Complete Question
In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
Answer:
The speed of the helicopter is 
Explanation:
From the question we are told that
The height at which he let go of the brief case is h = 130 m
The time taken before the the brief case hits the water is t = 6 s
Generally the initial speed of the briefcase (Which also the speed of the helicopter )before the man let go of it is mathematically evaluated using kinematic equation as
Here s is the distance covered by the bag at sea level which is zero
=>
=> 
=>
Answer:128 N
Explanation:
Sample of 3 cm and 4 mm diameter found to break under a minimum force of 128 N .
If sample is 1.5 cm long with same cross-sectional area then minimum force required to break is also 128 N because the applied force is same for any length and diameter of tendon.
Kinetic energy is calculated through the equation,
KE = 0.5mv²
At initial conditions,
m₁: KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J
m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J
Due to the momentum balance,
m₁v₁ + m₂v₂ = (m₁ + m₂)(V)
Substituting the known values,
(0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)
V = 0.2977 m/s
The kinetic energy is,
KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
KE = 0.03146 J
The difference between the kinetic energies is 0.0473 J.
Energy is calculated as power*time, so give the wattage of 1200 W (equivalent to 1200 Joules/second) and time of 30 seconds, multiplying these gives 36000 J or 36 kJ of electrical energy.
If electrical charge or current is needed: Power = voltage * current, so given the power of 1200 watts and voltage of 120 V, current is 1200 W / 120 V = 10 Amperes. Charge is calculated by multiplying 10 A*30 s = 300 C.
Answer:
given,
mass of copper = 100 g
latent heat of liquid (He) = 2700 J/l
a) change in energy
Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (300 - 4)
Q = 11153.63 J
He required
Q = m L
11153.63 = m × 2700
m = 4.13 kg
b) Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (78 - 4)
Q = 2788.41 J
He required
Q = m L
2788.41 = m × 2700
m = 1.033 kg
c) Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (20 - 4)
Q = 602.90 J
He required
Q = m L
602.9 = m × 2700
m =0.23 kg
