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kykrilka [37]
2 years ago
5

4. Which of the following substance will cool off the fastest?

Physics
1 answer:
madreJ [45]2 years ago
6 0

Answer:

I'm not sure but I think its aluminum

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What is x-rays ? and how it's formed ?​
tia_tia [17]

X - rays are highly penetrating electromagnetic radiations that are formed when the cathode rays strikes a dense metal.

It has many characteristics such as :

  • bears no charge.

  • travels with the speed of light.

  • can ionize gases.
3 0
2 years ago
A machinist turns the power on to a grinding wheel, at rest at tome t=0 s. the wheel accelerates uniformly for 10 s and reaches
Assoli18 [71]
To convert 2030 rad into rev, divide 2030 by 2pie. So final answer will be 2030/2 pie =323.08 revolutions.

3 0
3 years ago
(1.08×10-3)(9.3×10-4)
ira [324]
First you do the first parenthesis, (1.08 x 10 - 3) and you do it in the order of operations! (parenthesis, exponents, multiplication/division, add/subtract) to get 7.8. Then you take the second parenthesis (9.3 x 10 - 4) and do the same thing to get 89! You then times 7.8 by 89 to get 694.2! If it needs more elaboration just ask ^.^
4 0
3 years ago
A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb.
lana [24]

In this problem we have the electric field intensity E:

E = 6.5 × 10^4 newtons/coulomb

We have the magnitude of the load:

q = 6.4 × 10 ^{-19} coulombs

We also have the distance d that the load moved in a direction parallel to the field 1.2 × 10^{-2} meters.

We know that the electric potential energy (PE) is:

PE = qEd

So:

PE = (6.4 × 10^{-19})(6.5 × 10^4)(1.2 × 10^{-2})

PE = 5.0 x 10^{-16} joules

None of the options shown is correct.

6 0
3 years ago
A capacitor C is fully charged by connecting it to battery of V Volt. Then it is disconnected from battery. If the separation be
vodomira [7]

Answer:

Explanation:

i )

When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2

So charged stored in it will remain unchanged .

ii )

Potential difference = charge / capacitance

in the first case potential difference = Q / C

in the second case potential difference = Q / 2C

So potential difference becomes half .

iii ) electric field = potential diff / plate separation

in the first case electric field = Q / (d x C )

in the second case electric field = 2 Q / (d x 2C)

= Q / (d  x C )

So electric field remains unchanged .

iv)

energy stored in first case = Q² / 2C

In the second case energy stored = Q² / 2x2C

so energy stored becomes half .

4 0
3 years ago
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