Answer:
The answer is 1.0 N
Explanation:
inclination of tray=12^{\circ}
gravitational Force=5 N
Now this gravitational force has two component i.e.
5\sin \theta is parallel to the tray =1.039 N
5\cos \theta is perpendicular to the tray =4.890 N
Limited resources: resources that take a long time to replenish
Example: coal, oil, nuclear gas
Non- limited resource: resources that are constantly being replenished
Example: soil, wind, water
Answer:
The potential difference across the plates is 226 V.
Explanation:
Given;
area of the capacitor plate, A = 0.2 m²
separation, d = 0.1 mm = 0.1 x 10⁻³ m
charge on each plate, Q = 4 x 10⁻⁶ C
Charge on the capacitor is given by;
Q = CV
Where;
C is the capacitance of the capacitor, given as;
C = ε₀A / d
Then, the potential difference across the plates is given by;
![V = \frac{Q}{C} = \frac{Qd}{\epsilon_o A} = \frac{(4*10^{-6})(0.1*10^{-3})}{(8.85*10^{-12})(0.2)}\\\\V = 226 \ V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7BQ%7D%7BC%7D%20%3D%20%5Cfrac%7BQd%7D%7B%5Cepsilon_o%20A%7D%20%3D%20%5Cfrac%7B%284%2A10%5E%7B-6%7D%29%280.1%2A10%5E%7B-3%7D%29%7D%7B%288.85%2A10%5E%7B-12%7D%29%280.2%29%7D%5C%5C%5C%5CV%20%3D%20226%20%5C%20V)
Therefore, the potential difference across the plates is 226 V.
Answer:
The same pendulum could be adjusted to have the same period, in the equator must have a length of 3.949m.
Explanation:
Tnp= 4 sec
gnp= 9.83 m/sec²
Lnp= 3.97m
Tequ= 4 sec
gequ= 9.78 m/sec²
Lequ=?
Lequ= (Lnp* gequ) / gnp
Lequ= 3.949 m
Explanation:
It is given that,
The angular acceleration of the basketball, ![\alpha=10\ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%3D10%5C%20rad%2Fs%5E2)
Time taken, t = 3 seconds
We need to find the ball’s final angular velocity if the ball starts from rest. It can be calculated using definition of angular acceleration i.e.
![\alpha=\dfrac{\omega_f-\omega_i}{t}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cdfrac%7B%5Comega_f-%5Comega_i%7D%7Bt%7D)
![\omega_i=0\ (rest)](https://tex.z-dn.net/?f=%5Comega_i%3D0%5C%20%28rest%29)
![\omega_f=\alpha t](https://tex.z-dn.net/?f=%5Comega_f%3D%5Calpha%20t)
![\omega_f=10\ rad/s^2\times 3\ s](https://tex.z-dn.net/?f=%5Comega_f%3D10%5C%20rad%2Fs%5E2%5Ctimes%203%5C%20s)
![\omega=30\ rad/s](https://tex.z-dn.net/?f=%5Comega%3D30%5C%20rad%2Fs)
So, the ball's final angular velocity is 30 rad/s. Hence, this is the required solution.