):
The parallel component of the weight is:
where m is the object mass and
is the angle of the inclined plane.
The frictional force is
where
is the coefficient of static friction.
Equalizing the two forces, we have
from which we find
and so, in our problem the coefficient of static friction must be
The real world examples that I can give is the everyday
things that we do and some of these are:
I am walking to the end of the room holding three textbooks.
Playing tug of war with friends.
Getting the groceries from the car and put it inside of your house.
Answer:
4.24nm
0.385eV
Explanation:
Maximum wavelength (λmax) :
λmax = ( hc) /Φ
h = plancks constant = 6.63 * 10^-34
c = speed of light = 3*10^8
1ev = 1.6 * 10^-19
Φ = 2.93eV = 2.93* (1.6*10^-19) = 4.688*10^-19
λmax = [(6.63 * 10^-34) * (3 * 10^8)] / 4.688*10^-19
λmax = 19.89 * 10^-26 / 4.688*10^-19
λmax = 4.242 * 10^-7 m
λmax= 4.24nm
B.)
E = hc / eλ eV
λ = 3.75nm = 3.75 * 10^-7m = 375 *10^-9
E = (6.63 * 10^-34) * (3 * 10^8) / (1.6 * 10^-19) * (375 * 10^-9)
E = 19.89 * 10^-26 / 600 * 10^-28
E = 0.03315 * 10^-26 + 28
E = 0.03315 * 10^2
E = 3.315 eV
Stopping potential : (3.315 eV - 2.93eV) = 0.385eV
