Answer:
It depends on the size of the star but I think it's B
Explanation:
If the star is massive, it will eventually explode (supernova) and if it is a star with a high mass, the core of it will form a neutron star and if it is very massive the core will turn into a blackhole
A couple of things, if the shuttle burned all of it's fuel before entering Earth's atmosphere then that means that the shuttle was accelerating towards Earth until it ran out of fuel. At that point, there is little to no air resistance (friction) by the lack of an atmosphere so it keeps accelerating due to Earth's gravitational force. The closer the shuttle gets to Earth the stronger the gravitational pull the shuttle experiences. Note that, once the shuttle reaches Earth's atmosphere it will cause significant amount of friction and thus will cause the shuttle to slow down.
Answer : The average power delivered to the circuit is, 42.6 watt
Explanation :
The relation between power, voltage and current is:
where,
P = power of circuit = ?
I = current = 6.0 A
V = voltage = 7.1 V
Now put all the given values in the above formula, we get the average power delivered to the circuit.
Thus, the average power delivered to the circuit is, 42.6 watt
Comment
The only reason you can do this is that the charges are the same. If they were not, the problem would not be possible.
Equation
The field equation is, in its simplest form,
E = kq/r^2
So each of the charges are pulling / pushing in the same direction. The equation becomes.
kq/r^2 - (-kq/r^2) = Field magnitude in N/C
Givens
- K = 9 * 10^9 N m^2 / c^2
- E = 45 N/C
- r = 7.5/2 = 3.75 cm * ( 1 m / 100 cm) = 0.0375 m
- Find Q
Solution
k*q/0.0375 ^2 - (-kq/0.0375^2) = 45 N/C Combine
2*k*q / 0.0375^2 = 45 N/C Divide by 2
kq /(0.0375^2) = 22.5 N/C Multiply by 0.0375^2
kq = 22.5 * 0.0375 ^2 Find d^2
kq = 22.5 * 0.001406 Combine
kq = 0.03164 N/C * m^2 Divide by k
q = 0.03164 N * m^2 /C / 9*10^9 N m^2 / c^2
q = 2.84760 * 10 ^8 C
I've left the cancellation of the units for you. Notice that only 1 C is left and it is in the numerator as it should be.
What part of the bacterial cell helps it stick to surfaces
