1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
bixtya [17]
4 years ago
14

When connected to a battery, a lightbulb glows brightly. If the battery is reversed and reconnected to the bulb, the bulb will g

low
brighter.
dimmer.
with the same brightness.
not at all.
Physics
1 answer:
BartSMP [9]4 years ago
5 0
If the battery is reversed and reconnected to the bulb, the bulb will glow <span>with the same brightness. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that this is the answer that you were looking for and it has actually come to your help.</span>
You might be interested in
Megan wants to see if mice that are fed food A or B grow at a faster rate. She measures the mass of the mice on day 1, then divi
diamong [38]
<span>No, there is no control group because each group is treated under test conditions.</span>
7 0
3 years ago
Read 2 more answers
Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is
saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

\dot{m_{i}} = \rho_{i} v_{i}A_{i}

Where:

v_{i}: is the initial velocity = 20 m/s

A_{i}: is the inlet area of the nozzle = 60 cm²  

\rho_{i}: is the density of entrance = 2.21 kg/m³

\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}

0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}

A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!                                                                  

5 0
3 years ago
Read 2 more answers
Mary walked north from her home to Sheila's home, which is 4.0 kilometers away. Then she turned right and walked another 3.0 kil
bixtya [17]
The answer is speed: 4.7 km/h, velocity: 3.3 km/h.

Distances and time are given:
d1 =  4 km
d2 = 3 km
d3 = 5 km
t = 1.5 h

The speed can be expressed as a distance (d) divided by time (t). The average speed (s) is total distance travelled divided by time:
s = (d1 + d2)/t = (4+3)/1.5 = 7/1.5 = 4.7 km/h

The average velocity (v) is total displacement (d₁) from the starting point divided by time. Since Mary's starting point was home, and she walked to the supermarket, which is 5.0 kilometers from her own home, her displacement is 5 km:
v = d₁/t = 5/1.5 = 3.3 km/h

8 0
4 years ago
State two ways of reducing the drag forces on a bicycle
Eddi Din [679]
I am quite sure the first one is Friction, but I am not sure about the second one. Is it wind?
7 0
3 years ago
Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur
dimulka [17.4K]

Answer:

-1.7908787542\times 10^{-9}\ C

3.4260289211\times 10^{-9}\ C

1.7908787542\times 10^{-9}\ C

1.6351501669\times 10^{-9}\ C

Explanation:

r = Radius

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Electric field is given by

E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C

The charge is -1.7908787542\times 10^{-9}\ C

Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C

The charge is 3.4260289211\times 10^{-9}\ C

The charge inside will have the polarity changed

q=+1.7908787542\times 10^{-9}\ C

Outside the charge will be

3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C

3 0
3 years ago
Other questions:
  • Mary is 52kgs and standing on a ladder. What forces are acting on her?
    14·1 answer
  • Which of the following changes occurs naturally in a system over time?
    14·1 answer
  • Rita measured distances that a small boat traveled over 30 s. She wants to create a distance-time graph to show her data. Which
    6·1 answer
  • the idea that god created the first governments and selected those of royal birth to rule is referred to as_____.
    6·2 answers
  • A concrete pump of 120-cy/hr max. output is used to place 250 cy of concrete. It is supplied by 8 cy transit-mix trucks arriving
    15·1 answer
  • What is the net force of a 10 kg box with a velocity of 2 meters a second?
    14·2 answers
  • In the first figure here, a sled is held on an inclined plane by a cord pulling directly up the plane. The sled is to be on the
    14·1 answer
  • The si unit of average speed is m.s. True or False. anyone who is right I will give him brainlist if you are sure​
    6·1 answer
  • A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 65.5-gram mass is attached at the 21.0-cm
    13·1 answer
  • A circular loop of radius 13 cm carries a current of 13 A. A flat coil of radius 0.94 cm, having 58 turns and a current of 1.9 A
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!