Answer:Force on -7 uC charge due to charge placed at x = - 10cm
now we will have
towards left
similarly force due to -5 uC charge placed at x = 6 cm
now we will have
towards left
Now net force on 7 uC charge is given as
towards left
Explanation:
The period T of a pendulum is given by:
where L is the length of the pendulum while
is the gravitational acceleration.
In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is
. Using this data, we can solve the previous formula to find L:
Answer:
d ) is the answer.
Explanation:
Let M be the mass and R be the radius of each of ball , hoop and disc.
kinetic energy of sphere - 1/2 MV² + 1/2 I ω² ,ω is angular velocity and
V = ωR
kinetic energy of sphere - 1/2 MV² + 1/2 x 2/5 MR² ω²
= 1/2 MV² + 1/5 MR² ω²
MV² ( 1/2 + 1/5 )
= .7 MV²
kinetic energy of Disk - 1/2 MV² + 1/2 I ω² ,ω is angular velocity and
V = ωR
kinetic energy of Disk - 1/2 MV² + 1/2 x 1/2 MR² ω²
= 1/2 MV² + 1/4 MR² ω²
MV² ( 1/2 + 1/4 )
= .75 MV²
kinetic energy of Hoop - 1/2 MV² + 1/2 I ω² ,ω is angular velocity and
V = ωR
kinetic energy of hoop - 1/2 MV² + 1/2 MR² ω²
= 1/2 MV² + 1/2 MR² ω²
MV² ( 1/2 + 1/2 )
= MV²
Kinetic energy is largest in case of hoop and least in case of sphere . So hoop will go up to the highest point and sphere will go to a height which will be least among the three.
Answer:
A. Ahmed has a greater tangential speed than Jacques.
D. Jacques and Ahmed have the same angular speed.
Explanation:
Kinematics of the merry-go-round
The tangential speed of the merry-go-round is calculated using the following formula:
v = ω*R
Where:
v is the tangential speed in meters/second (m/s)
ω is the angular speed in radians/second (rad/s)
R is the angular speed in meters (m)
Data
dA = RA : Ahmed distance to the axis of rotation
dJ = RJ : Jacques distance to the axis of rotation
Problem development
We apply the formula (1)
v = ω*R
vA= ω*RA : Ahmed tangential speed
vJ= ω*RJ : Jacques tangential speed
Ahmed is at a greater distance from the axis of rotation than Jacques, then,
RA ˃ RJ and Ahmed and Jacques have the same speed ω, then:
vA ˃ vJ
The acceleration due to gravity on a large body is given by:
acceleration = (gravitational constant * mass of body) / radius²
For b, this is:
g = GM/r²
M = gr²/G
For a,
g(a) = 1/6 g
r(a) = 1/4 r
M(a) = g(a)r(a)²/G
M(a) = 1/6 g * (1/4 r)²/G
M(a) = gr²/96G
The fraction is given as:
M(a) / M = (gr²/96G) / (gr²/G)
M(a) / M = 1/96
The mass of planet a is 1/96 of the mass of planet b.