The speed of light is 3.00×108m/s. How long does it take for light to travel from Earth to the Moon and back again? Express your answer using two significant figures.

Answer:

U. With no variation.

Explanation:

Note- since temperature remains constant when pressure becomes twice and volume becomes half, and internal energy of ideal gas is function of only temperature so it remains constant. The internal energy is independent of the variables stated in the exercise.

Final answer:

The displacement is 100 m to the east.

Explanation:

The displacement can be calculated using the formula:

Displacement = Velocity × Time

In this case, the velocity is 10 m/s to the east and the time is 10 seconds.

So, Displacement = 10 m/s × 10 s = 100 m to the east.

Learn more about Displacement here:

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Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

1)

• Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
• There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

• Replacing by the givens, and solving for f, we get:

⇒     f = 8.6 Ghz (with two significative figures)

2)

• Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

       where v= c= speed of light in vacuum = 3*10⁸ m/s

       d= distance between mountaintops = 52 km = 52*10³ m

• Solving for t, we get:

       ⇒  t = 0.2 ms (with two significative figures)

Answer:

0.158 A.

Explanation:

Mass of gold deposited = 3 x 10^-3 kg

= 3 g

Molar mass = 196 g/mol

Number of moles = 3/196

= 0.0153 mol.

Faraday's constant,

1 coloumb = 96500 C/mol

Quantity of charge, Q = 96500 * 0.0153

= 1477.04 C.

Remember,

Q = I * t

t = 2.59 hr

= 2.59 * 3600 s

= 9324 s

Current, I = 1477.04/9324

= 0.158 A.

Answer:

0.158A

Explanation:

Using Faraday's first law of electrolysis which states that the mass(m) of a substance deposited or liberated at any electrode is directly proportional to the quantity of charge or electricity (Q) passed. i.e

m ∝ Q

m = Z Q

Where;

Z is the proportionality constant called electrochemical equivalent.

Faraday also observed that when 1 Faraday of electricity is equivalent to 96500C of charge.

Also,

Quantity of charge (Q), which is the product of current (I) passing through and the time taken (t) for the electrolysis, is given by;

Q = I x t;         ----------------------(i)

With all of these in place, now let's go answer the question.

Since the gold ions carry one elementary unit of positive charge, now let's write the cathode-half reaction for gold (Au) as follows;

Au⁺ + e⁻ = Au  ---------------------(ii)

From equation (ii) it can be deduced that when;

1 Faraday (96500C) of electricity is passed, 1 mole of Au forms ( = 197 grams of Au)   [molar mass of Au = 197g]

Then, 3.00 x 10⁻³ kg (= 3 g of Au) will be formed by 3g x 96500C / 197g = 1469.5C

Therefore, the quantity of charge (Q) deposited is 1469.5C

Substitute this value (Q = 1469.5C)  and time t = 2.59h (= 2.59 x 3600 s) into equation (i);

Q = I x t

1469.5  = I x 2.59 x 3600

1469.5  = I x 9324

Solve for I;

I = 1469.5 / 9324

I = 0.158A

Therefore, the current in the cell during that period is 0.158A

Note:

1 mole of gold atoms = 176g

i.e the molar mass of gold (Au) is 176g

(a) It's the force of (static) friction that keeps the car on the road and prevents it from skidding, and this friction is directed toward the center of the curve.

Recall that centripetal acceleration has a magnitude a of

a = v ² / R

where

v = tangential speed

R = radius of the curve

so that

a = (35 m/s)² / (215 m) ≈ 5.69767 m/s² ≈ 5.7 m/s²

Parallel to the road, the only force acting on the car is friction. So by Newton's second law, we have

∑ F = Fs = ma

where

Fs = magnitude of static friction

m = mass of the car

Then

Fs = (950 kg) (5.7 m/s²) ≈ 5412.79 N ≈ 5400 N

(b) Perpendicular to the road, the car is in equilbrium, so its weight and the normal force of the road on the car are equal in magnitude. By Newton's second law,

N - W = 0

where

N = magnitude of normal force

W = weight

so that

N = W = m g = (950 kg) (9.8 m/s²) = 9310 N

Friction is proportional to the normal force by a factor of µ, the coefficient of static friction:

Fs = µN

Assuming 35 m/s is the maximum speed the car can travel without skidding, we find

µ = Fs / N = (5400 N) / (9310 N) ≈ 0.581395 ≈ 0.58

Answer:

v = 4,244,699 m/s = (4.245 × 10⁶) m/s

Explanation:

The electric force on the proton is given by

F = qE

where q = charge on the proton = (1.602 × 10⁻¹⁹) C

E = Electric field = 720,000 N/C

F = (1.602 × 10⁻¹⁹ × 720000)

F = (1.153 × 10⁻¹³) N

But this force will accelerate the proton in this magnetic field in a form of trajectory motion.

We can obtain the acceleration using Newton's first law of motion relation

F = ma

m = mass of a proton = (1.673 × 10⁻²⁷) kg

a = (F/m)

a = (1.153 × 10⁻¹³)/(1.673 × 10⁻²⁷)

a = 68,944,411,237,298 m/s²

a = (6.894 × 10¹³) m/s²

This acceleration directs the proton from the positive plate to the negative plate, covering a distance of y = 0.006 m (the distance between the plates)

Using Equations of motion, we can obtain the time taken for the proton to move from the rest at the positive plate to the negative one.

u = initial velocity of the proton = 0 m/s

y = vertical distance covered by the proton = 0.006 m

a = acceleration of the proton in this direction = (6.894 × 10¹³) m/s²

t = time taken for the proton to complete this distance = ?

y = ut + (1/2) at²

0.006 = 0 + [(1/2)×(6.894 × 10¹³)×t²]

0.006 = (3.447 × 10¹³) t²

t² = (0.006)/(3.447 × 10¹³)

t² = 1.741 × 10⁻¹⁶

t = (1.32 × 10⁻⁸) s

Then we can then calculate the minimum speed to navigate the entire length of the plates without hitting the plates.

v = ?

x = 0.056 n

t = (1.32 × 10⁻⁸)

v = (x/t)

v = (0.056)/(1.32 × 10⁻⁸)

v = 4,244,699 m/s = (4.245 × 10⁶) m/s

Hope this Helps!!!

Answer:

v = 9.09×10⁵m/s

Explanation:

Given

d = the distance between plates = 0.6cm = 0.006

E = Electric field strength = 720000N/C

m =mass of the proton = 1.67 ×10-²⁷ kg

The

Electric potential energy of the field is converted into the the kinetic energy of the proton.

So

qV = 1/2mv²

But V = Ed

So q(Ed) = 1/2mv²

v² = 2qEd/m

v = √(2qEd/m)

v = √(2×1.6×10-¹⁹×720000×0.006/1.67×10-²⁷)

v = 9.09×10⁵m/s