Label the longitudinal wave

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Answer 1

Answer:

Answer:

???

Explanation:

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How long would it take a 500. W electric motor to do 15010 J of work?
What is it called when a Rock forms due to heat and pressure in the earth?
A parallel-plate vacuum capacitor has 7.72 J of energy stored in it. The separation between the plates is 3.30 mm. If the separation is decreased to 1.45 mm, For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Stored energy. Part A what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed

If the velocity of a pitched ball has a magnitude of 41.0 m/s and the batted ball's velocity is 50.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

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The change in momentum is (91 m/s) multiplied by the mass of the ball (which you neglected to mention).
That's exactly the impulse delivered by the bat.

(a) A woman climbing the Washington Monument metabolizes 6.00×102kJ of food energy. If her efficiency is 18.0%, how much heat transfer occurs to the environment to keep her temperature constant? (b) Discuss the amount of heat transfer found in (a). Is it consistent with the fact that you quickly warm up when exercising?

Answers

Answer:

492 kJ

Consistent

Explanation:

Q = Heat stored by woman from food = 600 k J

η = Efficiency of woman = 18% = 0.18

Q' = heat transferred to the environment

heat transferred to the environment is given as

Q' = (1 - η) Q

Inserting the values

Q' = (1 - 0.18) (600)

Q' = 492 kJ

Yes the amount of heat transfer is consistent. The process of sweating produces the heat and keeps the body warm

Final answer:

A woman climbing the Washington Monument metabolizes food energy with 18% efficiency, meaning 82% of the energy is lost as heat. When we calculate this value, we find that 492 kJ of energy is released as heat, which is consistent with the fact that people quickly warm up when exercising.

Explanation:

The woman climbing the Washington Monument metabolizes 6.00×10² kJ of food energy with an efficiency of 18%. This implies that only 18% of the energy consumed is used for performing work, while the remaining (82%) is lost as heat to the environment.

To calculate the energy lost as heat:

Determine the total energy metabolized, which is 6.00 × 10² kJ.
Multiply this total energy by the percentage of energy lost as heat (100% - efficiency), which gives: (6.00 × 10² kJ) * (100% - 18%) = 492 kJ.

The released heat of 492 kJ is consistent with the fact that a person quickly warms up when exercising, because a significant portion of the body's metabolic energy is lost as heat due to inefficiencies in converting energy from food into work.

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A proton and an alpha particle (helium nucleus consisting of two protons and two neutrons) are accelerated from rest across the same potential difference. Assume the proton mass and the neutron mass are roughly the same and neglect any relativistic effect. Compared to the final speed of the proton, the final speed of the alpha particle is?1. less by a factor of 22. less by a factor of √ 23. less by a factor of 44. greater by a factor of 25. the same

Answers

Answer:

option B

Explanation:

we know,

change in energy is equal to

$W = (1)/(2)m(v^2 - u^2)$

$W = (1)/(2)m(v^2 - 0^2)$

$W = (1)/(2)m v^2$

$q = (1)/(2)m v^2$

proton mass and the neutron mass are roughly the same

so,

$q \alpha m v^2$

now,

$(q_p)/(q_(\alpha)) = (m_p v_p^2)/(m_(\alpha)v_(\alpha)^2)$

$(q_p)/(q_(\alpha)) = (m_p v_p^2)/(2 m_pv_(\alpha)^2)$

we know,

mass of alpha particle is four times mass of the mass of proton.

mα = 4 m_p

$(e)/(2e) = ( v_p^2)/(4 v_(\alpha)^2)$

$( v_p^2)/(v_(\alpha)^2) = 2$

$v_(\alpha)^2 =( v_p^2)/(2)$

$v_(\alpha)=( v_p)/(√(2))$

less by a factor of √2

Hence, the correct answer is option B

Electrons with energy of 25 eV have a wavelength of ~0.25 nm. If we send these electrons through the same two slits (d = 0.16 mm) we use to produce a visible light interference pattern what is the spacing (in micrometer) between maxima on a screen 3.3 m away?

Answers

Answer:

The spacing is 5.15 μm.

Explanation:

Given that,

Electron with energy = 25 eV

Wave length = 0.25 nm

Separation d= 0.16 mm

Distance D=3.3 m

We need to calculate the spacing

Using formula of width

$\beta=(\lambda D)/(d)$

Put the value into the formula

$\beta=(0.25*10^(-9)*3.3)/(0.16*10^(-3))$

$\beta=5.15*10^(-6)\ m$

$\beta=5.15\ \mu m$

Hence, The spacing is 5.15 μm.

Final answer:

To calculate the spacing between maxima in a double slit interference pattern, we use the formula x = L * λ / d. Converting the given units to meters and plugging the values into the formula, we find that the spacing between maxima on the screen is approximately 5.14 micro meters.

Explanation:

To calculate the spacing between maxima, we can utilize the formula for double slit interference, θ = λ/d where λ represents the wavelength of the electron, d is the distance between the two slits, and θ is the angle of diffraction. Considering the small angle approximation for tan θ ≈ θ, we get x = L * λ / d, where x is the distance between maxima on the screen, and L is the distance from the slits to the screen.

Firstly, the electron's wavelength needs to be converted from nm to m, resulting in λ = 0.25 * 10^-9 m. Similarly, the slit separation d should be converted from mm to m, giving d = 0.16 * 10^-3 m. Inserting these values into the formula along with L = 3.3 m, we can solve for x.

x = (3.3 m * 0.25 * 10^-9 m) / 0.16 * 10^-3 m =~ 5.14 μm

So, the spacing between maxima on the screen is approximately 5.14 micrometers.

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Two charged metallic spheres of radii, R = 10 cms and R2 = 20 cms are touching each other. If the charge on each sphere is +100 nC, what is the electric potential energy between the two charged spheres?

Answers

Answer:

$200* 10^(-6)j$

Explanation:

We have given the radius of first sphere is 10 cm and radius of second sphere is 20 cm

So the potential of first sphere will be greater than the potential of the second sphere, so charge will flow from first sphere to second sphere

Let q charge is flow from first sphere to second sphere and then potential become same

So $V=(K(100-q))/(r_1)=(K* 100)/(r_2)$

200-100=2q+q

$q=(100)/(3)=33.33nC$

So $V=(K(100-q))/(r_1)=(9* 10^(9)* (100-33.33)* 10^(-9))/(10* 10^(-2))=6003V$

We know that potential energy U=qV $=33.33* 10^(-9)* 6003=200* 10^(-6)j$

Answer:

The electric potential energy between the two charged spheres is $199.9*10^(-6)\ J$

Explanation:

Given that,

Radius of first sphere $R_(1)=10\ cm$

Radius of second sphere $R_(2)=10\ cm$

Charge Q= 100 nC

We know charge flows through higher potential to lower potential.

Using formula of potential

$V=(k(Q-q))/(R_(1))$ ...(I)

$V=(k(Q+q))/(R_(2))$ ...(II)

From equation (I) and (II)

$(k(Q-q))/(R_(1))=(k(Q+q))/(R_(2))$

Put the value into the formula

$((100-q))/(10*10^(2))=((100+q))/(20*10^(-2))$

$(100-q)*20*10^(-2)=(100+q)*10*10^(-2)$

$q=(1000)/(30)$

$q=(100)/(3)$

$q=33.33\ nC$

So, the potential at R₁ and R₂

Using formula of potential

$V=(k(Q-q))/(R_(1))$

Put the value into the formula

$V=(9*10^(9)(100-33.33)*10^(-9))/(10*10^(-2))$

$V=6000.3\ Volt$

We need to calculate the electric potential energy between the two charged spheres

Using formula of the electric potential energy

$U=qV$

$U=33.33*10^(-9)*6.0003*10^(3)$

$U=199.9*10^(-6)\ J$

Hence, The electric potential energy between the two charged spheres is $199.9*10^(-6)\ J$

You are an electrician installing the wiring in a new home. The homeowner desires that a ceiling fan with light kits be installed in five different rooms. Each fan contains a light kit that can accommodate four 60-watt lamps. Each fan motor draws a current of 1.8 amperes when operated on high speed. It is assumed that each fan can operate more than three hours at a time and therefore must be considered a continuous-duty device. The fans are to be connected to a 15-ampere circuit. Because the devices are continuous-duty, the circuit current must be limited to 80% of the continuous connected load. How many fans can be connected to a single 15-ampere circuit

Answers

Answer:

3 fans per 15 A circuit

Explanation:

From the question and the data given, the light load let fan would have been

(60 * 4)/120 = 240/120 = 2 A.

Next, we add the current of the fan motor to it, so,

2 A + 1.8 A = 3.8 A.

Since the devices are continuos duty and the circuit current must be limited to 80%, then the Breaker load max would be

0.8 * 15 A = 12 A.

Now, we can get the number if fans, which will be

12 A/ 3.8 A = 3.16 fans, or approximately, 3 fans per 15 A circuit.

Final answer:

The total power draw of each fan is 3.8 amperes. Thus, considering a limit of 80% usage of 15 amperes, only 3 fans can be connected to a single circuit to keep the total power draw below 12 amperes.

Explanation:

The question is asking how many ceiling fans, each with a certain power draw, can be connected on a single 15-ampere circuit, considering that each fan is a continuous-duty device. The power draw of each fan when the motor is operated at high speed and the light kit is fully loaded is the sum of the power draw of the motor and the light kit. As the power draw of each motor is 1.8 amperes and the light kit is 240 watts or 2 amperes (calculated using the formula Power = Voltage x Current; assuming a voltage of 120 volts), the total power draw of each fan is 3.8 amperes. Considering the limit of 80% of the continuous load, only 12 amperes (80% of 15) can be used. Thus, 3 fans can be connected to the circuit as it reaches 11.4 amperes, close enough to the 12 amperes limit.

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