Answer:
Period of oscillation = 1.33 seconds
Explanation:
The period of oscillation is given by:
T = 2π√[I/(MgL)]
for I = 2MR² and L = R,
Given: L = 0.22m = R
T = 2π√[2R/g]
T = 2 × 3.142 Sqrt[( 2 × 0.22)/ 9.8]
T = 6.284 Sqrt(0.44/9.8)
T = 6.284 Sqrt(0.0449)
T = 6.284 × 0.2119
T = 1.33 sec
Answer:
Angle of reflection of light is 34 degree
Explanation:
As per law of reflection of light we know that
angle of incidence of light = angle of reflection of light
So here we know that
angle of incidence on the surface of oil is given as
so we know that
so here we can say that reflection angle of light will be same as angle of incidence
Since the ladder is standing, we know that the coefficient
of friction is at least something. This [gotta be at least this] friction
coefficient can be calculated. As the man begins to climb the ladder, the
friction can even be less than the free-standing friction coefficient. However,
as the man climbs the ladder, more and more friction is required. Since he
eventually slips, we know that friction is less than what's required at the top
of the ladder.
The only "answer" to this problem is putting lower
and upper bounds on the coefficient. For the lower one, find how much friction
the ladder needs to stand by itself. For the most that friction could be, find
what friction is when the man reaches the top of the ladder.
Ff = uN1
Fx = 0 = Ff + N2
Fy = 0 = N1 – 400 – 864
N1 = 1264 N
Torque balance
T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)
N2 = 439 N
Ff = 439= u N1
U = 440 / 1264 = 0.3481
Answer:
for the body to float, the density of the body must be less than or equal to the density of the liquid.
Explanation:
For a block to float in a liquid, the thrust of the liquid must be greater than or equal to the weight of the block.
Weight is
W = mg
let's use the concept of density
ρ_body = m / V
m = ρ_body V
W = ρ_body V g
The thrust of the body is given by Archimedes' law
B = ρ_liquid g V_liquid
as the body floats the submerged volume of the liquid is less than or equal to the volume of the block
ρ_body V g = ρ_liquid g V_liquid
ρ_body = ρ liquid Vliquido / V_body
As we can see, for the body to float, the density of the body must be less than or equal to the density of the liquid.
Answer:
It represents the change in charge Q from time t = a to t = b
Explanation:
As given in the question the current is defined as the derivative of charge.
I(t) = dQ(t)/dt ..... (i)
But if we take the inegral of the equation (i) for the time interval from t=a to
t =b we get
Q =∫_a^b▒〖I(t) 〗 dt
which shows the change in charge Q from time t = a to t = b. Form here we can say that, change in charge is defiend as the integral of current for specific interval of time.
