Answer:
(a) 8.362 rad/sec
(b) 6.815 m/sec
(c) 9.446 ![rad/sec^2](https://tex.z-dn.net/?f=rad%2Fsec%5E2)
(d) 396.22 revolution
Explanation:
We have given that diameter d = 1.63 m
So radius ![r=\frac{d}{2}=\frac{1.63}{2}=0.815m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bd%7D%7B2%7D%3D%5Cfrac%7B1.63%7D%7B2%7D%3D0.815m)
Angular speed N = 79.9 rev/min
(a) We know that angular speed in radian per sec
![\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec](https://tex.z-dn.net/?f=%5Comega%20%3D%5Cfrac%7B2%5Cpi%20N%7D%7B60%7D%3D%5Cfrac%7B2%5Ctimes%203.14%5Ctimes%2079.9%7D%7B60%7D%3D8.362rad%2Fsec)
(b) We know that linear speed is given by
![v=r\omega =0.815\times 8.362=6.815m/sec](https://tex.z-dn.net/?f=v%3Dr%5Comega%20%3D0.815%5Ctimes%208.362%3D6.815m%2Fsec)
(c) We have given final angular velocity ![\omega _f=675rev/min](https://tex.z-dn.net/?f=%5Comega%20_f%3D675rev%2Fmin)
And ![\omega _i=79.9rev/min](https://tex.z-dn.net/?f=%5Comega%20_i%3D79.9rev%2Fmin)
Time t = 63 sec
Angular acceleration is given by ![\alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cfrac%7B%5Comega%20_f-%5Comega%20_i%7D%7Bt%7D%3D%5Cfrac%7B675-79.9%7D%7B63%7D%3D9.446rad%2Fsec%5E2)
(d) Change in angle is given by
![\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev](https://tex.z-dn.net/?f=%5CTheta%20%3D%5Cfrac%7B1%7D%7B2%7D%28%5Comega%20_i%2B%5Comega%20_f%29t%3D%5Cfrac%7B1%7D%7B2%7D%28675%2B79.9%29%5Ctimes%201.05%3D396.22rev)
Answer:
a. 1.75 Nm²/C
b. Yes.
Explanation:
a. Electric Flux is given as:
Φ = E*A*cosθ
Where E = electric flux
A = Surface area
Φ = 14 * 0.25 * cos60
Φ = 1.75 Nm²/C
b. Yes, the shape of the sheet will affect the Flux through it. This is because flux is dependent on area of the surface and the area is dependent on the shape of the surface.
Answer:
2.45 J
Explanation:
The following data were obtained from the question:
Mass (m) = 0.5 kg
Height (h) = 1 m
Kinetic energy (KE) =?
Next, we shall determine the velocity of the rock after it has fallen half way. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 1/2 = 0.5 m
Final velocity (v) =?
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 0.5)
v² = 9.8
Take the square root of both side
v = √9.8
v = 3.13 m/s
Finally, we shall determine the kinetic energy of the rock after it has fallen half way. This can be obtained as follow:
Mass (m) = 0.5 kg
Velocity (v) = 3.13 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.5 × 3.13²
KE = 0.25 × 9.8
KE = 2.45 J
Therefore, the kinetic energy of the rock after it has fallen half way is 2.45 J
As per energy conservation we know that
Energy enter into the bulb = Light energy + Thermal energy
so now we have
energy enter into the bulb = 100 J
Light energy = 5 J
now from above equation we have
![100 = 5 + heat](https://tex.z-dn.net/?f=100%20%3D%205%20%2B%20heat)
![Heat = (100 - 5) J](https://tex.z-dn.net/?f=Heat%20%3D%20%28100%20-%205%29%20J)
![Heat = 95 J](https://tex.z-dn.net/?f=Heat%20%3D%2095%20J)
Weow that’s cool but what is your question