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3 years ago

At what temperature is Fahrenheit scale reading equal to twice of celsius?

1 answer:

5 0

At -24.613 F<span> temperature is the Fahrenheit scale reading equal to twice that of the Celsius and at </span>160 F<span> temperatuar is the Fahrenheit scale reading equal to half that of the Celsius? So, 160 degrees Celsius will equal 320 degrees Fahrenheit.</span>

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Explain what 'vibrating' means.

drek231 [11]

Answer:

Vibrating means to move quickly to and fro.

Explanation:

6 0

3 years ago

Incident rays parallel to the principle axis of a concave mirror will reflect _____. parallel to the principle axis through the

Svetradugi [14.3K]

I’m not sure if its correct but I think it’s focal Ray point

For concave mirrors, some generalizations can be made to simplify ray construction. They are: An incident ray traveling parallel to the principal axis will reflect and pass through the focal point. An incident ray traveling through the focal point will reflect and travel parallel to the principal axis.

8 0

3 years ago

Read 2 more answers

A point charge with a charge q1 = 4.00 μC is held stationary at the origin. A second point charge with a charge q2 = -4.10 μC mo

GrogVix [38]

Answer:

W = -0.480 J

Explanation:

given,

q₁ = 4 μC

q₂ = -4.10 μC

$W = kq_1q_2(\dfrac{1}{a}+\dfrac{1}{b})$

$b = \sqrt{(0.27-0)^2+(0.27-0)^2}$

b = 0.381

k = 8.99 × 10⁹ Nm²/C²

$W = 8.99\times 10^9\times 4\times 10^{-6}\times (-4.1 \times 10^{-6})(\dfrac{1}{0.17}+\dfrac{1}{0.381})$

$W = [-147.436\times (5.88-2.62)\times 10^{-3}]J$

W = -0.480 J

Work done by the electric force W = -0.480 J

4 0

3 years ago

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$