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2 years ago

What objects do balanced forces act on?

Physics

1 answer:

leva [86]2 years ago

4 0

Answer: Stationary or constant velocity

Explanation:

Objects with balanced forces acting on them experience no change in motion, or no acceleration. So these objects could either be stationary at rest or have a constant velocity. These include a hanging object, a floating object, an object on a table that doesn't move, and a car moving at a constant 10 mph

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The type of energy that depends on position is called

attashe74 [19]

The type of energy that depends on position is called
kinetic energy

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3 years ago

Most metals are good electrical conductors because (5 points) they hold their electrons they have a large amount of free electro

Hunter-Best [27]

Answer:

Metals are good conductors of electricity <u>because they have free electrons</u> which help in the transfer of charge from one point to another.

6 0

3 years ago

Complete the passage.

nataly862011 [7]

Answer:

Answer: Sound waves and some earthquake waves are longitudinal waves. Ocean, light and other earthquake waves are transverse waves.

Explanation:

There are 2 types of waves:

1. Longitudinal waves: These waves are defined as the waves in which the particles of the medium move in the direction of the wave. This requires a medium to travel. For Example: Sound Waves.

2. Transverse wave: These waves are defined as the waves in which the particles of the medium travel perpendicularly to the direction of the wave. This does not require a medium to travel. These can travel in vacuum also. For Example: Light waves.

Hence, Sound waves and some earthquake waves are longitudinal waves. Ocean, light and other earthquake waves are transverse waves

8 0

2 years ago

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 2.00 mm. If a

ValentinkaMS [17]

Answer:

a. 11.5kv/m

b.102nC/m^2

c.3.363pF

d. 77.3pC

Explanation:

Data given

$area=7.60cm^{2}\\ distance,d=2mm\\voltage,v=23v$

to calculate the electric field, we use the equation below

V=Ed

where v=voltage, d= distance and E=electric field.

Hence we have

$E=v/d\\E=\frac{23}{2*10^{-3}} \\E=11.5*10^{3} v/m\\E=11.5Kv/m$

b.the expression for the charge density is expressed as

σ=ξE

where ξ is the permitivity of air with a value of 8.85*10^-12C^2/N.m^2

If we insert the values we have

$8.85*10^{-12} *11500\\1.02*10^{-7}C/m^{2} \\102nC/m^{2}$

from the expression for the capacitance

$C=eA/d$

if we substitute values we arrive at

$C=\frac{8.85*10^{-12}*7.6*10^{-4}}{2*10^{-3} } \\C=\frac{6.726*10^{-15} }{2*10^{-3} } \\C=3.363*10^{-12}F\\C=3.363pF$

d. To calculate the charge on each plate, we use the formula below

$Q=CV\\Q=23*3.363*10^{-12}\\ Q=7.73*10^{-12}\\ Q=77.3pC$

8 0

3 years ago

two astronauts are taking a spacewalk outside the International Space Station the first astronaut has a mass of 64 kg the second

Fittoniya [83]

Answer:

Approximately $0.88\; {\rm m \cdot s^{-1}}$ to the right (assuming that both astronauts were originally stationary.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Since momentum of this system (of the astronauts) conserved:

$\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\end{aligned}$ .

Assuming that both astronauts were originally stationary. The total initial momentum of the two astronauts would be $0$ since the velocity of both astronauts was $0\!$ .

Therefore:

$\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}$ .

The final momentum of the first astronaut ( $m = 64\; {\rm kg}$ , $v = 0.8\; {\rm m\cdot s^{-1}}$ to the left) would be $p_{1} = m\, v = 64\; {\rm kg} \times 0.8\; {\rm m\cdot s^{-1}} = 51.2\; {\rm kg \cdot m \cdot s^{-1}}$ to the left.

Let $p_{2}$ denote the momentum of the astronaut in question. The total final momentum of the two astronauts, combined, would be $(p_{1} + p_{2})$ .

$\begin{aligned} & p_{1} + p_{2} \\ &= (\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}$ .

Hence, $p_{2} = (-p_{1})$ . In other words, the final momentum of the astronaut in question is the opposite of that of the first astronaut. Since momentum is a vector quantity, the momentum of the two astronauts magnitude ( $51.2\; {\rm kg \cdot m \cdot s^{-1}}$ ) but opposite in direction (to the right versus to the left.)

Rearrange the equation $p = m\, v$ to obtain an expression for velocity in terms of momentum and mass: $v = (p / m)$ .

$\begin{aligned}v &= \frac{p}{m} \\ &= \frac{51.2\; {\rm kg \cdot m \cdot s^{-1}}}{64\; {\rm kg}} && \genfrac{}{}{0}{}{(\text{to the right})}{} \\ &\approx 0.88\; {\rm m\cdot s^{-1}} && (\text{to the right})\end{aligned}$ .

Hence, the velocity of the astronaut in question ( $m = 58.2\; {\rm kg}$ ) would be $0.88\; {\rm m \cdot s^{-1}}$ to the right.

5 0

2 years ago