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Anit [1.1K]
2 years ago
13

A circuit contains two resistors in series. The voltage drop across the first is 10 V. The voltage drop across the second is als

o 10
V. What is the voltage provided by the power supply in this circuit?

A. 10 V

B. -10 V

C. 20 V

D. -20 V
Physics
1 answer:
Norma-Jean [14]2 years ago
6 0

Hi there!

Voltage in a series can be expressed by the following:

V_T = V_1 + V_2

In words, the total voltage is equal to the sum of the individual voltage drops in a SERIES circuit.

We can solve for the total voltage:

V_T = 10 + 10 = \boxed{\text{ D. 20 V}}

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A 7-hp (shaft) pump is used to raise water to an elevation of 19 m. if the mechanical efficiency of the pump is 82 percent, dete
Hoochie [10]
Convert the units of power,W = 7 hp = 7 * 745.69 = 5219.83 WCalculate the power input to the pump using the efficiency of the pump equationn=Wpump/Wshaft
Substitute 0.82 for n and 5219.83 for Wshaft0.82=Wpump/5219.83   Wpump=0.82*5219.83=4280.26 WCalculate the mass flow ratein=Wpump/(gz_2 )Where g is the acceleration due to gravity, and z_2 is the elevation of water. Substitute 4280.26 for Wpump, 9.81 m/s^2 for g, and 19m for z_2in = 4280.26 / 9.81 * 19 = 22.9640 m^3/sCalculate the volume flow rate of waterV=m/ρWhere ρ is the density of water. Substitute 22.9640  m^3/s for in and 1000  m^3/kg for ρ, we get V = 22.9640 / 1000 = 0.0230 kg/sTherefore, the volume flow rate of water is 0.0230 kg/s
7 0
3 years ago
a runner covers the last straight stretch of the race in 4 s. durning that time, he speeds up from 5 m/s to 9m/s what is the run
Sergio039 [100]

Answer:

The runner's acceleration was 1\ m/s^2

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

v_f=v_o+at

Solving for a:

\displaystyle a=\frac{v_f-v_o}{t}

The runner speeds up from vo=5 m/s to vf=9 m/s in t=4 seconds, thus:

\displaystyle a=\frac{9-5}{4}

\displaystyle a=\frac{4}{4}=1

The runner's acceleration was 1\ m/s^2

5 0
3 years ago
A wheelbarrow is a good example of a second-class lever. True or False
olganol [36]

Answer:

true

Explanation:

a wheelbarrow has its load situated between the fulcrum and the force the wheel Barrow is 2nd class because of its resistance between the force and the axis

3 0
3 years ago
A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan
Nesterboy [21]

Answer;

The temperature change for the second pan will be lower compared to the temperature change of the first pan

Explanation;

-The quantity of heat is given by multiplying mass by specific heat and by temperature change.

That is; Q = mcΔT

This means; the quantity of heat depends on the mass, specific heat capacity of a substance and also the change in temperature.

-Maintaining the same quantity of heat, with another pan of the same mass and greater specific heat capacity would mean that the change in temperature would be much less lower.

7 0
3 years ago
Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at
Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

v = \dfrac{2\pi R}{T}

v = \dfrac{2\pi\times 8}{4.30}

  v = 11.69 m/s

now, Force does the ring push on her at the top

- N - m g = \dfrac{-mv^2}{R}

N + m g = \dfrac{mv^2}{R}

N = \dfrac{mv^2}{R}- m g

N = m(\dfrac{v^2}{R}- g)

N = 58\times (\dfrac{11.69^2}{8}- 9.8)

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0
3 years ago
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