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Zinaida [17]
3 years ago
5

The circuit shown in the picture can be broken in four places. Which places can the circuit be broken and still have at least on

e light stay lit? How do you know?

Physics
1 answer:
myrzilka [38]3 years ago
3 0

Answer:

C & D

Explanation:

The C & D circuit places can be opened, in this way the bulbs in 1 & 2 will work. Since they would be connected in parallel with the battery.

The bulbs each have their respective resistance for this reason when they are connected to the battery a current is generated that flows through each of the bulbs.

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If I = 2.0 A in the circuit segment shown below, what is the potential difference VB - VA?
Tom [10]

Answer:

10 V

Explanation:

The potential difference between two points is the amount of work required to carry a unit charge from one point to the other point. This would result in a potential difference between this two points.

The difference between the potential across two points B and A is V_{BA}=V_B-V_A

From the image attached:

V_B-V_A=10-10I+20\\\\But\ I = 2A,hence:\\\\V_B-V_A=10-10(2)+20\\\\V_B-V_A=10-20+20\\\\V_B-V_A=10\ V

7 0
3 years ago
A generator converts
Ivan

Answer:

Mechanical; electrical

4 0
3 years ago
Read 2 more answers
An expensive vacuum system can achieve a pressure as low as 1.00×10^{-7} N/m^2 at 20ºC . How many atoms are there in a cubic cen
marta [7]

Answer:

24.70818432141\times 10^7\ atoms

Explanation:

P = Pressure = 1\times 10^{-7}\ N/m^2

V = Volume = 1 cm³

n = Amount of substance

N = Number of atoms

N_A = Avogadro's constant = 6.022\times 10^{23}\ /mol

R = Gas constant = 8.314 J/k mol

T = Temperature = 273.15+20 = 293.15 K

From the ideal gas law

PV=nRT\\\Rightarrow n=\frac{PV}{RT}

n=\frac{N}{N_A}

\frac{N}{N_A}=\frac{PV}{RT}\\\Rightarrow N=N_A\times \frac{PV}{RT}\\\Rightarrow N=\frac{1\times 10^{-7}\times 1\times 10^{-6}}{8.314\times 293.15}\times 6.022\times 10^{23}\\\Rightarrow N=24708184.32141\ atoms=24.70818432141\times 10^7\ atoms

The number of atoms is 24.70818432141\times 10^7\ atoms

8 0
3 years ago
The distance from home plate to the pitchers mound is 18.5 meters. For a pitcher capable of throwing at 3.85m/s(86mi/hr), how mu
Taya2010 [7]

Given that:

Distance , s = 18.5 m

Velocity , v = 3.85 m/s

Time , t =?

Since,

Velocity = distance/time

or

Time= distance/velocity

time= 18.5/ 3.85

time= 4.8 s

So the time elapse between the release of the ball and the ball passing home plate is 4.8 seconds.

7 0
3 years ago
A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.
labwork [276]

Answer:

9.4 m/s

Explanation:

According to the work-energy theorem, the work done by external forces on a system is equal to the change in kinetic energy of the system.

Therefore we can write:

W=K_f -K_i

where in this case:

W = -36,733 J is the work done by the parachute (negative because it is opposite to the motion)

K_i = 66,120 J is the initial kinetic energy of the car

K_f is the final kinetic energy

Solving,

K_f = K_i + W=66,120+(-36,733)=29387 J

The final kinetic energy of the car can be written as

K_f = \frac{1}{2}mv^2

where

m = 661 kg is its mass

v is its final speed

Solving for v,

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s

4 0
3 years ago
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