Answer:
Explanation:
A )
The ball floats with half of it exposed above the water level . So it must have density half that of water . In other words its density must have been 500 kg / m³
B )
Tension in the ball will be equal to net force acting on the ball
Net force on the ball = buoyant force - weight .
4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1000 - 893 )
= 40.65 x 10⁻⁶ N .
C )Tension in the 3 rd ball will be equal to net force acting on the ball
Net force on the ball = weight - buoyant force
= 4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1320 - 1000 )
= 121.6 x 10⁻⁶ N .
2. Wheel and axle
3.fulcrum
6.wheel and axle (I think) hope this helps!
The punching bag applies the same amount of force to the boxer’s hand. (C)
Work done in moving a proton = potential difference×Charge of a proton
= 164×1.6×10⁻¹⁹ = 2.624×10⁻¹⁷ J
This work done should be equal to change in kinetic energy.
Initial speed of proton is zero therefore K.E initial will be zero.
Work done = final kinetic energy = 2.624×10⁻¹⁷ J
K.E = mv²/2
v² = 2(2.624×10⁻¹⁷)/1.6×10⁻²⁷ = 3.28×10¹⁰ m/s
∴ v = 1.811×10⁵ m/s
Answer:
0.012 m
Explanation:
m = mass of the marble = 0.0265 kg
M = mass of the pendulum = 0.250 kg
v = initial velocity of the marble before collision = 5.05 m/s
V = final velocity of marble-pendulum combination after the collision = ?
using conservation of momentum
m v = (m + M) V
(0.0265) (5.05) = (0.0265 + 0.250) V
V = 0.484 m/s
h = height gained by the marble-pendulum combination
Using conservation of energy
Potential energy gained by the combination = Kinetic energy of the combination just after collision
(m + M) gh = (0.5) (m + M) V²
gh = (0.5) V²
(9.8) h = (0.5) (0.484)²
h = 0.012 m