Answer:
Work done = 0.3142 Nm
Explanation:
Mass of Object is 50 g
Circular path of radius is 10 cm ⇒ 0.1 m
Work done = Force × Distance = ?
*Distance moved (circular path) ⇒ Circumference of the circular path
2πr = 2 × 3.142 × 0.1 ⇒ 0.6284 m
*Force that is enough to move a 50 g must be equal or more than its weight.
therefore convert 50 grams to newton = 0.5 N
Recall that; work done is force times distance
∴ 0.5 N × 0.6284 m
Work done = 0.3142 Nm
The maximum range of the golf ball is 18.6 m.
The given parameters;
height reached by the ball, h = 3 m
distance of the ball, x = 14
the initial velocity of the ball, u = 13.5 m/s
The maximum range of the ball is calculated as follows;
![R = \frac{u^2 sin(2 \theta)}{g}](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7Bu%5E2%20sin%282%20%5Ctheta%29%7D%7Bg%7D)
at maximum range, the angle of projection, ![\theta = 45^0](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2045%5E0)
![R_{max} = \frac{u^2}{g} \\\\R_{max} = \frac{13.5^2}{9.8} \\\\R_{max} = 18.6 \ m](https://tex.z-dn.net/?f=R_%7Bmax%7D%20%20%3D%20%5Cfrac%7Bu%5E2%7D%7Bg%7D%20%5C%5C%5C%5CR_%7Bmax%7D%20%20%3D%20%5Cfrac%7B13.5%5E2%7D%7B9.8%7D%20%5C%5C%5C%5CR_%7Bmax%7D%20%20%3D%20%2018.6%20%5C%20m)
Thus, the maximum range of the golf ball is 18.6 m.
Learn more here: brainly.com/question/24293822
Friction is caused by two solids rubbing together quickly.
Answer:
The power output of this engine is ![P = 17.5 W](https://tex.z-dn.net/?f=P%20%3D%20%2017.5%20W)
The the maximum (Carnot) efficiency is ![\eta_c = 0.7424](https://tex.z-dn.net/?f=%5Ceta_c%20%20%3D%200.7424)
The actual efficiency of this engine is ![\eta _a = 0.46](https://tex.z-dn.net/?f=%5Ceta%20_a%20%20%3D%200.46)
Explanation:
From the question we are told that
The temperature of the hot reservoir is ![T_h = 1250 \ K](https://tex.z-dn.net/?f=T_h%20%3D%201250%20%5C%20K)
The temperature of the cold reservoir is ![T_c = 322 \ K](https://tex.z-dn.net/?f=T_c%20%20%3D%20%20322%20%5C%20K)
The energy absorbed from the hot reservoir is ![E_h = 1.37 *10^{5} \ J](https://tex.z-dn.net/?f=E_h%20%20%3D%201.37%20%2A10%5E%7B5%7D%20%5C%20J)
The energy exhausts into cold reservoir is ![E_c = 7.4 *10^{4} J](https://tex.z-dn.net/?f=E_c%20%20%3D%207.4%20%2A10%5E%7B4%7D%20J)
The power output is mathematically represented as
![P = \frac{W}{t}](https://tex.z-dn.net/?f=P%20%20%3D%20%20%5Cfrac%7BW%7D%7Bt%7D)
Where t is the time taken which we will assume to be 1 hour = 3600 s
W is the workdone which is mathematically represented as
![W = E_h -E_c](https://tex.z-dn.net/?f=W%20%3D%20%20E_h%20%20-E_c)
substituting values
![W = 63000 J](https://tex.z-dn.net/?f=W%20%3D%2063000%20J)
So
![P = \frac{63000}{3600}](https://tex.z-dn.net/?f=P%20%3D%20%20%5Cfrac%7B63000%7D%7B3600%7D)
![P = 17.5 W](https://tex.z-dn.net/?f=P%20%3D%20%2017.5%20W)
The Carnot efficiency is mathematically represented as
![\eta_c = 1 - \frac{T_c}{T_h}](https://tex.z-dn.net/?f=%5Ceta_c%20%20%3D%20%201%20-%20%5Cfrac%7BT_c%7D%7BT_h%7D)
![\eta_c = 1 - \frac{322}{1250}](https://tex.z-dn.net/?f=%5Ceta_c%20%20%3D%20%201%20-%20%5Cfrac%7B322%7D%7B1250%7D)
![\eta_c = 0.7424](https://tex.z-dn.net/?f=%5Ceta_c%20%20%3D%200.7424)
The actual efficiency is mathematically represented as
![\eta _a = \frac{W}{E_h}](https://tex.z-dn.net/?f=%5Ceta%20_a%20%20%3D%20%20%20%5Cfrac%7BW%7D%7BE_h%7D)
substituting values
![\eta _a = \frac{63000}{1.37*10^{5}}](https://tex.z-dn.net/?f=%5Ceta%20_a%20%20%3D%20%20%5Cfrac%7B63000%7D%7B1.37%2A10%5E%7B5%7D%7D)
![\eta _a = 0.46](https://tex.z-dn.net/?f=%5Ceta%20_a%20%20%3D%200.46)