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tino4ka555 [31]
3 years ago
15

If one branch of a 120-v power lines is protected by a 20-A fuse, will the fuse carry an 8-Ώ load

Physics
1 answer:
Mrrafil [7]3 years ago
7 0

Answer:

No I won't.

It will carry 6ohm load.

Explanation:

It obeys ohms law therefore V=IR

120=20R

R=120/20

R= 6 ohms

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Explain why liquids solidify when they are cooled.
Setler [38]
Because the molecules that move freely begin to compact closer together, with less heat, means less molecular activity. 
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3 years ago
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An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below
grin007 [14]
<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let P be the pressure at a point.

Let p be the density fluid at a point.

Let v be the velocity of fluid at a point.

Bernoulli's equation states that P+\frac{1}{2}pv^{2}+pgh=constant for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let p_{up} be the pressure of a point just above the wing.

Let p_{do} be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}

So,p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa

Force is given by the product of pressure difference and area.

Given that area is 27ms^{2}.

So,lifting force is 16398.48\times 27=442758.96N

6 0
3 years ago
How might the climate be affected if you were people drove automobiles and more people road bikes or use public transportation l
Hoochie [10]
The level of greenhouse gases in our atmosphere would decrease, due to less automobiles.
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3 years ago
The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard
gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

|\Delta E_{ij}| = |E_i-E_j|

Our states are given by

E_1 = -11.7eV

E_2 = -4.2eV

E_3 = -3.3eV

In this way the energy balance for the states would be given by,

|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\

|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV

|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

8 0
3 years ago
A stretched string has a mass per unit length of 4.87 g/cm and a tension of 16.7 N. A sinusoidal wave on this string has an ampl
Burka [1]

Answer:

Explanation:

Given that,

Mass per unit length is

μ = 4.87g/cm

μ=4.87g/cm × 1kg/1000g × 100cm/m

μ = 0.487kg/m

Tension

τ = 16.7N

Amplitude

A = 0.101mm

Frequency

f = 71 Hz

The wave is traveling in the negative direction

Given the wave form

y(x,t) = ym• Sin(kx + ωt)

A. Find ym?

ym is the amplitude of the waveform and it is given as

ym = A = 0.101mm

ym = 0.101mm

B. Find k?

k is the wavenumber and it can be determined using

k = 2π / λ

Then, we need to calculate the wavelength λ using

V = fλ

Then, λ = V/f

We have the frequency but we don't have the velocity, then we need to calculate the velocity using

v = √(τ/μ)

v = √(16.7/0.487)

v = 34.29

v = 5.86 m/s

Then, we can know the wavelength

λ = V/f = 5.86 / 71

λ = 0.0825 m

So, we can know the wavenumber

k = 2π/λ

k = 2π / 0.0825

k = 76.18 rad/m

C. Find ω?

This is the angular frequency and it can be determined using

ω = 2πf

ω = 2π × 71

ω = +446.11 rad/s

D. The angular frequency is positive (+) because the direction of propagation of wave is in the negative direction of x

5 0
3 years ago
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