Answer:
### SI units

The **required **value is required in SI units.

The **required **answer is

The SI unit of **mass**, **length **and **time **is **kg, m** and **s **respectively.

In order to **convert **one **unit **into another it has to be **multiplied **or **divided **by the **conversion **factors.

A definite **magnitude **which has some **quantity **which is defined by **convention **or **law **is called a **unit**.

The** conversion factors **are

1 min = 60 s

So,

Learn more about** SI units:**

Answer:
Hi hi hi hi hi hi hi

In an evironmental system of subsystem, the mass balance equation is:__________.

In this example we will use pendulum motion to actually measure the acceleration of gravity on a different planet. An astronaut on the surface of Mars measures the frequency of oscillation of a simple pendulum consisting of a ball on the end of a string. He finds that the pendulum oscillates with a period of 1.5 s. But the acceleration due to gravity on Mars is less than that on earth, gMars=0.38gearth. Later, during a journey to another planet, the astronaut finds that his simple pendulum oscillates with a period of 0.92 s. What planet is he now on?SOLUTIONSET UP Each planet has a different value of the gravitational acceleration g near its surface. The astronaut can measure g at his location, and from this he can determine what planet he's on. First we use the information about Mars to find the length L of the string that the astronaut is swinging. Then we use that length to find the acceleration due to gravity on the unknown planet.

A laser (electromagnetic wave) has the maximum electric field strength of 1.0x1011 V/m. What is the force the laser applies on a mirror (totally reflective) of 5.0 mm2 area? A. 2.76 x105N B. 1.21 x106N C. 1.94 x106N D.4.43 x105 N E. 7.82 x104N

Which of these objects are constantly in motion? Select all that apply.A.EarthB.PlanesC.TrainsD.BloodE.SunF.Cars

1. A surfboarder rides a wave for 23.7 m at a constant rate of 4.1 m/s. How long did his triptake?

In this example we will use pendulum motion to actually measure the acceleration of gravity on a different planet. An astronaut on the surface of Mars measures the frequency of oscillation of a simple pendulum consisting of a ball on the end of a string. He finds that the pendulum oscillates with a period of 1.5 s. But the acceleration due to gravity on Mars is less than that on earth, gMars=0.38gearth. Later, during a journey to another planet, the astronaut finds that his simple pendulum oscillates with a period of 0.92 s. What planet is he now on?SOLUTIONSET UP Each planet has a different value of the gravitational acceleration g near its surface. The astronaut can measure g at his location, and from this he can determine what planet he's on. First we use the information about Mars to find the length L of the string that the astronaut is swinging. Then we use that length to find the acceleration due to gravity on the unknown planet.

A laser (electromagnetic wave) has the maximum electric field strength of 1.0x1011 V/m. What is the force the laser applies on a mirror (totally reflective) of 5.0 mm2 area? A. 2.76 x105N B. 1.21 x106N C. 1.94 x106N D.4.43 x105 N E. 7.82 x104N

Which of these objects are constantly in motion? Select all that apply.A.EarthB.PlanesC.TrainsD.BloodE.SunF.Cars

1. A surfboarder rides a wave for 23.7 m at a constant rate of 4.1 m/s. How long did his triptake?

**Answer:**

**a) 8.99*10³ V b) 4.5*10⁻² J c) 0 d) 0**

**Explanation:**

**a)**

- The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
- For a point charge, it can be expressed as follows:

- As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
- This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
- In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at any other corner, as follows:

- The potential at point C is 8.99*10³ V

**b)**

- The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:

- The work needed is 0.045 J.

**c)**

- If we replace one of the charges creating the potential at the point C, by one of the same magnitude, but opposite sign, we will have the following equation:

- This means that the potential due to both charges is 0, at point C.

**d) **

- If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.

m

(b) If each holds one end of a rope, and the man pulls on the rope so that he moves 1.3 m, how far from the woman will he be now?

m

(c) How far will the man have moved when he collides with the woman?

m

**Answer:**

Given that

m₁ = 50 kg

m₂=80 kg

d= 12 m

a)

We know that center of mass given as

X = (x₁m₁+x₂m₂)/(m₁+m₂)

Lets take distance of CM from woman is X

So now by putting the value

X = (0 x 50+12 x 80)/(50+80)

**x=7.38 m**

b)

There is no any external force so the CM will not move.

So we can say that

x₁m₁+x₂m₂ = 0

50(x) - 80(1.3)=0

x=2.08

So the distance move by woman d=12-2.08-1.3=8.62 m

**d=8.62 m**

c) lets take distance move by man is x

50 (x) - 80 (12-x) =0

x=7.38

So the distance move by woman d=12-7.38

**d=4.62 m**

2)(b) If you throw your 1.08-kg boot with an average force of 391 N, and the throw takes 0.576 s (the time interval over which you apply the force), what is the magnitude of the force that the boot exerts on you? (Assume constant acceleration.)

391 N

3)(c) How long does it take you to reach shore, including the short time in which you were throwing the boot?

Just number 3

Answer:

1a) The direction to throw the boot is directly away from the closest shore.

2b) The magnitude of the force that the thrown boot exerts on the engineer = 391 N

3c) Time taken to reach shore = 8.414 s

Explanation:

1a) Newton's third law of motion explains that for every action, there is an equal and opposite reaction.

The force generated by throwing the boot in one direction is exerted back on the engineer as recoil in the opposite direction.

Hence, the best direction to throw the boot is opposite the direction that the engineer intends to move towards.

2b) Just as explained in (1a) above, the force exerted in one direction always has a reaction of the same magnitude in the opposite direction.

Hence, the force exerted by the boot on the engineer is equal to the force exerted by the engineer on the boot = 391 N.

3c) For this part, we analyze the total motion of the engineer.

The force exerted by the boot on the engineer initially accelerates the engineer until the engineer reaches a constant velocity dictated the impulse of the initial force (since impulse is equal to change in momentum), this constant velocity then takes the engineer all the way to shore, since the ice surface is frictionless.

The weight of the engineer = W = 588 N

W = mg

Mass of the engineer = (W/g) = (588/9.8) = 60 kg

Force exerted on the engineer by the thrown boot = F = 391 N

F = ma

Initial acceleration of the engineer = (F/m) = (391/60) = 6.52 m/s²

We can then calculate the distance covered during this acceleration

X₁ = ut + ½at₁²

u = initial velocity of the engineer = 0 m/s (the engineer was initially at rest)

t₁ = time during which the force acts = 0.576 s

a = acceleration during this period = 6.52 m/s²

X₁ = 0 + 0.5×6.52×0.576² = 1.08 m

For the second part of the engineer's motion, the velocity becomes constant.

So, we first calculate this constant velocity

Impulse = Change in momentum

F×t = mv - mu

F = Force causing motion = 391 N

t = time during which the force acts = 0.576 s

m = mass of the engineer = 60 kg

v = final constant velocity of the engineer = ?

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

391 × 0.576 = 60v

v = (391×0.576/60) = 3.7536 m/s.

The distance from the engineer's initial position to shore is given as 30.5 m

The engineer covers 1.08 m during the time the force causing motion was acting.

The remaining distance = X₂ = 30.5 - 1.08 = 29.42 m

We can then calculate the time taken to cover the remaining distance, 29.42 m at constant velocity of 3.7536 m/s

X₂ = vt₂

t₂ = (X₂/v) = (29.42/3.7536) = 7.838 s

Time taken to reach shore = t₁ + t₂ = 0.576 + 7.838 = 8.414 s

Hope this Helps!!!

**Answer:**

**Explanation:**

Frequency( ν ) 4667 x 10³ Hz = 4.667 x 10⁶ Hz

Energy of one photon = hν [ h is plank's constant ]

= 6.6 x 10⁻³⁴ x 4.667 x 10⁶ = 30.8 x 10⁻²⁸ J

Power = 84 x 10³ J/s

No of photons emitted = Power / energy of one photon

= 84 x 10³ / 30.8 x 10⁻²⁸ =2.727 x 10³¹ per second .

**Answer:**

**Explanation:**

The force experienced by the moving electron in the magnetic field is expressed as **F = qvBsinθ** where;

q is the charge on the electron

v is the velocity of the electron

B is the magnetic field strength

θ is the angle that the velocity of the electron make with the magnetic field.

**Given parameters**

F = 1.40*10⁻¹⁶ N

q = 1.6*10⁻¹⁹C

v = 3.94*10³m/s

B = 1.23T

**Required**

Angle that the velocity of the electron make with the magnetic field

Substituting the given parameters into the formula:

1.40*10⁻¹⁶ = 1.6*10⁻¹⁹ * 3.94*10³ * 1.23 * sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁹⁺³sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁶sinθ

sinθ = 1.40*10⁻¹⁶/7.75392 * 10⁻¹⁶

sinθ = 1.40/7.75392

sinθ = 0.1806

θ = sin⁻¹0.1806

**θ₁ = 10.4⁰**

Since sinθ is positive in the 1st and 2nd quadrant, θ₂ = 180-θ₁

θ₂ = 180-10.4

**θ₂ = 169.6⁰**

**Hence, the angle that the velocity of the electron make with the magnetic field are 10.4⁰ and 169.6⁰**

**Answer:**

**300 clicks...**

**Explanation:**

Output on 3 clicks = 10 dB

Increasing 10 by a factor of 100 equals 1000 dB so,

Its simple math, clicks will also increase in the same ratio and it shall take 300 clicks to increase the volume by a factor of 100.