Answer:
Magnitude of impulse, |J| = 4 kg-m/s
Explanation:
It is given that,
Mass of cart 1, 
Mass of cart 2, 
Initial speed of cart 1, 
Initial speed of cart 2, 
(stationary)
The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.
V = 1 m/s
The magnitude of the impulse exerted by one cart on the other is given by:
J = -4 kg-m/s
or
|J| = 4 kg-m/s
So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.
Refer to the diagram shown below.
i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω
Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10
Also,
R₂*i = 9.5 (2)
Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A
From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)
Answer: 0.08 Ω
Answer:
The time to boil the water is 877 s
Explanation:
The first thing we must do is calculate the external resistance (R) of the circuit, from the description we notice that it is a series circuit, by which the resistors are added
V = i (r + R)
We replace we calculate
r + R = V / i
R = v / i - r
R = 10/12 -0.04
R = 0.793 Ω
We calculate the power supplied
P = V i = I² R
P = 12² 0.793
P = 114 W
This is the power dissipated in the external resistance
We use the relationship, that power is work per unit of time and that work is the variation of energy
P = E / t
t = E / P
t = 100 10³/114
t = 877 s
The time to boil the water is 877 s
Force, newtons 3rd law of motion stated for every action there is an equal and opposite reaction
