What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50×10^4v ?
Answer:
Explanation:
A and B are in series , Total resistance = Ra + Rb
This resistance is in parallel with single resistor C
Equivalent resistance Re = Rc x ( Ra + Rb ) / [Rc + ( Ra + Rb )]
Now this combination is in series in single resistance D .
Total resistance = Rd + Re
= Rd + { Rc x ( Ra + Rb ) / [Rc + ( Ra + Rb )] }
Answer:
i dont know
Explanation:
i dont know since you didn't provide something to base off of
Answer:
distance = 33.124 meters
Explanation:
To solve this question, we will use one of the equations of motion which is:
s = ut + 0.5a * t^2
where:
s is the distance that we want to get
u is the initial velocity = 0
a is the acceleration due to gravity = 9.8 m/sec^2
t is the time = 2.6 sec
Substitute with the givens in the equation to get the distance as follows:
s = ut + 0.5a * t^2
s = (0)(2.6) + 0.5(9.8)(2.6)^2
s = 33.124 meters
Hope this helps :)
In general, the quantity of heat energy, Q, required to raise a mass m kg of a substance with a specific heat capacity of <span>c </span>J/(kg °C), from temperature t1 °C to t2 °C is given by:
<span>Q </span>= <span>mc(t</span><span>2 </span><span>– t</span>1<span>) joules</span>
<span>So:</span>
(t2-t1) =Q / mc
<span>As we know:
Q = 500 J </span>
<span>m = 0.4 kg</span>
<span>c = 4180 J/Kg </span>°c
<span>We can take t1 to be 0</span>°c
t2 - 0 = 500 / ( 0.4 * 4180 )
t2 - 0 = 0.30°c
