Answer
given,
mass of steel ball, M = 4.3 kg
length of the chord, L = 6.5 m
mass of the block, m = 4.3 Kg
coefficient of friction, μ = 0.9
acceleration due to gravity, g = 9.81 m/s²
here the potential energy of the bob is converted into kinetic energy
v = 11.29 m/s
As the collision is elastic the velocity of the block is same as that of bob.
now,
work done by the friction force = kinetic energy of the block
d = 7.23 m
the distance traveled by the block will be equal to 7.23 m.
Answer:
Explanation:
a ) speed of passenger = circumference / time
= 2π R / Time
= 2 x 3.14 x 50 / 60
= 5.23 m /s
b )
centrifugal force = m v² /R
= (882 /9.8 ) x 5.23² / 50
= 77.47 N
Apparent weight at the highest point
real weight - centrifugal force
= 882 - 77.47
= 804.53 N
Apparent weight at the lowest point
real weight + centrifugal force
= 882 +77.47
= 959.47 N
c ) if the passenger’s apparent weight at the highest point were zero
centrifugal force = weight
mv² /R = mg
v² = gR
= 9.8 X 50
v = 22.13 m /s
d )
apparent weight
mg - mv² / R
= 882 - (882 / 9.8 )x 22.13²/50
= 882 + 882
= 1764 N
=
Answer:
f = 485.62 N
Explanation:
Since, the bag is moving with some acceleration. Hence, the unbalanced force will be given as:
Unbalanced Force = Horizontal Component Applied Force - Frictional Force
Unbalanced Force = Fx - f
But, from Newtons Second Law of Motion:
Unbalanced Force = ma
comparing the equations:
ma = Fx - f
f = F Cos θ - ma
where,
f = frictional force = ?
F = Applied force = 593 N
m = mass of person = 49 kg
a = acceleration = 0.57 m/s²
θ = Angle with horizontal = 30°
Therefore,
f = (593 N)(Cos 30°) - (49 kg)(0.57 m/s²)
f = 513.55 N - 27.93 N
<u>f = 485.62 N</u>
Answer:
Let I and j be the unit vector along x and y axis respectively.
Electric field at origin is given by
E= kq1/r1^2 i + kq2/r2^2j
= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)
= (2.88i + 1.44j)*10^-3 N/C
Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3
F= (1.382 i + 0.691 j) *10^-21
Goodluck
Explanation:
It is strong enough to penetrate through flesh but not bone so we can see if there are fractures or breaks in our skeleton
