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3 years ago

TRUE or FALSE: Feeling "weightless" can be the result of accelerating downwards towards Earth.

1 answer:

4 0

I believe it would be false because in outer space, everything is weightless due to the weak gravitational force which makes things float in the air. Weightless means no weight to it and if we’re accelerating down that means the gravitational is working but in this case it would probably be false

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A 5.00-V battery charges the parallel plates in a capacitor, with a plate area of 865 mm2 and an air-filled separation of 3.00 m

Westkost [7]

Answer:

W = 3.21x10⁻¹¹ J

Explanation:

The work required to separate the plates can be calculated using the following equation:

$W = U_{2} - U_{1} = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2})$

Where:

U₂: is the final stored energy

U₁: is the initial stored energy

C₂: is the final capacitance

C₁: is the initial capacitance

V₁: is the initial potential difference = 5.00 V

V₂: is the final potential difference

The initial and final capacitance is:

$C_{1} = \epsilon_{0}*\frac{A}{d_{1}}$

Where:

ε₀: is the vacuum permittivity = 8.85x10⁻¹² C²/(N*m²)

d: is the initial distance = 3.00 mm = 3.00x10⁻³ m

A: is the plate area = 865 mm² = 8.65x10⁻⁴ m²

$C_{1} = \epsilon_{0}*\frac{A}{d_{1}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 \cdot 10^{-3} m} = 2.55 \cdot 10^{-12} F$

Similarly, C₂ is:

$C_{2} = \epsilon_{0}*\frac{A}{d_{2}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 + 3.00 \cdot 10^{-3} m} = 1.28 \cdot 10^{-12} F$

Now, V₂ can be calculated by finding the initial charge (q₁):

$q_{1} = C_{1}V_{1} = 2.55 \cdot 10^{-12} F*5.00 V = 1.28 \cdot 10^{-11} C$

Since, q₁ is equal to q₂, V₂ is:

$V_{2} = \frac{q_{2}}{C_{2}} = \frac{1.28 \cdot 10^{-11} C}{1.28 \cdot 10^{-12} F} = 10 V$

Finally, we can find the work:

$W = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2}) = \frac{1}{2}(1.28 \cdot 10^{-12} F*(10 V)^{2} - 2.55 \cdot 10^{-12} F(5.00 V)^{2}) = 3.21 \cdot 10^{-11} J$

Therefore, the work required to separate the plates is 3.21x10⁻¹¹ J.

I hope it helps you!

4 0

3 years ago

Betty is six years old and often asks her parents many questions. Her parents respond to her questions however trivial they may

Brums [2.3K]

This behavior helps Betty in intellectual development.

5 0

2 years ago

Read 2 more answers

The vibrations along a transverse wave move in a direction _________.

GrogVix [38]

Answer: perpendicular to it oscillations.

Explanation: A transverse wave is a wave whose oscillations is perpendicular to the direction of the wave.

By perpendicular, we mean that the wave is oscillating on the vertical axis (y) of a Cartesian plane and the vibration is along the horizontal axis (x) of the plane.

Examples of transverse waves includes wave in a string, water wave and light.

Let us take a wave in a string for example, you tie one end of a string to a fixed point and the other end is free with you holding it.

If you move the rope vertically ( that's up and down) you will notice a kind of wave traveling away from you ( horizontally) to the fixed point.

Since the oscillations is perpendicular to the direction of wave, it is a transverse wave

5 0

2 years ago

I. What is the initial velocity of the car?

qaws [65]

Answer:

I. 0 m/s

II. 20 m/s

III. Part BC

Explanation:

I. Determination of the initial velocity.

From the diagram given above,

The motion of the car starts from the origin. This implies that the car start from rest and as such, the initial velocity of the car is 0 m/s

II. Determination of the maximum velocity attained.

From the diagram given above, we can see clearly that the maximum velocity is 20 m/s.

III. Determination of the part of the graph that represents zero acceleration.

It important that we know the meaning of zero acceleration.

Zero acceleration simply means the car is not accelerating. This can only be true when the car is moving with a constant velocity.

From the graph given above, the car has a constant velocity between B and C.

Therefore, part BC illustrates zero acceleration.

6 0

3 years ago

The photeselestric effect is observed when light of a sufficiently high frequency is focused onto a polished metal surface, emit

Helga [31]

Answer:

$3.4\cdot 10^{-19} J$

Explanation:

In order to convert the work function of cesium from electronvolts to Joules, we must use the following conversion factor:

$1 eV = 1.6 \cdot 10^{-19} J$

In our problem, the work function of cesium is

$E=2.1 eV$

so, we can convert it into Joules by using the following proportion:

$1 eV : 1.6\cdot 10^{-19} J = 2.1 eV : x\\x=\frac{(1.6\cdot 10^{-19} J)(2.1 eV)}{1 eV}=3.4\cdot 10^{-19} J$

8 0

3 years ago