An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?PLEASE SORT QUESTION BELOWSort the following quantities as known or unknown. Take the horizontal direction to be along the x axis.mQ: the mass of the quarterbackmB: the mass of the football(vQx)i: the horizontal velocity of quarterback before throwing the ball(vBx)i: the horizontal velocity of football before being thrown(vQx)f: the horizontal velocity of quarterback after throwing the ball(vBx)f: the horizontal velocity of football after being thrown

Answer:

B. less

Explanation:

acceleration due to gravity on Earth, g = 9.8 m/s²

acceleration due to gravity on Moon, g = 1.6 m/s²

Given mass of the object as, m = 5 kg

Weight of an object is given as, W = mg

Weight of the object on Earth, W = 5 x 9.8 = 49 N

Weight of the object on Moon, W = 5 x 1.6 = 8 N

Therefore, the object weighs less on the moon compared to its weight on Earth.

The correct option is "B. less"

When the object is in equilibrium so the statement i.e. not true should be that object should be at rest.

In the case when the net force of an object should be zero at the time when it is in equilibrium. In the case when the net force should be zero the acceleration should also be zero. And, if the acceleration if zero so the speed and velocity should remain the same. So for maintaining the equilibrium, the object should not have to be only at rest.

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Answer:

7.7 km 26°

Explanation:

The total x component is:

x = 2.5 cos(35°) + 5.2 cos(22°) = 6.87

The total y component is:

y = 2.5 sin(35°) + 5.2 sin(22°) = 3.38

The magnitude is:

d = √(x² + y²)

d = 7.7 km

The direction is:

θ = atan(y/x)

θ = 26°

Answer:

The correct magnitude of the coil's magnetic field= =50μT

Explanation :

The magnetic field takes place as a result of movement of charge I e current, can also occurs from magnetised material, magnetic field as a result of charge movement can be deducted using right hand grip rule.

At the equator magnetic field lines are parallel towards the earth's surface and the angle of inclination of the magnetic lines of force at the horizontal position is referred to as the angle of dip at the point.

As the current is produced then the varying magnetic field is opposed ,then there is induced current when the coil is positioned at varying magnetic field.

Given from the question,

angle below horizontal θ=60-degree

The Earth's magnetic field B=50μT

The horizontal magnetic field can be expressed in terms of the formula below;

BH=Bcos⁡θ

B(H) = earth's horizontal component of magnetic field

Ø is the angle between coil's field

B=the magnetic field in Tesla

Then,

BH=50μT×cos60∘

=50μT× 0.5

As the current is passed through the coil to produce a field , when combined with the earth's field, which creates a net field with the same strength and dip angle (60 degrees below horizontal) as the earth's field.

It can be deducted that B has the same magnitude and angle which makes the those vertical component to cancel each other since they are the same.

For the magnetic field to be pointed out at North direction, we can calculate the corrected magnetic field using the formula below

Bc=2BH

Bc=2×50μT× 0.5=50μT

The correct magnitude of the coil's magnetic field= =50μT

Answer:

The field lines go out of Earth near Antarctica, enter Earth in northern Canada, and are not aligned with the geographic poles.

Explanation:

Answer: the field lines go out of earth near the north pole, enter earth in the south pole, and are not aligned with the geographic poles.

Explanation: just took the test on edge 2020.

Answer:

h = 31.46 m

Explanation:

We have,

Initial speed of a projectile is 37.7 m/s

It was projected at an angle of 41.2° above the horizontal on a long flat firing range.

It is required to find the maximum height reached by the projectile. The formula used to find it is given by :

[tex]h=\dfrac{u^2\sin^2\theta}{2g}[/tex]

Plugging all the known values,

[tex]h=\dfrac{(37.7)^2\times \sin^2(41.2)}{2\times 9.8}\\\\h=31.46\ m[/tex]

So, the maximum height reached by the projectile is 31.46 m.

The so-called "force" of friction always acts exactly opposite to the direction in which an object is moving.  So for this question, since the box is moving to the right, the "force" of friction is acting toward the left.

There's not enough information given in the question to determine its magnitude.

Answer:

Lead > Aluminium > Wood > Styrofoam

Explanation:

Buoyant force is described by the Archimedes's Principle, which states that buoyant force is equal to the weight of the fluid displaced by the submerged object. By Newton's Laws, the buoyant force is represented by the following equation of equilibrium:

[tex]\Sigma F = F_{D} - W_{cube} = 0[/tex]

[tex]F_{D} = W_{cube}[/tex]

[tex]F_{D} = \rho_{cube} \cdot g \cdot V_{cube}[/tex]

Where:

[tex]\rho_{cube}[/tex] - Density of the cube, measured in kilograms per cubic meter.

[tex]g[/tex] - Gravitational constant, measured in meters per square second.

[tex]V_{cube}[/tex] - Volume of the cube, measured in cubic meters.

Let suppose that volume of the cube is known. Given that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the buoyant force is computed for each material:

Lead ([tex]\rho_{cube} = 11,300\,\frac{kg}{m^{3}}[/tex])

[tex]F_{D} = \left(11,300\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]

[tex]F_{D} = 110,819.1V_{cube}[/tex]

Aluminium ([tex]\rho_{cube} = 2,700\,\frac{kg}{m^{3}}[/tex])

[tex]F_{D} = \left(2,700\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]

[tex]F_{D} = 26478.9V_{cube}[/tex]

Wood ([tex]\rho_{cube} = 800\,\frac{kg}{m^{3}}[/tex])

[tex]F_{D} = \left(800\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]

[tex]F_{D} = 7845.6V_{cube}[/tex]

Styrofoam ([tex]\rho_{cube} = 50\,\frac{kg}{m^{3}}[/tex])

[tex]F_{D} = \left(50\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]

[tex]F_{D} = 490.35V_{cube}[/tex]

Therefore, the buoyant forces that the water exerts on the cubes from largest to smallest corresponds to: Lead > Aluminium > Wood > Styrofoam.

Answer:

A and B are true

Explanation:

C and D are false

Answer:

The answer is the length, cross-sectional area, and the amount of electric current passing through them.

So,

a

b

d

are the correct answers.

Explanation:

I hope this helps future Physics students who are also struggling...

If you have any questions let me know!!

This answer is 100% correct

HEYA!!

Here's your Answer:

Energy is the power we use for transportation, for heat and light in our homes and for the manufacture of all kinds of products. There are two sources of energy: renewable and non-renewable energy

RENEWABLE ENERGY: Renewable energy, often referred to as clean energy, comes from natural sources or processes that are constantly replenished. For example, (sunlight or wind keep shining and blowing), even if their availability depends on time and weather.

NON-RENEWABLE ENERGY:Non-renewable energy comes from sources that will run out or will not be replenished in our lifetimes—or even in many, many lifetimes.

Most non-renewable energy sources are (fossil fuels: coal, petroleum, and natural gas).

HOPE IT HELPS!!!

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

     ∑  F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

          cos 41 = T₁ₓ / T1

          sin 41 = T₁y / T1

          T₁ₓ = T₁  cos 41

          T₁y= T₁  sin 41

for cable gold

           cos 63 = T₂ / T₂

           sin 63 = [tex]T_{2y}[/tex] / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

              T₃ - W = 0

              T₃ = W

              T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

   - T₁ₓ + T₂ₓ = 0

  T₁ₓ = T₂ₓ

   T1 cos 41 = T2 cos 63

   T1 = T2 cos 63 / cos 41                (1)

y Axis

      [tex]T_{1y}[/tex] + T_{2y} - T3 = 0

       T₁ sin 41 + T₂ sin 63 = T₃          (2)

to solve the system we substitute equation 1 in 2

        T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

         T₂ (cos 63 tan 41 + sin 63) = W

         T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

          T₂ = 200 / (cos 63 tan 41 + sin 63)

          T₂ = 200 / 1,2856

           T₂ = 155.6 N

we substitute in 1

            T₁ = T₂ cos 63 / cos 41

             T₁ = 155.6 cos63 / cos 41

             T₁ = 93.6 N

therefore the tension in each cable is

            T₁ = 93.6 N

             T₂ = 155.6 N

             T₃ = 200 N

Answer:

Explanation:

Initially skier is at a height of 351 m . Her kinetic enery will be nil because she is at rest . Her potential energy will be calculated as follows

potential energy = mgh where m is mass , h is height and g is acceleration due to gravity

potential energy = 58 x 9.8 x 351

= 199508 .4 J

Total mehanical energy = potential energy + kinetic energy

= 199508.4 J

According to conservation of mechanical energy , at the height of 142 m also total mechanical energy will be same . At this height some potential energy will be converted into kinetic energy but total of potential and kinetic energy will be same.

Answer:

[tex]v = \sqrt{20h} [/tex]

Explanation:

The potential energy (PE) we are looking here is gravitational potential energy (GPE).

GPE= mgh,

where m is the mass of an object,

g is the gravitational field strength

h is the height of the object

KE= ½mv²,

where m is the mass and v is the velocity

loss in GPE= gain in KE

mgh= ½mv²

gh= ½v² (divide by m throughout)

Assuming that the object is on earth, then g= 10N/Kg

½v²= 10h (substitute g=10)

v²= 20h (×2 on both sides)

v= √20h (square root both sides)

Answer:

  0.180 m/s

Explanation:

Solving the equations for conservation of momentum and energy for an elastic collision gives ...

  v₁' = ((m₁ -m2)v₁ +2m₂v₂)/(m₁ +m₂) . . . . v₁' is the velocity of m₁ after collision

__

Here, we have (m₁, m₂, v₁, v₂) = (20 g, 40 g, 0.540 m/s, 0 m/s).

Substituting these values in to the equation for v₁', we have ...

  v₁' = ((20 -40)(0.540) +2(40)(0))/(20 +40) = (-20/60)(0.540)

  v₁' = -0.180 . . . m/s

The speed of the 20 g block after the collision is 0.180 m/s to the left.

Answer:

a) 10.6V

b) E = 4.9V/m, +x direction

c) E = 4.9V/m, +x direction

Explanation:

You have the following function:

[tex]V=a+bx[/tex]  (1)

for the potential in a region between x=0 and x=6.00 m

a = 10.6 V

b = -4.90V/m

[tex]V=10.6V-4.90\frac{V}{m}x[/tex]

a) for x=0 you obtain for V:

[tex]V=10.6V-4.90\frac{V}{m}(0m)=10.6V[/tex]

b) The relation between the potential difference and the electric field can be written as:

[tex]E=-\frac{dV}{dx}=-b=-(-4.9V/m)=4.9V/m[/tex]  (2)

the direction is +x

c) The electric field is the same for any value of x between x=0 and x=6m.

Hence,

E = 4.9V/m, +x direction

Answer:

6.0621 m/s

Explanation:

we all know that horizontal component of the velocity in projectile motion is

= u cosα

u= initial speed of 7.00 m/s of the gun.

α = angle of the initial velocity with the horizontal = 30°

therefore, horizontal component of the ball's initial velocity = 7×cos30°

[tex]7\times\frac{\sqrt{3} }{2} \\=6.0621 \text{m/s}[/tex]

Answer:

Explanation:

We shall show all given data in vector form and calculate the direction of force with the help of following formula

force F = q ( v x B )

q is charge , v is velocity and B is magnetic field.

Given B = - Bk ( i is  right  , j is  upwards  and k is straight up the page  )

v = v j

F = q ( vj x - Bk )

= -Bqvi

The direction is towards left .

a ) If velocity is down

v = - v j

F = q ( - vj x - bk )

= qvB i

Direction is right .

b ) v = v i

F = q ( vi x - Bk )

= qvB j

force is upwards

c ) v = - vi

F = q ( -vi x - Bk )

= -qvBj

force is downwards

d ) v = - v k

F = q( - vk x -Bk  )

= 0

No force will be created

e ) v =  v k

F = q(  vk x -Bk  )

= 0

No force will be created  

Answer:

33 m/s

Explanation:

v = √(T / (m/L))

where v is the velocity,

T is the tension,

and m/L is the mass per length.

v = √(200 N / (0.055 kg / 0.30 m))

v = 33 m/s

Answer:

The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s is  J.

Explanation:

Thus the kinetic energy of the ball is 25 J. The unit Joule (J) is the same as kgm^2/s^2 and is the SI unit for kinetic energy.

Hope this helps.

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

[tex]v_{avg}=\frac{x_{all}}{t}[/tex]

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

[tex]v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}[/tex]

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

Answer:

The average velocity of the object is 0.1cm/s

Explanation:

Given that the object travels from point 15cm to 165cm and back to 25cm within 100 seconds

The average velocity is calculated as thus.

Average Velocity = ∆D/t

Where ∆D represent the displacement.

The displacement is calculated as follows.

∆D = End point - Start Point.

From the question, the end and start point are 25cm and 15cm respectively.

Hence,

∆D = 25cm - 15cm

∆D = 10cm.

t = 100 seconds

So, Average Velocity = 10cm/100s

Average Velocity = 0.1cm/s

Hence, the average velocity of the object is 0.1cm/s

Answer:

The normal force is the force that the floor does as a reaction of the gravitational force that an object does against the floor (is the resistance that objects have when other objects want to move trhough them, and the force comes by the 3rd Newton's law, and this is specially used in cases where the first object is fixed, like walls or the floor). With this in mind, the point in where the normal force will be greater is the point that is closer to the center of mass of the object (the point with more mass)

If the wheels are in the extremes of the object, and the center of mass is in the middle of the object, the normal force will be equal. Now if for example, you put a little mass in one end of the object, now the center of weight displaces a little bit and is not centered, and the side is where you put the weight on will receive a bigger normal force from the floor than the other side.

The force on the front and rear wheels can be determined by finding the

force acting under equilibrium conditions.

The normal reaction between the floor and the front wheel is larger. The

correct option is c) The force on the front wheels must be greater than the

force on the rear wheels.

Reasons:

The forces acting on the wheels are;

Vertical forces acting on the cart:

The weight of the cart acting on the floor

The normal reaction of the floor on the wheels

The horizontal acting on the cart:

Pushing force acting horizontally

Friction force acting in the reverse direction

At equilibrium, the cart does not tip over

Sum of moment about the rear wheel = [tex]\mathbf{W \times \dfrac{d}{2} + F \times h - N_{front} \times d} = 0[/tex]

Where;

h = The height of the cart

d = depth of the cart

[tex]N_{front}[/tex]  = The normal reaction at the front wheels

Therefore;

[tex]\displaystyle N_{front} = \frac{W \times \dfrac{d}{2} + F \times h}{d} = \mathbf{ \frac{W}{2} + \frac{F \times h}{d}}[/tex]

Sum of moment about the front wheel = [tex]\mathbf{ F \times h + N_{rear} \times d - W \times \dfrac{d}{2} } = 0[/tex]

Therefore;

[tex]\displaystyle N_{rear} = \frac{W \times \dfrac{d}{2} - F \times h}{d} = \mathbf{ \frac{W}{2} - \frac{F \times h}{d}}[/tex]

Which gives;

[tex]\displaystyle N_{front} = \mathbf{ \displaystyle N_{rear} + 2 \times \frac{F \times h}{d}}[/tex]

[tex]\displaystyle 2 \times \frac{F \times h}{d} > 0[/tex]

Therefore;

[tex]\displaystyle N_{front} > \displaystyle N_{rear}[/tex]

The correct option is c) The force on the front wheels must be greater than

the force on the rear wheels.

Learn more here:

https://brainly.com/question/17574217

Answer:

Explanation:

Time to cover first 100 km = 1 hour.

time remaining = 3.15 - 1 = 2.15 hour .

Time to cover next 42 km = 1 hour .

Time remaining = 2.15-1 = 1.15 hour.

Distance to be covered = 310 - 142

= 168 km

least speed needed = distance remaining / time remaining

= 168 / 1.15

= 146.08 km / h .

A natural force of attraction exerted by the earth upon objects, that pulls objects towards earth's center is called Gravitational force .

Answer:

0.824m/s

Explanation:

To calculate the average velocity we have to find the total distance and the total time

We have to find the distance and time in each motions

FIRST MOTION

The values given are

Speed= 2m/s , t = 60minutes

The time has to be converted to seconds. 60×60 = 3600seconds

Distance= speed×time

= 2× 3600

= 7200m

In the first motion the distance is 7200m and the time is 3600seconds

SECOND MOTION

The values given are

Distance= 3000m

Time= 25 mins to seconds

= 25×60

= 1500 seconds

In the second motion distance is 3000m and the time is 1500 seconds

The total distance can be calculated by applying the formular ( d1-d2) since she moved in an opposite direction

Total distance= 7200-3000

= 4200m

The total time (t1+t2) = 1500+3600

= 5100 seconds

Therefore, average velocity is calculated by applying the formular

Total distance/ Total time

= 4200/5100

= 0.824m/s South

Hence the average velocity is 0.824m/s South.

Answer:

The woman's average velocity during her entire motion is 2 m/s

Explanation:

Given;

initial speed of the woman, u =  2.00 m/s

initial time taken, t₁ = 60 minutes = 3600 seconds

initial displacement of the woman, x₁ = ?

final displacement of the woman, x₂ = 3000 m north

final time taken , t₂ = 25.0 minutes = 1500 seconds

The woman's average velocity during her entire motion:

initial displacement of the woman, x₁ = u x t₁  = 2.00 m/s x 3600 seconds

                                                                           = 7200 m South

[tex]Average \ velocity = \frac{\delta X}{\delta t} = \frac{X_1-X_2}{t_1-t_2} \\\\V_{avg.} = \frac{7200-3000}{3600-1500} = \frac{4200}{2100} = 2 \ m/s[/tex]

Therefore, the woman's average velocity during her entire motion is 2 m/s

Answer:

A

Explanation:

Firstly, recall the equation for the circumference of a circle (C=2*pi*r). Then apply the formula speed = distance/time.

Answer:

i see they did already i am just supporting them

Answer:

A

Explanation:

Firstly, recall the equation for the circumference of a circle (C=2*pi*r). Then apply the formula speed = distance/time.

Answer:

i think its 1-10 ´ 10-15 m

Explanation:

It is found that nuclear radii range from 1-10 ´ 10-15 m. This radius is much smaller than that of the atom, which is typically 10-10 m. Thus, the nucleus occupies an extremely small volume inside the atom. The nuclei of some atoms are spherical, while others are stretched or flattened into deformed shapes.

Answer:

ΔQ = 0.1 kJ

[tex]\mathbf{v_f = 1.445*10^{-3} m^3}[/tex]

[tex]\mathbf{P_f = 156.5 \ kPa}[/tex]

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

Explanation:

GIven that:

Diameter of the piston-cylinder = 12 cm

Pressure of the piston-cylinder = 100 kPa

Temperature =24 °C

Length of the piston = 20 cm

Boundary work ΔW = 0.1 kJ

The gas is compressed and The temperature of the gas remains constant during this process.

We are to find ;

a. How much heat was transferred to/from the gas?

According to the first law of thermodynamics ;

ΔQ = ΔU + ΔW

Given that the temperature of the gas remains constant during this process; the isothermal process at this condition ΔU = 0.

Now

ΔQ = ΔU + ΔW

ΔQ = 0 + 0.1 kJ

ΔQ = 0.1 kJ

Thus; the amount of heat that was transferred to/from the gas is : 0.1 kJ

b. What is the final volume and pressure in the cylinder?

In an isothermal process;

Workdone W = [tex]\int dW[/tex]

[tex]W = \int pdV \\ \\ \\W = \int \dfrac{nRT}{V}dv \\ \\ \\ W = nRt \int \dfrac{dv}{V} \\ \\ \\ W = nRT In V |^{V_f} __{V_i}} \\ \\ \\ W = nRT \ In \dfrac{V_f}{V_i}[/tex]

Since the gas is compressed ; then [tex]v_f< v_i[/tex]

However;

[tex]W =- nRT \ In \dfrac{V_f}{V_i}[/tex]

[tex]W =- P_1V_1 \ In \dfrac{V_f}{V_i}[/tex]

The initial volume for the cylinder is calculated as ;

[tex]v_1 = \pi r^2 h \\ \\ v_1 = \pi r^2 L \\ \\ v_1 = 3.14*(6*10^{-2})^2*(20*10^{-2}) \\ \\ v_1 = 2.261*10^{-3} \ m^3[/tex]

Replacing over values into the above equation; we have :

[tex]100 = - ( 100*10^3 *2.261*10^{_3}) In (\dfrac{v_f}{v_i}) \\ \\ - In (\dfrac{v_f}{v_i})= \dfrac{100}{(100*10^3*2.261*10^{-3})} \\ \\ - In \ v_f + In \ v_i = \dfrac{100}{226.1} \\ \\ - In \ v_f = - In \ v_i + \dfrac{100}{226.1} \\ \\ - In \ v_f = - In (2.261*10^{-3} + \dfrac{100}{226.1 } \\ \\ - In \ v_f = 6.1 + 0.44 \\ \\ - In \ v_f = 6.54 \\ \\ - In \ v_f = -6.54 \\ \\ v_f = e^{-6.54} \\ \\ \mathbf{v_f = 1.445*10^{-3} m^3}[/tex]

The final pressure can be calculated by using :

[tex]P_1V_1 = P_2V_2 \\ \\ P_iV_i = P_fV_f[/tex]

[tex]P_f =\dfrac{P_iV_i}{V_f}[/tex]

[tex]P_f =\dfrac{100*2.261*10^{-3}}{1.445*10^{-3}}[/tex]

[tex]P_f = 1.565*10^2 \ kPa[/tex]

[tex]\mathbf{P_f = 156.5 \ kPa}[/tex]

c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?

The change in entropy of the gas is given  by the formula:

[tex]\Delta S=\dfrac{\Delta Q}{T}[/tex]

where

T =  24 °C = (24+273)K

T = 297 K

[tex]\Delta S=\dfrac{-100 \ J}{297 \ K}[/tex]

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

Answer:

a) 4.681*10^10 electrons

b) 3.67*10^12 electrons

Explanation:

The amount of electrons in a charge of 1C is:

[tex]1C=6.2415*10^{18}\ electrons[/tex]

You use the previous equality as a conversion factor.

a) The sing of the charge is not important in the calculation of the number electrons, so, you use the absolute value of the charge

[tex]7.50nC=7.50*10^{-9}C*\frac{6.2415*10^{18}}{1C}=4.681*10^{10}\ electrons[/tex]

In 7.50nC there are 4.61*10^18 electrons

b)

[tex]0.580\mu C=0.580*10^{-6}C*\frac{6.2415*10^{18}}{1C}=3.67*10^{12}\ electrons[/tex]

To obtain a charge of  0.580 µC in a neutral object you need to take out 3.67*10^12 electrons