Practice finding the volume of a sphere.

Compare a series circuit powered by six 1.5-volt batteries to a series circuit powered by a single 9-volt battery. Make sure the

lana [24]

Answer:

Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.

Explanation:

We are asked to compare two series circuits having equal number of light bulbs.

1st circuit is powered by 6 batteries each having a voltage of 1.5V

2nd circuit is powered by a single battery having a voltage of 9V.

The six batteries in the 1st circuit can be connected together in series or in parallel.

When the batteries are connected in series (positive terminal of one battery connected to negative terminal of another battery) their voltage gets added which means

Voltage of pack = number of batteries*voltage of each battery

Voltage of pack = 6*1.5

Voltage of pack = 9 volts

But the current remains same in the series connection since there is only path for the current to flow.

On the other hand, when the batteries are connected in parallel, the voltage remains same but the current increases.

Circuit 1:

In this circuit, we have 6 batteries each of 1.5 volts connected in series to provide a voltage of 9 volts.

We have connected 2 bulbs in this series circuit.

The voltage will be equally divided between two bulbs if both bulbs are identical in construction.

So there will be 4.5 volts across each bulb and both bulbs will have same brightness.

Circuit 2:

In this circuit, we have 1 battery which provide a voltage of 9 volts.

We have connected 2 bulbs in this series circuit just like in circuit 1.

The voltage will be equally divided between two bulbs if both bulbs are identical in construction.

So there will be 4.5 volts across each bulb and both bulbs will have same brightness.

Conclusion:

Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.

3 0

3 years ago

Someone has suggested that the air-standard Otto cycle is more accurate if the two polytropic processes are replaced with isentr

omeli [17]

Answer:

q_net,in = 585.8 KJ/kg

q_net,out = 304 KJ/kg

n = 0.481

Explanation:

Given:

- The compression ratio r = 8

- The pressure at state 1, P_1 = 95 KPa

- The minimum temperature at state 1, T_L = 15 C

- The maximum temperature T_H = 900 C

- Poly tropic index n = 1.3

Find:

a) Determine the heat transferred to and rejected from this cycle

b) cycle’s thermal efficiency

Solution:

- For process 1-2, heat is rejected to sink throughout. The Amount of heat rejected q_1,2, can be computed by performing a Energy balance as follows:

W_out - Q_out = Δ u_1,2

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

c_v*(T_2 - T_L) = R*(T_2 - T_L)/n-1 - q_1,2

- Using polytropic relation we will convert T_2 = T_L*r^(n-1):

c_v*(T_L*r^(n-1) - T_L) = R*(T_1*r^(n-1) - T_L)/n-1 - q_1,2

- Hence, we have:

q_1,2 = T_L *(r^(n-1) - 1)* ( (R/n-1) - c_v)

- Plug in the values:

q_1,2 = 288 *(8^(1.3-1) - 1)* ( (0.287/1.3-1) - 0.718)

q_1,2= 60 KJ/kg

- For process 2-3, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

Q_in = Δ u_2,3

q_2,3 = u_3 - u_2

q_2,3 = c_v*(T_H - T_2)

- Again, using polytropic relation we will convert T_2 = T_L*r^(n-1):

q_2,3 = c_v*(T_H - T_L*r^(n-1) )

q_2,3 = 0.718*(1173-288*8(1.3-1) )

q_2,3 = 456 KJ/kg

- For process 3-4, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

q_3,4 - w_in = Δ u_3,4

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

c_v*(T_4 - T_H) = - R*(T_4 - T_H)/1-n + q_3,4

- Using polytropic relation we will convert T_4 = T_H*r^(1-n):

c_v*(T_H*r^(1-n) - T_H) = -R*(T_H*r^(1-n) - T_H)/n-1 + q_3,4

- Hence, we have:

q_3,4 = T_H *(r^(1-n) - 1)* ( (R/1-n) + c_v)

- Plug in the values:

q_3,4 = 1173 *(8^(1-1.3) - 1)* ( (0.287/1-1.3) - 0.718)

q_3,4= 129.8 KJ/kg

- For process 4-1, heat is lost from the system. The Amount of heat rejected q_4,1, can be computed by performing a Energy balance as follows:

Q_out = Δ u_4,1

q_4,1 = u_4 - u_1

q_4,1 = c_v*(T_4 - T_L)

- Again, using polytropic relation we will convert T_4 = T_H*r^(1-n):

q_4,1 = c_v*(T_H*r^(1-n) - T_L )

q_4,1 = 0.718*(1173*8^(1-1.3) - 288 )

q_4,1 = 244 KJ/kg

- The net gain in heat can be determined from process q_3,4 & q_2,3:

q_net,in = q_3,4+q_2,3

q_net,in = 129.8+456

q_net,in = 585.8 KJ/kg

- The net loss of heat can be determined from process q_1,2 & q_4,1:

q_net,out = q_4,1+q_1,2

q_net,out = 244+60

q_net,out = 304 KJ/kg

- The thermal Efficiency of a Otto Cycle can be calculated:

n = 1 - q_net,out / q_net,in

n = 1 - 304/585.8

n = 0.481

6 0

3 years ago

If a 9V battery produces a current of 3 A through a load, what is the resistance of the load

Elden [556K]

3 ohms hope this helps :D ❤

7 0

3 years ago

Read 2 more answers

t is desired to produce aligned carbon fiber-epoxy matric composite having a longitudinal tensile strength of 800 MPa. Given (a)

Nesterboy [21]

Answer:

The value of critical length = 3.46 mm

The value of volume of fraction of fibers = 0.43

Explanation:

Given data

$\sigma_{T}$ = 800 M pa

D = 0.017 mm

L = 2.3 mm

$\sigma_{f}$ = 5500 M pa

$\sigma_{m}$ = 18 M pa

$\sigma_c$ = 13.5 M pa

(a) Critical fiber length is given by

$L_{c} = \sigma_{f} (\frac{D}{2 \sigma_{c} } )$

Put all the values in above equation we get

$L_{c} =5500 (\frac{0.017}{(2) (13.5)} )$

$L_{c} = 3.46$ mm

This is the value of critical length.

(b).Since this critical length is greater than fiber length Than the volume fraction of fibers is given by

$V_{f} = \frac{\sigma_T - \sigma_m}{\frac{L\sigma_c}{D} - \sigma_m }$

Put all the values in above formula we get

$V_{f} = \frac{800-18}{\frac{(2.3)(13.5)}{0.017} - 18 }$

$V_{f}$ = 0.43

This is the value of volume of fraction of fibers.

3 0

3 years ago

If you are involved in a collision where there is injury, you must report the incident within .......

Misha Larkins [42]

Answer:

24 hours

Explanation:

you must exchange insurance details after a collision if someone is injured. Otherwise you must report the collision to us as soon as possible (and no later than 24 hours). Although you must report such a collision straight away you should always seek medical help in the first instance.

4 0

3 years ago