If one bulb goes out then all the others won't light up because electricity will be cut off. It's a disadvantage because in a parallel circuit if one bulb burns out all the others will still be on because they won't be affected. I hope I've helped you ☺
Acceleration of the both masses tied together= 6m/s²
Explanation:
The force is given by F= ma
so 5= m1 (8)
m1=0.625 Kg
for m2
5=m2 (24)
m2=0.208 kg
now total mass= m1+m2=0.625+0.208
Total mass=M=0.833 Kg
now F= ma
5= 0.833 (a)
a= 5/0.833
a=6m/s²
Answer:
L=55.9m
Explanation:
The equation for the period of a simple pendulum is:
![T=2\pi\sqrt{\frac{L}{g}}](https://tex.z-dn.net/?f=T%3D2%5Cpi%5Csqrt%7B%5Cfrac%7BL%7D%7Bg%7D%7D)
In our case what we know is the period and the acceleration of gravity, and we need to know the length of the pendulum, so we can write:
![L=(\frac{T}{2\pi})^2g](https://tex.z-dn.net/?f=L%3D%28%5Cfrac%7BT%7D%7B2%5Cpi%7D%29%5E2g)
Which for our values is:
![L=(\frac{15s}{2\pi})^2(9.81m/s^2)=55.9m](https://tex.z-dn.net/?f=L%3D%28%5Cfrac%7B15s%7D%7B2%5Cpi%7D%29%5E2%289.81m%2Fs%5E2%29%3D55.9m)
Answer: certificate authority
Explanation: The user submits Identification data and certificate request to the registration authority (RA). The RA validates this information and sends the certificate request to the certificate Authority (CA)
Answer:
The correct option is (B).
Explanation:
The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,
![T^2\propto a^3\\\\T^2=ka^3](https://tex.z-dn.net/?f=T%5E2%5Cpropto%20a%5E3%5C%5C%5C%5CT%5E2%3Dka%5E3)
It is mentioned that, an asteroid with an orbital period of 8 years. So,
![(8)^2=ka^3\\\\64=ka^3\\\\a=(64)^{\dfrac{1}{3}}\\\\a=4\ AU](https://tex.z-dn.net/?f=%288%29%5E2%3Dka%5E3%5C%5C%5C%5C64%3Dka%5E3%5C%5C%5C%5Ca%3D%2864%29%5E%7B%5Cdfrac%7B1%7D%7B3%7D%7D%5C%5C%5C%5Ca%3D4%5C%20AU)
So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.