Convert 7 (gcm^2)/(min^2) into a value in standard S.I. units. Be sure to use scientific notation if necessary. You do not need to answer units.


Answer 1

The required value is required in SI units.

The required answer is 1.94*10^(-10)\ \text{kg m}^2/\text{s}^2

SI units

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\ \text{g}=10^(-3)\ \text{kg}

1\ \text{cm}=10^(-2)\ \text{m}

1\ \text{cm}^2=10^(-4)\ \text{m}^2

1 min = 60 s

1\ \text{min}^2=60*60\ \text{s}^2


7\ \text{g cm}^2/\text{min}^2=7* (10^(-3)* 10^(-4))/(60* 60)\n =1.94*10^(-10)\ \text{kg m}^2/\text{s}^2

Learn more about SI units:

Answer 2
Answer: Hi hi hi hi hi hi hi

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Points A, B, and C are at the corners of an equilateral triangle of side 8 m. Equal positive charges of 4 mu or micro CC are at A and B. (a) What is the potential at point C? 8.990 kV * [2.5 points] 2 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 8.990 OK (b) How much work is required to bring a positive charge of 5 mu or micro CC from infinity to point C if the other charges are held fixed? .04495 J * [2.5 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] .04495 OK (c) Answer parts (a) and (b) if the charge at B is replaced by a charge of -4 mu or micro CC. Vc= kV [2.5 points] 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] W =



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



  • 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:

        V =(k*q)/(d)

  • 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:

        V = (2*q*k)/(d) = (2*8.99e9N*m2/C2*4e-6C)/(8m) =\n \n V= 8.99e3 V

  • The potential at point C is 8.99*10³ V


  • 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:

        W = V * q = 8.99e3 V* 5e-6C = 4.5e-2 J

  • The work needed is 0.045 J.


  • 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:

       V = (8.99e9N*m2/C2*(4e-6C))/(8m)  + ((8.99e9N*m2/C2*(-4e-6C))/(8m)) = 0

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


  • 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.

A 50 kg woman and an 80 kg man stand 12.0 m apart on frictionless ice.(a) How far from the woman is their CM?

(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?

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



Given that

m₁ = 50 kg

m₂=80 kg

d= 12 m


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


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


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


So the distance move by woman d=12-7.38

d=4.62 m

As a prank, your friends have kidnapped you in your sleep, and transported you out onto the ice covering a local pond. Since you're an engineer, the first thing you do when you wake up is drill a small hole in the ice and estimate the ice to be 6.7cm thick and the distance to the closest shore to be 30.5 m. The ice is so slippery (i.e. frictionless) that you cannot seem to get yourself moving. You realize that you can use Newton's third law to your advantage, and choose to throw the heaviest thing you have, one boot, in order to get yourself moving. Take your weight to be 588 N. (Lucky for you that, as an engineer, you sleep with your knife in your pocket and your boots on.)1)(a) What direction should you throw your boot so that you will most quickly reach the shore? away from the closest shore perpendicular to the closest shore straight up in the air at your friend standing on the closest shore
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



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


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!!!

A 4,667 kHz AM radio station broadcasts with a power of 84 kW. How many photons does the transmitting antenna emit each second.




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 .

An electron moving at 3.94 103 m/s in a 1.23 T magnetic field experiences a magnetic force of 1.40 10-16 N. What angle does the velocity of the electron make with the magnetic field? There are two answers between 0° and 180°. (Enter your answers from smallest to largest.)



10.4⁰ and 169.6⁰


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


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

The volume control on a stereo is designed so that three clicks of the dial increase the output by 10 dB. How many clicks are required to increase the power output of the loudspeakers by a factor of 100?



300 clicks...


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.