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3 years ago

Two pebbles, Pebble A and Pebble B, are thrown horizontally with the same force. Pebble A's mass is 3 times the

Physics

1 answer:

Hitman42 [59]3 years ago

5 0

Answer:

Pebble A has 1/3 the acceleration as pebble B.

Explanation:

F = m×a

mass of a = 3 × mass of b (m_a = 3 × m_b)

Same starting force, F

m_a = mass of a

m_b = mass of b

a_a = acceleration of a

a_b = acceleration of b

F = m_a × a_a = m_b × a_b

3 × m_b × a_a = m_b × a_b

3 × a_a = a_b

a_a = a_b / 3

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If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?

Paraphin [41]

Answer:

Option C. 70 Ω

Explanation:

Data obtained from the question include:

Resistor (R) = 20 Ω

From diagram given ABOVE, we observed the following

1. R and R are in parallel connections.

2. 2R and 2R are in parallel connections.

3. 4R and 4R are in parallel connections.

Next, we shall determine the equivalent resistance in each case.

This is illustrated below:

1. Determination of the equivalent resistance for R and R parallel connections.

R = 20 Ω

Equivalent R = (R×R) /(R+R)

Equivalent R = (20 × 20) /(20 + 20)

Equivalent R = 400/40

Equivalent R = 10 Ω

2. Determination of the equivalent resistance for 2R and 2R parallel connections.

R = 20 Ω

2R = 2 × 20 = 40 Ω

Equivalent 2R = (2R×2R) /(2R+2R)

Equivalent 2R = (40 × 40) /(40 + 40)

Equivalent 2R = 1600/80

Equivalent 2R = 20 Ω

3. Determination of the equivalent resistance for 4R and 4R parallel connections.

R = 20 Ω

4R = 4 × 20 = 80 Ω

Equivalent 4R = (4R×4R) /(4R+4R)

Equivalent 4R = (80 × 80) /(80 + 80)

Equivalent 4R = 6400/160

Equivalent 4R = 40 Ω

Thus, the equivalence of R, 2R and 4R are now in series connections. We can obtain the equivalent resistance in the circuit as follow:

Equivalent of R = 10 Ω

Equivalent of 2R = 20 Ω

Equivalent of 4R = 40 Ω

Equivalent =?

Equivalent = Equivalent of (R + 2R + 4R)

Equivalent = 10 + 20 + 40

Equivalent = 70 Ω

Therefore, the equivalent resistance between point A and B is 70 Ω.

6 0

3 years ago

Which graph would you use to show the percentage of cookies sold by each

tatuchka [14]

Answer:

You would use a bar graph

4 0

3 years ago

An internal combustion engine uses fuel, of energy content 44.4 MJ/kg, at a rate of 5 kg/h. If the efficiency is 28%, determine

Fudgin [204]

Efficiency = Power Output / Power Input

Power Input = Rate of Energy input = 44.4 MJ/kg * 5 kg/h

= 222 MJ/h

But 1 hour = 3600seconds

222 MJ/h = 222 MJ/3600s = 0.061667 MW J/s = Watts

Power input = 0.061667 MW = 61 667 W

From Efficiency = Power Output / Power Input

   28% = Power Output / 61667

   Power Output = 0.28 * 61667

   Power Output = 17266.76 W

Power Output = 17 267 W

Rate of heat Rejection = Power input - Power output

      = 61667 - 17267 = 44400 W

Rate of heat Rejection = 44 400 W.

C- Copyright.

5 0

3 years ago

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of t

Maslowich

Answer:

a) The time taken to travel from 0.18 m to -0.18m when the amplitude is doubled = 2.76 s

b) The time taken to travel from 0.09 m to -0.09 m when the amplitude is doubled = 0.92 s

Explanation:

a) The period of a simple harmonic motion is given as T = (1/f) = (2π/w)

It is evident that the period doesn't depend on amplitude, that is, it is independent of amplitude.

Hence, the time it would take the block to move from its amplitude point to the negative of the amplitude point (0.09 m to -0.09 m) in the first case will be the same time it will take the block to move from its amplitude point to negative of the amplitude point in the second case (0.18 m to -0.18 m).

Hence, time taken to travel from 0.18 m to -0.18m when the amplitude is doubled is 2.76 s

b) Now that the amplitude has been doubled, the time taken to move from amplitude point to the negative amplitude point in simple harmonic motion, just like with waves, is exactly half of the time period.

The time period is defined as the time taken to complete a whole cycle and a while cycle involves movement from the amplitude to point to the negative amplitude point then fully back to the amplitude point. Hence,

0.5T = 2.76 s

T = 2 × 2.76 = 5.52 s

Note that the displacement of a body undergoing simple harmonic motion from the equilibrium position is given as

y = A cos wt (provided that there's no phase difference, that is, Φ = 0)

A = amplitude = 0.18 m

w = (2π/5.52) = 1.138 rad/s

When y = 0.09 m, the time = t₁₂ = ?

0.09 = 0.18 Cos 1.138t₁ (angles in radians)

Cos 1.138t₁ = 0.5

1.138t₁ = arccos (0.5) = (π/3)

t₁ = π/(3×1.138) = 0.92 s

When y = -0.09 m, the time = t₂ = ?

-0.09 = 0.18 Cos 1.138t₂ (angles in radians)

Cos 1.138t₂ = -0.5

1.138t₂ = arccos (-0.5) = (2π/3)

t₂ = 2π/(3×1.138) = 1.84 s

Time taken to move from y = 0.09 m to y = -0.09 m is then t = t₂ - t₁ = 1.84 - 0.92 = 0.92 s

Hope this Helps!!!

3 0

3 years ago

Chemists use a wide array of techniques for determining the exact composition and structure of a compound. One of the most robus

attashe74 [19]

Answer:

Δμ = hΔf/B

Explanation:

If the photon energy , ΔE = hΔf where Δf = small frequency shift and since the potential energy change of the magnetic dipole moment μ in magnetic field B from parallel to anti-parallel state is ΔU = ΔμB. where Δμ = small shift in magnetic moment.

Since the magnetic energy change equals the photon energy,

ΔE = ΔU

hΔf = ΔμB

Δμ = hΔf/B

5 0

3 years ago