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Setler79 [48]
2 years ago
9

What is the speed of light in a vacuum?

Physics
2 answers:
Novosadov [1.4K]2 years ago
7 0

Answer:

300,000 km/sec

Explanation:  The first one.

vivado [14]2 years ago
6 0

Answer:

The first one is the right answer.

You might be interested in
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
3 years ago
Convert to the fractional equivalent and reduce 21.12
nignag [31]

The decimal point is placed after two digits starting from the end. For each decimal place, we can write the number divided by 100.

21.12 can be written as \frac{2112}{100}.

Divide the numerator and denominator by 2:

\frac{2112}{100}= \frac{156}{50}

The numerator and denominator can be divided by 2 again:

\frac{78}{25}

There is no other common factor between numerator and denominator other than 1. Hence, it is the reduced form.




3 0
3 years ago
Read 2 more answers
A very long conducting cylinder (length L) of radius R(R&lt;R and (b)r
mina [271]

Answer:

Explanation:

We have to find electric potential V at a distance r.

a) For r>R,

The electric field in the cylinder is given by

E.A equating it to the other electric field given by

б.A/ε₀

Here the area of cylinder is given by= 2*3.14*r*L

While for the outside, the area= 2*3.14*R*L

Equating both, we get

E= бR/rε₀

Now,

The potential difference is given as:

ΔV= -бR/rε₀ and integrating right side with respect to dr under limits r and R.

Where ΔV= V₀-V

So solving we get

V₀=V-бR/ε₀ln (r/R)

b) For r<R i.e. inside the cylinder

There will be no electric field produced as E=0

So ultimately Vin= V

c) V=0 at r= infinity.

4 0
3 years ago
Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The
Galina-37 [17]

Answer:

220 A

Explanation:

The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.

So, F = BI₁L

F = (μ₀I₂/2πd)I₁L

F = μ₀I₁I₂L/2πd

Given that the current in the rods are the same, I₁ = I₂ = I

So,

F = μ₀I²L/2πd

Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²

So, F = W

μ₀I²L/2πd = mg

making I subject of the formula, we have

I² = 2πdmg/μ₀L

I = √(2πdmg/μ₀L)

substituting the values of the variables into the equation, we have

I = √(2π × 0.011 m × 0.1 kg × 9.8 m/s²/[4π × 10⁻⁷ H/m × 1.1 m])

I = √(0.01078 kgm²/s²/[2 × 10⁻⁷ H/m × 1.1 m])

I = √(0.01078 kgm²/s²/[2.2 × 10⁻⁷ H])

I = √(0.0049 × 10⁷kgm²/s²H)

I = √(0.049 × 10⁶kgm²/s²H)

I = 0.22 × 10³ A

I = 220 A

7 0
3 years ago
A race car has a centripetal acceleration of 13.33 m/s^2 as it goes around a curve. if the curve is a circle with a radius 30 m
anzhelika [568]

Answer:

The speed of the car, v = 19.997 m/s

Explanation:

Given,

The centripetal acceleration of the car, a = 13.33 m/s²

The radius of the curve, r = 30 m

The centripetal force acting on the car is given by the formula

                                   F = mv²/r

Where    v²/r is the acceleration component of the force

                                       a = v²/r

Substituting the values in the above equation

                                        13.33 = v²/30

                                         v² = 13.33 x 30

                                         v² = 399.9

                                         v = 19.997 m/s

Hence, the speed of the car, v = 19.997 m/s

3 0
3 years ago
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