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Dafna1 [17]
3 years ago
15

Consider the sections of two circuits illustrated above. Select True or False for all statements. After connecting c and d to a

battery, the current through R3 always equals the current through R4. Rcd is always less than or equal to R3. After connecting a and b to a battery, the voltage across R1 always equals the voltage across R2. Rab is always less than or equal to R1.
Physics
1 answer:
Basile [38]3 years ago
5 0

Answer:

Follows are the solution to the given question:

Explanation:

In this question the missing file of the circuit is not be which is defined in attached file please find it.

In Option 1, this statement is true because the current is on R_3 \  and \ R_4, that is the same.

In option 2, this statement is false because Rcd=R_3+R_4 therefore it implies that Rcd is always larger then R_3.

In option 3,  this statement is true because the voltage of R_1 \ and \ R_2 is always equal.

In option 4, this statement is true because Rab= \frac{1}{(\frac{1}{R1}+\frac{1}{R2})}=\frac{R1R2}{(R1+R2)}. \frac{R2}{(R1+R2)}is always smaller then 1 therefore,R_1 \times (\frac{R2}{(R1+R2)}) is always equal to R1.

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The gravitational force between two objects that are 2.1 times 10^-1 m apart is 3.2 times 10^-6 N. If the mass of one object is
Shtirlitz [24]

Answer:

Mass of the other object is 38.45 kg.

Explanation:

Given:

The gravitational force between two objects is, F=3.2\times 10^{-6} N

Mass of one object is, m_{1}=55\ kg

Distance between the objects is, r=2.1\times 10^{-1}\ m

Gravitational constant is, G =6.674\times 10^{-11}\ m^3 kg^{-1}s^{-2}

Let the mass of the other object be m_{2}\ kg

Gravitational force is given as:

F=\frac{Gm_1m_2}{r^2}

Plug in the given values and solve for m_2. This gives,

3.2\times 10^{-6}=\frac{6.674\times 10^{-11}\times 55\times m_2}{(2.1\times 10^{-1})^2}\\3.2\times 10^{-6}=\frac{6.674\times 10^{-11}\times 55\times m_2}{0.0441}\\3.2\times 10^{-6}=8.3236\times 10^{-8}m_2\\m_2=\frac{3.2\times 10^{-6}}{8.3236\times 10^{-8}}= 38.45\ kg

Therefore, the mass of the other object is 38.45 kg.

3 0
4 years ago
What is the equivalent resistance between points A and C if R1=1430, R2=1350, R3=1100, R4=1350, and R5=1150.
Marianna [84]

R1 + R4 = 1430 + 1350 = 2780 = R14    series combination of R1 & R4

R2 + R5 = 1350 + 1150 = 2500 = R25

The circuit has been reduced to 3 resistors in parallel

R314 = 2780 * 1100 / (2780 + 1100) = 788  this is the resistance of the parallel combination of R14 and R3

R31425 = 2500 * 788 / (2500 + 788) = 599 which is the equivalent of the circuit  - you can also use the formula for 3 resistors in parallel but this seems simpler

7 0
3 years ago
A 40 kg shell is flying at a speed of 72 km/h. It explodes into two
viva [34]

Answer: This is true

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3 years ago
When running on its 11.4 V battery, a laptop computer uses 8.3 W. The computer can run on battery power for 8.5 h before the bat
AveGali [126]

Answer:

The laptop battery is capable of supplying 253980 J.

Explanation:

Given;

voltage of the battery, V = 11.4 V

power consumed by the laptop, P = 8.3 W

duration of the battery before depletion, t = 8.5 hours

Determine the amount of energy supplied by the laptop battery within 8.5 hours.

Energy, E = Power x time

Energy, E = 8.3 W x 8.5 h = 70.55 Wh

Energy, E in joules = 70.55 wh x 3600 s/h = 253980 J

Therefore, the amount of energy supplied by the laptop battery within 8.5 hours is 253980 J.

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