Answer:
= 14.88 N
Explanation:
Let's begin by listing out the given variables:
M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,
g = 9.8 m/s²
At equilibrium, the sum of all external torque acting on an object equals zero
τ(net) = 0
Taking moment about we have:
(M + m) g * 0.5L - 
(L - d) = 0
⇒ = [(M + m) g * 0.5L] ÷ (L - d)
= [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)
= 59.535 ÷ 2.4
= 24.80625 N ≈ 24.81 N
Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N
Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N
Using sum of equilibrium in the vertical direction, we have:
+ = W + w ------- Eqn 1
Substituting T2, W & w into the Eqn 1
+ 24.81 = 26.46 + 13.23
= <u>14.88</u> N
Answer: 50 gram superball that strikes the wall at 1 m/s and bounces away at 0.8 m/s has greater change in kinetic energy.
Explanation:
50 gram superball that strikes the wall at 1 m/s and bounces away at 0.8 m/s has the greater change in kinetic energy because the collision is elastic in nature that is bodies separates after collision and doesn't lose any kinetic energy.
Also for an elastic collision, both the momentum and energy of the bodies are conserved compare to inelastic collision where only momentum is conserved but not the kinetic energy(this is attributed to bodies that sticks together after collision).
Answer:
B
Explanation:
A body has kinetic energy that is moving
Answer:
The magnetic field strength due to current flowing in the wire is9.322 x 10⁻⁶ T.
Explanation:
Given;
electric current, I = 21.3 A
distance of the magnetic field from the wire, R = 45.7 cm = 0.457 m
The strength of the resulting magnetic field at the given distance is calculated as;
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A
Therefore, the magnetic field strength due to current flowing in the wire is 9.322 x 10⁻⁶ T.
Answer: 1872 N
Explanation:
This problem can be solved by using one of the Kinematics equations and Newton's second law of motion:
(1)
(2)
Where:
is the bullet's final speed (when it leaves the muzzle)
is the bullet's initial speed (at rest)
is the bullet's acceleration
is the distance traveled by the bullet before leaving the muzzle
is the force
is the mass of the bullet
Knowing this, let's begin by isolating from (1):
(3)
(4)
(5)
Substituting (5) in (2):
(6)
Finally:
