575 - Physics Questions Answers
A body rotating with uniform angular acceleration covers 100 π (radian) in the first 5 s after the start. Its angular speed at the end of 5 s ( in radian/s) is
initial angular velocity = 0
so use θ = 1/2 αt2
and ω = αt
if A + B + C = 0 then A x B is
It is not given that in which terms we have to get answer so many answers are possible
A possible solution
A + B + C = 0
A+B = -C
B X (A+B) = - B X C
BXA + 0 = - B X C
AXB = BXC ( ANS)
This answer is given by SARIKA and modified by admin only.
a small block slides down a smooth inclined plane , starting from rest at time t = 0 .let Sn be the distance travelled by the block in time interval t = n - 1 to t = n. then the ratio Sn / Sn + 1 is ??
distance travelled in n seconds = 1/2 an2 (a is the acceleration in the direction of incline)
distance travelled in n-1 second = 1/2 a(n-1)2
distance travelled in n+1 second = 1/2 a(n+1)2
so according to given condition Sn / Sn+1 = [1/2 an2 - 1/2 an2] / [1/2 a(n+1)2 - 1/2 an2]
now solve
NAME THE OUTERMOST LAYER OF SUN ????
i think its corona
Two identical particles each of mass M & charge Q are placed some distance apart . If they are in equilm under mutual gravitational & electric force then calculate the order of signature of charge Q/M in SI system.
ANS = 10-10 C/kg
on comparing gravitational force and electrical force,
we get 9*109 * q2/r2 = 6.67*10-11* m2/r2
so q/m = 10-10 approx
1. A mass "m" moving with a velocity "u" hits a surface at an angle "Q" with the normal at the point of hitting . How much force does it exerts, if no energy is lost ?
2. A thin cicular loop of radius "R" rotates about its vertical diameter with angular frequency "W" . show that a small bead on the wire loop remains at its lowermost point for W ≤ (root of g/R) . what is the angle made by the radius vector joining the center to the bead with the vertical downwards direction for W= ( roots of 2g/R) . neglect friction .
3. A rear side of a truck is open and a box of 40 kg is placed 5m away from the open end . the coefficient of friction b/w box and surface in 0.15 on a straight road , the truck starts from rest and acclerates with 2m/s2. at what distance from the starting point does the box fall off the truck .
4. Straight from rest , a mass "m" slides down on inclined plane "Q" in a time "n times " the time to slide down the same lenght in absence of friction . Find the coefficient of friction .
1) angle is measured from vertical so vertical component of velocity = ucosθ
and since no energy is lost so returning velocity in vertical direction will be same
so impulse = (mucosθ) - (-mucosθ)
force cant be calculated without time of impact.
2)
at the position given in diagram mrω2cosθ = mgsinθ for equilibrium
so rω2cosθ = gsinθ and r = Rsinθ
so Rsinθω2cosθ = gsinθ so Rω2cosθ = g or ω = √[g/Rcosθ] so ω ≥ √[g/R]
thus for W ≤ (root of g/R), bead will remain at the lowermost point.
for second part compare W= ( roots of 2g/R) and ω = √[g/Rcosθ]
3) pseudo in opposite direction = ma
frictional force in the direction of motion of truck = 0.15mg
so equation of motion of box in opposite direction is ma - 0.15mg = ma'
so 2-1.5 = a' or a' = 0.5
distance = 5m
so time taken upto fall of box can be obtained by s = 1/2 a' t2
after calculating t, use again s = 1/2 (2) t2 for finding distance travelled by truck.
4) in first case with friction s = 1/2 (gsinθ-μgcosθ) (nt)2
in second case s = 1/2 (gsinθ) t2
now solve.
Projectile motions Plzz help!!!!?
2)If α = 30 degrees and β = 30 degrees and 'a'= 4.9 m, find the initial velocity of projection.
component of initial velocity along OB = ucos30
component of final velocity along OB = 0
component of g along OB = -gsin30
now use v = u + at
in second part clearly explain "a"
s1 = 0*10+1/2*a*100 = 50a
s1+s2 = 0*20+1/2*a*400 = 200a
now solve
a fixed uniformly charged ring of radius 3m has a positive linear charge density 50/3µC/m. a point charge 5μC is moving towards the ring along its axis such that its kinetic energy 4m away is 5J. Its kinetic energy at the centre of the ring will be?
compare total energy at the two points and get answer
a body is projected vertically upward with speed 32 m/s. total time taken in its complete journey is
suppose it travelled h height thus from bottom to top it will covered h height in time t1 & at top its vel will be zero thus from eqn v= u-gt => 0= 32- 10t u can find t1= 3.2s , same time particle will take in downward journey so total time = 6.4 sec.
This answer is given by sarika and is correct answer, we made some modifications only