Answer:
The velocity of the gun relative to the ground is 19.66 m/s
Explanation:
Given data,
The mass of the gun, M = 15.0 kg
The mass of the bullet, m = 50 g
The velocity of the train, v = 75 km/h
= 20.83 m/s
The velocity of bullet relative to train, V' = 350 m/s
The velocity of bullet relative to ground, V = 350 + 20
= 370 m/s
According to the law of conservation of momentum,
Mv' + mV' = 0
= -1.17 m/s
Therefore, the velocity of the gun with,
v₀ = V + v'
= 20.83 - 1.17
= 19.66 m/s
Hence, the velocity of the gun relative to the ground is 19.66 m/s
It's not the potential energy. It's just the potential.
It's greatest at the positive terminal of the battery or power supply.
Yes it depends on the column on the periodic table
Answer:
The extension of the wire is 0.362 mm.
Explanation:
Given;
mass of the object, m = 4.0 kg
length of the aluminum wire, L = 2.0 m
diameter of the wire, d = 2.0 mm
radius of the wire, r = d/2 = 1.0 mm = 0.001 m
The area of the wire is given by;
A = πr²
A = π(0.001)² = 3.142 x 10⁻⁶ m²
The downward force of the object on the wire is given by;
F = mg
F = 4 x 9.8 = 39.2 N
The Young's modulus of aluminum is given by;
Where;
Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²
Therefore, the extension of the wire is 0.362 mm.
Answer:
(a) 1.21 m/s
(b) 2303.33 J, 152.27 J
Explanation:
m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s
(a) Let their velocity after striking is v.
By use of conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v
v = ( - 356.25 + 607.94) / 208 = 1.21 m /s
(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2
= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)
= 0.5 (1335.94 + 3270.7) = 2303.33 J
Kinetic energy after collision = 1/2 (m1 + m2) v^2
= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J
