Answer:
See below
Explanation:
With switches open, the circuit is a simple series circuit ....the ammeters will have the same readings
V = IR
I = V/R = 5 / (10+5+5) = .25 A
b) With S1 closed 5 ohm and 10 ohm in parallel become = 5 *10 / (5+10) = 3.33 ohm
then the series circuit current becomes
5 v / ( 10 + 3.33 + 5 ) = ammeter 1 = .273 amps
ammeter 2 will get a portion of this ...the smaller resistor will get 2/3 ...the 10 ohm resistor will get 1/3 .273 * 10 / 15 =.182 amps
We know that AD is the shortest way but ABCD is connected in series
Equivalent Resistance =R1+R2+R3
=(10+10+10)ohm
=30ohm
Since BC and AD is connected in parallel
1/Req=1/R1 + 1/R2
=1/30 + 1/10
=3+1/30
=4/30
So, Req=30/4 ohm
Now,Current drawn from the battery
=> I = V/R
=> I = 30/4 ohm/3v
=> I = 30/4*3
So, I = 10/4 A
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A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)
<h3>What is power?</h3>
In physics, power (P) is the work (W) done over a period of time.
- Step 1. Calculate the work done by the bodybuilder each time.
The bodybuilder lifts a 215 kg (m) weight to a height of 0.90 m (h). Being the gravity (g) of 9.81 m/s², we can calculate the work done in each lift using the following expression.
W = m × g × h = 215 kg × 9.81 m/s² × 0.90 m = 1.9 × 10³ N
- Step 2. Calculate the work done by the bodybuilder over 10 times.
W = 10 × 1.9 × 10³ N = 1.9 × 10⁴ N
- Step 3. Calculate the power exerted by the bodybuilder.
The bodybuilder does a work of 1.9 × 10⁴ N in a 45-s span.
P = 1.9 × 10⁴ N/45 s = 421 W
A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)
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I haven't worked on Part-A, and I don't happen to know the magnitude of the gravitational force that the Sun exerts on the Earth.
But whatever it is, it's exactly, precisely, identical, the same, and equal to the magnitude of the gravitational force that the Earth exerts on the Sun.
I think that's the THIRD choice here, but I'm not sure of that either.
Answer:
less frequency in the energy