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Butoxors [25]
3 years ago
8

Paraffin wax is used to make candlestrue or false​

Physics
1 answer:
tangare [24]3 years ago
7 0

Answer:

true

Explanation:

PLS MAKE ME AS BRAINLIST

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What is a plasma?A.A superheated liquidB.A solid in which the particles vibrate very quicklyC.A gas in which the atoms have star
Mandarinka [93]

plasma is a superheated liquid

So, a would-be the correct option.

5 0
1 year ago
0/2 File Limit
slamgirl [31]

Answer:

Speed at which it will reach the ground is given as

v_f = 46.8 m/s

Total time for which it will remain in air is given as

t = 6.3 s

Explanation:

As we know that the object is projected upwards with speed

v_i = 15 m/s

g = - 9.81 m/s^2

now when it will reach the ground then we have

y = v_y t + \frac{1}{2} at^2

so we have

-100 = 15 t - \frac{1}{2}(-9.81) t^2

4.905 t^2 - 15 t - 100 = 0

so we have

t = 6.3 s

Now speed of the object when it reaches the ground is given as

v_f = v_i + at

v_f = -15 + (9.81)(6.3)

v_f = 46.8 m/s

8 0
3 years ago
Compute the expected shell-model quadrupole moment of 209Bi () and compare with the experimental value, - 0.37 b
Over [174]

Answer:

0.22 b

Explanation:

Quadrupole moment of the nucleon is,

Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2}

And also,

R^{2}=R^{2} _{0}A^{\frac{2}{3} }

And, R _{0}=1.2\times 10^{-15}m

Now,

Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }

For Bismuth j=\frac{9}{2} and A is 209.

Q=-\frac{2\frac{9}{2} -1}{2(\frac{9}{2} +1)}\frac{3}{5}(1.2\times 10^{-15}) ^{2}(209)^{\frac{2}{3} }\\Q=0.628\times 35.28\times 10^{-30} \\Q=22.15\times 10^{-30} m^{2} \\Q=0.2215\times 10^{-28} m^{2} \\Q=0.22 barn

Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b

3 0
3 years ago
A bike travels at a constant speed for 4.00 m/s for 5.00 seconds. How far does it go
kicyunya [14]

Answer:

20 metres

Explanation:

<em>Speed</em><em> </em><em>=</em><em> </em><em>distance</em><em> </em><em>÷</em><em> </em><em>time</em>

<em> </em>

<em>s =  \frac{d}{t}</em>

If we substitute the values:

4 =  \frac{d}{5}

<em>20 = d</em>

3 0
3 years ago
Read 2 more answers
Two identical speakers are emitting a constant tone that has a wavelength of 0.50 m. Speaker A is located to the left of speaker
Marianna [84]

Answer:

a) and c).

Explanation:

For a complete destructive interference occur, it must be met the following condition relating the wavelength, and the difference in the paths taken by the sound emitted by the sources until arriving to the listening point:

d = |dA- dB| = (2n-1)*(λ/2)

For n= 1,  d = λ/2 = 0.25 m, it doesn't meet any of the cases.

For n=2, d= 3*(λ/2) = 0.75 m

In the case a) we have dA = 2.15 m and dB = 3.00 m, so dB-dA = 0.75 m, which means that in the location stated by case a) a complete destructive interference would occur.

For n=3, d= 5*(λ/2) = 5*0.25 m = 1.25 m.

This is just the case c) because we have dA = 3.75 m and dB = 2.50 m, so dA-dB = 1.25 m, which means that in the location stated by case c) a complete destructive interference would occur also.

The remaining cases don't meet the condition stated above, so the statements found to be true are a) and c),

5 0
3 years ago
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