Answer:
Part a)
V = 15 Volts
Part b)
P = 180 Watt
Part c)
Rate = 720 Watt
Explanation:
Part a)
When battery is in charging then the potential difference of the terminals of cell is given by
here we know that
EMF = 12 volts
i = 60 A
r = 0.050 ohm
now we have
Part b)
Rate of thermal energy dissipated is the energy which is dissipating across the resistor
so here we have
P = 180 Watt
Part c)
Rate at which Energy stored inside the cell is the rate of electrical energy that is converted into the chemical energy
Answer:
ms⁻¹
Explanation:
Consider the motion of the bullet-block combination after collision
= mass of the bullet = 0.0382 kg
= mass of wooden block = 3.78 kg
= velocity of the bullet-block combination after collision
= spring constant of the spring = 833 N m⁻¹
= Amplitude of oscillation = 0.190 m
Using conservation of energy
Kinetic energy of bullet-block combination after collision = Spring potential energy gained due to compression of spring
ms⁻¹
= initial velocity of the bullet before striking the block
Using conservation of momentum for the collision between bullet and block
ms⁻¹
Answer:
Option a
Explanation:
The action of the rapid rubbing of the hands are a result of chemical reaction and responses that takes place inside our body.
When we rub our hands together, that shows the mechanical motion of the body or we can say that while rubbing hands chemical energy gets converted to mechanical energy (here, Kinetic energy).
While rubbing, due to friction, heat is produced and hence we can say that this mechanical energy has transformed to heat energy.
Thus the conversion here is from chemical to mechanical to heat energy.
Mass of the block = 1.4 kg
Weight of the block = mg = 1.4 × 9.8 = 13.72 N
Normal force from the surface (N) = 13.72 N
Acceleration = 1.25 m/s^2
Let the coefficient of kinetic friction be μ
Friction force = μN
F(net) = ma
μmg = ma
μg = a
μ = 
μ = 
μ = 0.1275
Hence, the coefficient of kinetic friction is: μ = 0.1275
Answer:
a) velocity v = 322.5m/s
b) time t = 19.27s
Explanation:
Note that;
ads = vdv
where
a is acceleration
s is distance
v is velocity
Given;
a = 6 + 0.02s
so,
Remember that
![v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bds%7D%7Bdt%7D%20%5C%5C%5Cfrac%7Bds%7D%7Bv%7D%20%3D%20dt%5C%5C%5Cint%5Climits%5Es_0%20%7B%5Cfrac%7Bds%7D%7B%5Csqrt%7B12s%2B0.02s%5E%7B2%7D%20%7D%20%7D%20%7D%20%5C%2C%20ds%20%3D%20%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt%20%5C%5Ct%3D%20%20%285%5Csqrt%7B2%7D%20%29%20ln%20%20%5Cfrac%7B%7C%20%5Bs%20%2B%20300%20%2B%20%5Csqrt%7B%28s%5E%7B2%7D%20%20%2B%20600s%29%7D%20%5D%20%7C%7D%7B300%7D%20.......2)
substituting s = 2km =2000m, into equation 1
v = 322.5m/s
substituting s = 2000m into equation 2
t = 19.27s
