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
maw [93]
3 years ago
9

Please Help Me!!!!!!!​

Physics
1 answer:
Elden [556K]3 years ago
5 0

Answer:

B

Explanation:

You might be interested in
A ball having a mass of 0.20 kilograms is placed at a height of 3.25 meters. If it is dropped from this height, what will be the
asambeis [7]
EC_1 + EP_1 = EC2 + EP_2

EC_2 = 0

EC_2 = EP_1 - EP_2

EC_2 = mg(H_1 - H_2) = 0.20 kg * 9.8 m/s^2 * (3.25 m - 1.5m) = 3.43 J
7 0
3 years ago
Read 2 more answers
As demonstrated by the Doppler Effect, why does sound increase in pitch as a sound source approaches you?
scZoUnD [109]
I think so it ans wold be d.because if the source approaches the observer more and more wavefront will pass and it get squeezed.so wavelength decreases and frequency increases.
3 0
2 years ago
Read 2 more answers
An automobile starter motor has an equivalent resistance of 0.0510 Ω and is supplied by a 12.0 V battery with a 0.0090 Ω interna
Alisiya [41]

Answer:

a) 200A

b) 10.2V

c) 2.04kW

d)

I=80A

V=4.08V

P=0.326kW

Explanation:

Here we have a circuit of one power source and two resistors in series, the first question is asking for the current, so according to Ohm's Law:

I=\frac{V}{R}

Where R is the equivalent resistance of the resistors in series

R=0.0510+0.0090=0.0600[ohm]

I=\frac{12.0}{0.0600}=200A

To calculate the voltage dropped by the motor we have to apply the voltage divider rule:

V_m=V*\frac{R_m}{R_m+R_s}\\V_m=12.0*\frac{0.0510}{0.0600}\\V_m=10.2V

The power dissipated supplied to the motor is given by:

P=I^2*R_m\\P=(200)^2*0.0510=2.04kW

now solving adding a 0.0900 ohm resistor:

I=\frac{12.0}{0.15}=80A

V_m=12.0*\frac{0.0510}{0.15}\\V_m=4.08V

P=I^2*R_m\\P=(200)^2*0.0510=0.326kW

5 0
3 years ago
The amount of energy that must be absorbed or lost to raise or lower the temperature of 1 g of liquid water by 1°c _____.
Fiesta28 [93]

Answer:

4.2 J

Explanation:

Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K

From specific heat capacity,

Q = cmΔt.............................. Equation 1

Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.

Given: m = 1 g = 0.001 kg, Δt = 1 °C

Constant : c = 4200 J/kg.°C

Substitute into equation 1

Q = 0.001×4200(1)

Q = 4.2 J.

Hence the energy absorbed or lost = 4.2 J

6 0
3 years ago
A wire with a resistance of 20 is connected to a 12 battery. What is the current flowing through the wire?
Dima020 [189]
15.49 should be the answer if that is 12 watt battery.
3 0
3 years ago
Other questions:
  • You have a 160-Ω resistor and a 0.430-H inductor. Suppose you take the resistor and inductor and make a series circuit with a vo
    10·1 answer
  • I need help with finding these?
    15·1 answer
  • How do you get derived units from derived quantities
    14·1 answer
  • A stream of warm water is produced in a steady-flow mixing process by combining 1.0 kg/s of cool water at 25 °C with 0.8 kg/s of
    5·1 answer
  • A typical land snail's speed is 12.2 meters per hour. How many miles will the snail travel in one day(24hrs)?
    9·1 answer
  • On that system is<br> In an isolated system, the net __<br> zero.
    5·1 answer
  • Un resorte se alarga 5 cm bajo la acción de una fuerza de 39,2 N. ¿Cuál es la constante del resorte? Si ahora la fuerza es 68,6
    10·1 answer
  • The phase velocity of transverse waves in a crystal of atomic separation a is given byy = csin(ka/2) pka/2 1. What is the disper
    14·1 answer
  • When a mule stops suddenly, the packages
    9·1 answer
  • Why are mud shoes better then high heels
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!