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Llana [10]
3 years ago
9

Don’t know how to do this can someone help

Physics
1 answer:
Nataly [62]3 years ago
8 0

Answer:

I cant see the paper very well

Explanation:

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If the moon disapared what affect would this have on earths tides
Free_Kalibri [48]
If the moon disapared what affect would this have on earths tides:
There would no longer be any tides.
The moon is what causes the push and the pull of waves.        
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3 years ago
The velocity of a motor car moving along a road increases from 10m/s to 50{1} ms in 8s. Find its avarege acceleration. RESULT
lesantik [10]

Answer:

hope it helps you................

7 0
4 years ago
Pls help i begg youuuuu
Anna [14]
Pitch is the sensation of certain frequencies to the ear. High frequency = high pitch, low frequency = low pitch. 

f = c(speed of the wave) /  <span>λ (wavelength)

1. 343m/s / 0.77955m = 439.99 Hz   
     This corresponds to pitch A 

2. 343m/s / 0.52028m = 659.26 Hz
</span>     This corresponds to pitch E 
<span>
3. 343m/s / 0.65552m = 523.349 Hz
    </span>This corresponds to pitch C

4. using f = c /  λ
  λ = c / f<span>
     = 343m/s / 587.33 = 0.583999 m = 0.584 m

</span>
3 0
3 years ago
When electrons are accelerated by 2450v in an electron microscope they will have wavelengths of
Sholpan [36]
I think the answer is A I’m
Not sure tho
5 0
2 years ago
One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i
kati45 [8]

Answer:

The force exerted on an electron is 7.2\times10^{-18}\ N

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance  = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}

E_{1}=1.0183\times10^{3}\ N/C

Using formula of electric field again

E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}

E_{2}=-1.064\times10^{3}\ N/C

We need to calculate the resultant electric field

Using formula of electric field

E=E_{1}+E_{2}

Put the value into the formula

E=1.0183\times10^{3}-1.064\times10^{3}

E=-0.045\times10^{3}\ N/C

We need to calculate the force exerted on an electron

Using formula of electric field

E = \dfrac{F}{q}

F=E\times q

Put the value into the formula

F=-0.045\times10^{3}\times(-1.6\times10^{-19})

F=7.2\times10^{-18}\ N

Hence, The force exerted on an electron is 7.2\times10^{-18}\ N

8 0
3 years ago
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