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Artyom0805 [142]
3 years ago
10

Power lines often carry high voltages. For example, a single power line may carry 100,000 volts. An outlet in a home only carrie

s 120-240 volts. What accounts for this difference in voltage?
- Voltage is decreased due to traveling long distances.
- Voltage is decreased by local transformers.
- Voltage is decreased as it is lost to the environment.
Physics
2 answers:
almond37 [142]3 years ago
8 0

Answer: The second one? C.

adoni [48]3 years ago
7 0

Answer:

B. - Voltage is decreased by local transformers.

Explanation:

edge 2021  beep boop

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A dog runs down his driveway with an initial speed of 5 m/s for 8 s, and then uniformly increases his speed to 10 m/s in 5 s.
MakcuM [25]
Acceleration is given by change in velocity divided by change in time, so his acceleration should just be (10-5)/5 which is [tex] \frac{5}{2} \frac{m}{s^{2}} [tex]

The driveway is 40 meters plus 225/4. You can do the math.
8 0
4 years ago
A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th
Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

T=\frac{1}{f}

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

T=\frac{1}{0.42 Hz}=2.38 s

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: x=0, so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

E=K=\frac{1}{2}mv^2

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

E=U=\frac{1}{2}kA^2

Since the total energy must be conserved, we have:

\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}kA^2 is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}mv_{max}^2 is the mechanical energy of the system when x=0, which is when the system has maximum velocity, v_{max}

Equalizing the two expressions, we can solve to find v_{max}, the maximum velocity:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m

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3 years ago
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What is potential difference​
lisabon 2012 [21]

Answer:

VOLTAGE

Explanation:

Potential difference is the difference in the amount of energy that charge carriers have between two points in a circuit. ... A potential difference of one Volt is equal to one Joule of energy being used by one Coulomb of charge when it flows between two points in a circuit

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Calculate the force of gravitation between two objects of masses 50 kg and 120 kg respectively, kept at a distance of 10 m from
NNADVOKAT [17]

Answer:

Answer

F=G×

d

2

m×M

m = 50 kg

M = 120 kg

Distance, d = 10 m

G=6.7×10

−11

Nm

2

/kg

2

F=6.67×10

−11

×

10

2

50×120

F=6.67×60×10

−11

F=4.02×10

9

N

6 0
3 years ago
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____ is the thermal energy transfer that occurs from the motion of fluid in a liquid or gas.
Marysya12 [62]
The answer is C. Convection.

Radiation is the transfer of energy through electromagnetic waves. Conduction, on the other hand, is energy transfer through solid material. Condensation occurs when warm gas makes contact with a cold surface or environment, which causes a change in state from gas to liquid. These definitions rule out choices A, B, and D. 

Convection is movement of particles in fluid or gas. As liquid/gas heats up, it becomes less dense and rises, moving away from the heat source. When it rises, it eventually cools down and sinks again. It is warmed up again and the cycle continues. This is why there is a circular motion when convection occurs. 
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