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3 years ago

Which type of mirror would you use to cook a hot dog and why?

Physics

1 answer:

alexandr1967 [171]3 years ago

8 0

As long as it’s a good mirror then any one of them is fine bc at the end of the day i’m getting a hot dog

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Very easy physics maths about current and other stuffs only one question please help me

Reil [10]

Answer:

The total voltage of the circuit is 18 volts.

Explanation:

We have, three identical resistors of resistance 3 ohms are connected in series in a circuit.

For a series combination, the equivalent resistance is given by the sum of individual resistances i.e.

$R_{eq}=R_1+R_2+R_3\\\\R_{eq}=3+3+3\\\\R_{eq}=9\ \Omega$

Let V is the voltage of the battery if the current in the circuit is 2 A. So,

$V=IR$

$V=2\times 9\\\\V=18\ V$

So, the total voltage of the circuit is 18 volts.

7 0

3 years ago

What would be a good reason to increase friction between surfaces? to allow objects to slide past each other easily to reduce we

Lelu [443]

So basically the objects would be sandpaper and smooth metal, the sandpaper can indirectly touch the metal since it’s so smooth and it won’t cause any temp change either

5 0

3 years ago

Read 2 more answers

41. Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 h

romanna [79]

Answer:

138.3 days

Explanation:

Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.

The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.

T^2 is proportional to r^3

That is,

T^2/r^3 = constant

98^2 / 62^3 = T^2 / 78^3

Make T^2 the subject of formula.

T^2 = 98^2 / 62^3 × 78^3

T^2 = 19123.2

T = sqrt ( 19123.2 )

T = 138.2867 days

Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.

4 0

3 years ago

(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$