Answer:
A heat engine is a device that converts internal energy into work. Internal energy is increased by the addition of heat. The efficiency of a heat engine is a measurement of how efficiently it works. Efficiency compares the amount of useful energy extracted from a process to the total energy input. The heat engine will be more efficient if the percentage is higher.
Explanation:
Answer:
The angle is 
Explanation:
From the question we are told that
The distance of the dartboard from the dart is 
The time taken is 
The horizontal component of the speed of the dart is mathematically represented as

where u is the the velocity at dart is lunched
so

substituting values

=> 
From projectile kinematics the time taken by the dart can be mathematically represented as

=> 


=> 
![\theta = tan^{-1} [0.277]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20tan%5E%7B-1%7D%20%5B0.277%5D)

Here We can use principle of angular momentum conservation
Here as we know boy + projected mass system has no external torque
Since there is no torque so we can say the angular momentum is conserved

now we know that
m = 2 kg
v = 2.5 m/s
L = 0.35 m
I = 4.5 kg-m^2
now plug in all values in above equation

![1.75 = [4.5 + 0.245]\omega](https://tex.z-dn.net/?f=1.75%20%3D%20%5B4.5%20%2B%200.245%5D%5Comega)


so the final angular speed will be 0.37 rad/s
Answer:
Explanation:
Given that, .
R = 12 ohms
C = 500μf.
Time t =? When the charge reaches 99.99% of maximum
The charge on a RC circuit is given as
A discharging circuit
Q = Qo•exp(-t/RC)
Where RC is the time constant
τ = RC = 12 × 500 ×10^-6
τ = 0.006 sec
The maximum charge is Qo,
Therefore Q = 99.99% of Qo
Then, Q = 99.99/100 × Qo
Q = 0.9999Qo
So, substituting this into the equation above
Q = Qo•exp(-t/RC)
0.9999Qo = Qo•exp(-t / 0.006)
Divide both side by Qo
0.9999 = exp(-t / 0.006)
Take In of both sodes
In(0.9999) = In(exp(-t / 0.006))
-1 × 10^-4 = -t / 0.006
t = -1 × 10^-4 × - 0.006
t = 6 × 10^-7 second
So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge
Ewan ko po ehh sorri po’ hahaha thanks me