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Oksanka [162]
2 years ago
13

An 8.0-ohm resistor and a 6.0-ohm resistor are connected in series with a battery. The potential difference across the 6.0-ohm r

esistor is measured as 12-V. (a) Find the total resistance of the circuit. (b) Find the current in the 6.0-ohm resistor. (c) Find the potential difference across the battery.
Please show work! :)
Physics
1 answer:
ipn [44]2 years ago
6 0

Answer:

a) 14 Ω

b) 2.0 A

c) 28 V

Explanation:

a) The total resistance of resistors in series is the sum:

R = R₁ + R₂

R = 8.0 Ω + 6.0 Ω

R = 14 Ω

b) The current in the 6.0 Ω resistor can be found with Ohm's law:

V = IR

12 V = I (6.0 Ω)

I = 2.0 A

c) Since the resistors are in series, they have the same current.  So the total voltage is:

V = IR

V = (2.0 A) (14 Ω)

V = 28 V

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Answer:

  • The gravity does a work of - 117.6 Joules.
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Explanation:

The work done by the gravity simply is the difference in gravitational potential energy multiplied by -1:

W_g = - \Delta E_p = - (mgh_f  - m g h_i)

where m is the mass of the ball, g is the acceleration due to gravity, h_f is the final height and h_i is the initial height.

So, if the radius is 2.00 m, then the difference of height will be 4 meters:

W_g = - mg (h_f - h_i)

W_g = - 3.00 \ kg \ 9.8 \frac{m}{s^2} \ 4 \m

W_g = - 117.6 Joules

As the tension is perpendicular to the velocity of the ball, the force is always perpendicular to the direction of motion. So, the differential of work will be:

dW = \vec{F}  d\vec{r} = 0

6 0
2 years ago
A charge q1= 3nC and a charge q2 = 4nC are located 2m apart. Where on the line passing through these charges is the total electr
Ipatiy [6.2K]

Answer:

Explanation:

Electric field due to a charge Q at a point d distance away is given by the expression

E = k Q / d , k is a constant equal to 9 x 10⁹

Field due to charge = 3 X 10⁻⁹ C

E = E = \frac{9\times 10^9\times3\times10^{-9}}{d^2}

Field due to charge = 4 X 10⁻⁹ C

E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}

These two fields will be equal and opposite to make net field zero

\frac{9\times 10^9\times3\times10^{-9}}{d^2} = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}[/tex]

\frac{3}{d^2} =\frac{4}{(2-d)^2}

\frac{2-d}{d} =\frac{2}{1.732}

d = 0.928

5 0
3 years ago
When students work in a chemistry lab, the location of which items would be the most important for each student to know?
irina [24]
The safety items such as eye washing station, fire blanket, and the class shower.
6 0
3 years ago
I’ll give brainliest!! please help and answer correctly! plsss answer quick
Rashid [163]

Answer: The motion of the object will remain the same

Explanation:

6 0
3 years ago
A body travels 10 meters during the first 5 seconds of its travel, and it travels a total of 30 meters over the first 10 seconds
Inessa05 [86]

Answer:

A body travels 10 meters during the first 5 seconds of its travel,and a total of 30 meters over the first 10 seconds of its travel

20miles / 5sec = 4miles /sec would be the average speed for the last 20 m

Explanation:

The answer is 4 m/s.

In the first 5 seconds, a body travelled 10 meters. In the first 10 seconds of the travel, the body travelled a total of 30 meters, which means that in the last 5 seconds, it travelled 20 meters (30m + 10m).

The relation of speed (v), distance (d), and time (t) can be expressed as:

v = d/t

We need to calculate the speed of the second 5 seconds of the travel:

d = 20 m (total 30 meters - first 10 meters)

t = 5 s (time from t = 5 seconds to t = 10 seconds)

Thus:

v = 20m / 5s = 4 m/s

PLEASE GIVE BRAINIEST!! HOPE THIS HELPS

5 0
3 years ago
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