A conducting sphere has a net charge of -4.8x10-17 C. What is the approximate number of excess electrons on the sphere?

Answer:

Explanation:

pV = nrT

n = PV/RT

n = (100*10^3)(.01)/(300*0.082057)

n = 40.62 moles

Answer:

1. 6.15m/s

2. 820N

Explanation:

The total upward force

= 410x2

= 820

g = 9.81

a = v²/r

= 2xT - msg = m x v²/r

= 820-37*9.81 = 37v²/3.06

= 820-362.97 = 37v²/3.06

= 457.03 = 12.09v²

To get v²

V² = 457.03/12.09

V² = 37.8

V = √37.8

V = 6.15m/s

B. We already have the answer to this question

The force exerted is simply gotten by this calculation

2x410

= 820N

Answer:

(d) 2.93 x 10⁻³ N

Explanation:

Given;

current in the wire, I = 11.4 A

angle of inclination, θ = 11.4⁰

magnetic field on the wire, B = 11. 4  x 10⁻³

length of the wire, L = 11.4 cm = 0.114 m

The magnitude of magnetic force on the wire is given by;

F = BILsinθ

F = (11.4 x 10⁻³)(11.4)(0.114)(sin 11.4°)

F = 0.00293 N

F = 2.93 x 10⁻³ N

Therefore, the correct option is "D"

Answer:

Q = 425 kJ

Explanation:

Given that,

Mass, m = 25 kg

The clown doll head that heats up from 25°C to 35°C

The specific heat is 1700 J/kg°C

We need to find the internal energy of it. The heat required to raise the temperature is given by the formula as follows :

[tex]Q=mc\Delta T\\\\Q=25\times 1700\times (35-25)\\\\Q=425000\ J\\\\Q=425\ kJ[/tex]

So, 425 kJ of thermal energy is severed.

Answer:

t = 1.32 s

Explanation:

We are given;. Frequency of C4 note; F_c = 262 Hz

In conversions, we know that 1 Hz = 1 cycle/s

Thus, F_c = 262 cycles/s

Now, we want to find out how much time it takes for 346 air pressure maxima to pass a stationary listener.

346 air pressure maxima denotes that the air pressure maxima is 346 cycles.

Thus, time will be;

t = 346 cycles/262 cycles/s

t = 1.32 s

The time taken for the musical note to pass the stationary listener is 1.32 s.

The given parameters:

The frequency of a sound wave is defined as the number of cycles completed per second by the wave.

[tex]F = \frac{n}{t}[/tex]

where;

The time taken for the musical note to pass the stationary listener is calculated as follows;

[tex]262 = \frac{n}{t} \\\\t = \frac{n}{262} \\\\t = \frac{346}{262} \\\\t = 1.32 \ s[/tex]

Thus, the time taken for the musical note to pass the stationary listener is 1.32 s.

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Answer:

2872.8 N

Explanation:

We have the following information

m =n72kg

Δy = 18m

t = 0.95s.

From here we use the equation

Δy=1/2at2 in order to solve for the acceleration.

So a

=( 2x 18m)/(0.95s²)

= 36/0.9025

= 39.9m/s2.

From there we use the equation

F = ma

F=(72kg) x (39.9)

= 2872.8N.

2872.8N is the average net force exerted on him in the barrel of the cannon.

Thank you!

Answer:

An experiment is a procedure carried out to support, refute, or validate a hypothesis. Experiments provide insight into cause-and-effect by demonstrating what outcome occurs when a particular factor is manipulated.

Explanation:

Explanation:

Speed is the rate of change of distance with time. It is a scalar quantity with magnitude both no direction.

  Speed  = [tex]\frac{distance}{time}[/tex]

The unit is m/s or km/hr or mile/hr

Distance covered is simply the length of the path traveled.

 The unit is m or km or miles

Time taken is the duration of an event.

  The unit is seconds or minutes or hour.

We know, by conservation of energy :

[tex]\dfrac{kx^2}{2}=mgh[/tex]

Therefore,

[tex]\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}[/tex]

Putting given values, we get :

[tex]\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm[/tex]

Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.

Hence, this is the required solution.

Answer:

it makes me wish I was a cartoon

Answer:

goofy and stupid and act like a kid

Explanation:

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the formula to be used here is

ω = 2π/T

Where ω is the angular frequency (in rad/s)

T is the period - the time taken for Block A to complete one oscillation and return to it's original position.

To solve for this period T, the formula below should be used

T = 2π√m/k

where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)

Answer:

The velocity is  [tex]v =2.455 \ m/s[/tex]

Explanation:

From the question we are told that  

   The diameter of the pipe is  [tex]d = 4.25 \ cm = 0.0425 \ m[/tex]

    The magnetic field is  [tex]B = 0.575 \ T[/tex]

     The hall voltage is  [tex]V_H = 60.0 mV = 60 *10^{-3} \ V[/tex]

Generally the average fluid velocity is mathematically represented as

          [tex]v = \frac{V}{ B * d }[/tex]

=>       [tex]v = \frac{60*10^{-3}}{ 0.575 * 0.0425 }[/tex]

=>       [tex]v =2.455 \ m/s[/tex]

Answer:

When traveling behind other vehicles, there should be at least a four second space between your vehicles. When the car in front of you passes a stationary object, slowly count to yourself. If you pass the object before the allotted time, you should back off. When traveling at night or inclement weather, these times should be doubled.

Don't talk on a cell phone while driving. Phones detract from your ability to concentrate on the road and increase your chance of a collision by nearly 400%. If you must use the phone, pull over to a safe, well-lit parking lot and place your call there. After completing your call you may continue on your way.bey all speed limits and signs.

Answer:

B R is constant V increases /increases

If one increase the length of the wire in the circuit, then, R increases, V decreases, I decreases. The correct option is D.

The resistance (R) of a wire increases as its length is increased in a circuit. This is because longer wires have more resistance because resistance increases with length.

Ohm's Law states that the current (I) in the circuit will drop if the resistance (R) rises while the voltage (V) stays constant. This is due to the inverse proportionality between current and resistance.

Thus, the correct option is D.

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Answer:  Please see answer in explanation column.

Explanation:

Given that

v≈(331 + 0.60T)m/s

where Temperature, T =  14°C

v≈(331 + 0.60 x 14)m/s

v =331+ 8.4 = 339.4m/s

In our solvings, note that

f= frequency

 λ=wavelength

L = length

v= speed of sound

a) Length of the pipe is calculated using the fundamental frequency formulae that

f=v/2L

Length = v/ 2f

= 339.4m/s/ 2 x 494Hz ( s^-1)= 0.3435m

b) wavelength of the fundamental standing wave in the pipe

L = nλ/2,

λ = 2L/ n

λ( wavelength )= 2 x 0.3435/ 1

= 0.687m

c) frequency of the fundamental standing wave in the pipe

F = v/  λ

= 339.4m/s/0.687m=

494.03s^-1 = 494 Hz

d) the frequency in the traveling sound wave produced in the outside air.

This is the same as the frequency in the open organ pipe = 494Hz

e)The wavelength of the travelling sound wave produced in the outside air is the same as the wavelength calculated in b above = 0.687m

f) To play D above middle c . the distance is given by

L =v/ 2 f

= 343/ 2 x 294

=0.583m

Answer:option B

Explanation:

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s  

Explanation:

1) We can find the speed of the object by conservation of energy:

[tex] E_{i} = E_{f} [/tex]

[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]                              

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s [/tex]

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]    

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]      

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s [/tex]      

3) When the object is at the equilibrium position, the speed of the object is:

[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]    

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]      

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s [/tex]

I hope it helps you!                                                                                        

(1) the speed of the object when compression of the spring is 8 cm is 0.449 m/s

(2) the speed of the object when compression of the spring is 5 cm is 0.648 m/s

(3) the speed of the object when the spring is at equilibrium is 0.748 m/s

Given that the mass of the object, m = 5 kg

spring constant, k = 280 N/m

compression of the spring , x = 10 cm = 0.1m

(i) the spring compression is at d = 8cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.08)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.2016[/tex]

v = 0.449 m/s

(ii) the spring compression is at d = 5cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.05)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.42[/tex]

v = 0.648 m/s

(iii) the spring is at equilibrium so compression is at d = 0cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.56[/tex]

v = 0.748 m/s

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Answer:

V' = V/2

Explanation:

The voltage across a parallel plate capacitor is given as follows:

V = Q/C

where,

V = Voltage across capacitor

Q = Charge on Capacitor

C = Capacitance of Capacitor = A∈₀/d

Therefore,

V = Qd/A∈₀

where,

A = Area of plate

d = distance between plates

∈₀ = permittivity of free space

FOR CAPACITOR 1:

Q = Q

d = d

A = A

V = V

Therefore,

V = Qd/A∈₀   --------------- equation (1)

FOR CAPACITOR 2:

V' = ?

Q' = Q

d' = d

A' = 2A

Therefore,

V' = Q'd'/A'∈₀

V' = Qd/2A∈₀

V' = (1/2)(Qd/A∈₀)

using equation (1):

V' = V/2

Answer:

The distance of the bar D = 1153 km

Explanation:

The electric force is the one that takes place between electric charges.

The electric force with which two point charges are attracted or repelled at rest is directly proportional to their product, inversely proportional to the square of the distance that separates them and acts in the direction of the line that joins them.

Recall that:

Electrical force(F) = I*B*L

where;

I = the current,

B = the magnetic field strength,

L = the length of the bar

However;

From the second equation of motion,

F = Ma

Since; (F) = I*B*L

Then,

Ma = IBL,

where;

M is the mass;

a is the acceleration

Making the acceleration (a) the subject of the formula, we have

a = IBL/M

Similarly;

From the third equation of motion;

v^2= u^2+2as,

where v and u are the final velocity and the initial velocity respectively

Here u = 0

Also; let distance s = D

Then

v^2 = 2aD

where;

a = IBL/M

Making the distance D  the subject of the formula, we get:

D = v^2/2a = v2*M/(2IBL)

D = 11200² × 20/(2×2300×0.86×0.55)

D = 1153047.155 m

D = 1153 km

Answer:

1612903nm

Explanation:

Doppler effect can be regarded as the change in frequency of a wave with respect toobserver that move relative to the wave source

We can expressed as

(λo - λs)/λs = v/s

λo= peak wavelength

λs= peak wavelength observed

C= speed of light

(λo -615×10^9 )/615×10^9 = 2,500,000/(3×10^8)

1.5375=3×10^8λo -184.5

1.5375+184.5=3×10^8λo

186=3×10^8λo

λo=1612903nm

Therefore the peak wavelength that is observed when the star is traveling away from the eart to the velocity given is 1612903nm

Answer: Natural science can be divided into two main branches

Explanation:

life science and physical science. Life science is alternatively known as biology, and physical science is subdivided into branches: physics, chemistry, astronomy and Earth science.

Answer:

lucky mauld mauldgomary was an british poet...

Answer:

Because mechanical energy (the sum of potential and kinetic energy) is conserved, as the kinetic energy increases, the potential energy decreases. The maximum kinetic energy is achieved when the pendulum passes through the lowest point, and the maximum potential energy is achieved at the highest point.

Answer:

B.

Explanation:

Answer:

You have an apple pen. :)

Answer:

I have a pen

I have a apple

apple pen

Explanation:

Answer:

[tex]3978873.58\ \text{Pa}[/tex]

[tex]0.00002122\ \text{m}[/tex]

Explanation:

F = Force = 5000 N

r = Radius of circular cross section = 2 cm

l = Length of disk = 0.8 cm

A = Area = [tex]\pi r^2[/tex]

Y = Young's modulus = [tex]1.5\times 10^9\ \text{N/m}^2[/tex]

Pressure is given by

[tex]P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{5000}{\pi (2\times 10^{-2})^2}\\\Rightarrow P=3978873.58\ \text{Pa}[/tex]

The pressure on the cross section is [tex]3978873.58\ \text{Pa}[/tex]

The change in length of the cross section is given by

[tex]\Delta L=\dfrac{PL}{Y}\\\Rightarrow \Delta L=\dfrac{3978873.58\times 0.8\times 10^{-2}}{1.5\times 10^9}\\\Rightarrow \Delta L=0.00002122\ \text{m}[/tex]

The deformation produced is [tex]0.00002122\ \text{m}[/tex]

Answer:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!  

The ball rotates 6.78 revolutions.

Explanation:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!        

At the bottom the ball has the following angular speed:

[tex] \omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s [/tex]

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

[tex] sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m [/tex]

To find the revolutions we need the time, which can be found using the following equation:                

[tex] v_{f} = v_{0} + at [/tex]  

[tex] t = \frac{v_{f} - v_{0}}{a} [/tex] (1)

So first, we need to find the acceleration:

[tex] v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L} [/tex]    (2)  

By entering equation (2) into (1) we have:

[tex] t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}} [/tex]

Since it starts from rest (v₀ = 0):  

[tex] t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s [/tex]

Finally, we can find the revolutions:  

[tex] \theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev [/tex]

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!                                                                                                                                                                                          

Answer:

C. lenses refract light; mirrors do not

This question involves the concepts of reflection and refraction.

The comparison of lenses and mirrors in their interaction with light is "C. Lenses refract light; mirrors do not.".

When it comes to the interaction with light, the key difference between lenses and mirrors is the difference of refraction and reflection. Reflection means the complete rebound of the light rays after striking on a surface without any absorption or transmission. On the other hand, refraction is the  bending of light rays, while passing through a medium, without any rebound or absorption.

Lenses are tansparent from both sides, so they refract the light rays. While, mirrors are coated opaque from one side, so they reflect back the light rays.

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Answer:

A. The current in the wire increases.

Explanation:

Increasing the speed at which a magnet moves in and out of a wire coil increases the current in the wire.

This phenomenon shows the inter-relationship between electricity and magnetic fields.