Question 8 of 10
When you see your reflection in a flat (plane) mirror, what type of image are
you seeing?
A. Smaller
B. Virtual
C. Larger
D. Real

Answer:

a) ρ = 2.33 10⁵ C / m³, b) INSULATING

Explanation:

a) For this exercise we use the important concept of density

            ρ = Q / V

The volume of a sphere is

           V = 4/3 πr3

            V = 4/3 π (1.27 10⁻²)³

            V = 8.58 10⁻⁶ m³

let's calculate

           ρ = 2 / 8.58 10⁻⁶

           ρ = 2.33 10⁵ C / m³

b) in a conductor the charges are mobile, therefore due to electoral repulsion all the charges must concentrate on the surface.

In an insulator the charges are fixed, therefore they are distributed throughout the volume

Consequently the sphere must be INSULATING

Answer:

19.6133 newton.

...?.......

Explanation:

don't know don't know don't know

Answer:

the value of the charge is   q₃ = [tex]- \frac{4}{9} q[/tex]- 4/9 q, in the position  x= l/3

Explanation:

The forces in this system are given by Coulomb's law

           F = [tex]k \frac{q_1 q_2}{ r^2}[/tex]

As the forces are vector, we must add them as a vector, also let us take that charges of the same sign repel and charges of the opposite sign attract, therefore for the system to be in equilibrium the third charge must be of the opposite sign, that is, NEGATIVE .

Let's analyze the situation for card charge, let's use the indexes 1 for the charge q₁ = + q located at the origin (x₁ = 0), index 2 for the face q₂ = + 4q located at x₂ = l and the index 3 for the third charge

Let's find the location of charge 3 so that it is in balance

               ∑ F = F₁₃ - F₂₃ = 0

               F₁₃ = F₂₃

we seek every force

              F₁₃ = k [tex]\frac{q_1q_3 }{x_{13}^2}[/tex]

              F₂₃ = k [tex]\frac{q_2q_3}{ x_{23}^2}[/tex]

the distances are

              x₁₃ = x-0 = x

              x₂₃ = l -x

we substitute

              k \frac{q_1q_3 }{x^2} = k \frac{q_2q_3 }{(l-x)^2}

we solve

             (l-x)² = [tex]\frac{q_2}{q_1}[/tex]   x²

              l² - 2lx + x² = \frac{4q}{q} x²

              3x² + 2l x - l² = 0

we solve the quadratic equation

              x = [-2l ± [tex]\sqrt{4l^2 + 4\ 3\ l^2}[/tex] / (2 3)

              x = [-2l + 2l [tex]\sqrt{1+3}[/tex]] / 6 = 2l [-1 ± 2] / 6

              x₁ = -l

              x₂ = l / 3

as charge 3 must be between the two charges the correct answer is

              x = l / 3

with charge 3 in this location it remains in equilibrium regardless of the value of its NEGATIVE charge.

now let's analyze the force on the other charges

charge 1 so that it is in balance

           ∑F = F₁₃ - F₁₂ = 0

           F₁₃ = F₁₂

          [tex]k \frac{q_1q_3}{(l/3)^2} = k \frac{q_1 q_2}{ l^2}[/tex]

           l² q₃ = q₂ (l/3)²

           q₃ = q₂ / 9

            q₃ = 4q / 9

            q₃ = [tex]\frac{4}{9} q[/tex]

in summary the value of the charge is   q₃ = [tex]- \frac{4}{9} q[/tex]- 4/9 q

in the position  x= l/3

Answer: the first is incorrect the second is correct the third is correct

Explanation: the fourth is correct the fifth is incorrect

Answer:

woh its really exist?

Explanation:

cakra

Answer:

2.92 m

Explanation:

As we know, frequency × Wavelength = Speed of light

so here frequency of 102.7 MHz can be written as 102.7× 10⁶ Hz..

So Lambda (wavelength) = 3×10⁸/ 102.7 × 10⁶ which gives 2.92 metres or 2.92 × 10¹⁰ Å

Answer:

We conclude that the quantity of  the charge that flows through the wire in 50 s will be 45 C.

Explanation:

Given

To determine

We need to find the quantity of  the charge that flows through the wire in 50 s.

Important Tip:

A current of 1 Ampere = 1 Coulomb  of charge flowing in 1 second

Using the formula involving charge and current

[tex]I=\frac{Q}{t}[/tex]

where:

now substituting I = 0.9 and t = 50 in the formula

[tex]I=\frac{Q}{t}[/tex]

[tex]0.9\:=\:\frac{Q}{50}[/tex]

switch sides

[tex]\frac{Q}{50}=0.9[/tex]

Multiply both sides by 50

[tex]\frac{50Q}{50}=0.9\cdot \:50[/tex]

Simplify

[tex]Q=45[/tex] C

Therefore, we conclude that the quantity of  the charge that flows through the wire in 50 s will be 45 C.

Answer:Number of waves =2.2

Explanation:

The frequency of A wave is defined as the number of waves moving  a fixed point per second which is given as

f=[tex]\frac{N}{t}[/tex]

F= Frequency

N= number of waves

t= time

We first find the frequency

Given velocity of the wave, v = 13m/s

and wavelength, λ as 6meters(distance between two consecutive crest)

, we fiind the frequency using

v= f  λ

frequency = v /  λ

= (13m/s ) / 6

=2.16s-1

f =  [tex]\frac{N}{t}[/tex]

2.16s₋¹=  [tex]\frac{N}{1s}[/tex]

Therefore N, Number of waves =  2.16 waves  rounded up to 2.2 waves

The answer is c.velocity is speed and direction

Answer:

Approximately [tex]1\; \rm kg[/tex]. (Assumption: [tex]g = 9.8\; \rm m \cdot s^{-2}[/tex].)

Explanation:

Let the mass of this ball be [tex]x\; \rm kg[/tex].

Initial gravitational potential energy of this ball: [tex]m \cdot g \cdot h \approx (39.2\, x) \; \rm J[/tex].

Just before the ball hits the ground, all [tex](39.2\, x)\; \rm J[/tex] of gravitational potential energy would have been converted to kinetic energy. Calculate the velocity of the ball at that moment:

[tex]\begin{aligned} \text{kinetic energy} = \frac{1}{2}\, m \cdot v^{2} \end{aligned}[/tex].

Therefore, right before hitting the ground, the velocity of the ball would be:

[tex]\begin{aligned} v&= \sqrt{\frac{2\, (\text{kinetic energy})}{m}} \\ &= \sqrt{\frac{2 \times (39.2\, x)}{x}}\\ & \approx \sqrt{2 \times 39.2} \approx 8.85\; \rm m \cdot s^{-1}\end{aligned}[/tex].

The momentum [tex]p[/tex] of an object of mass [tex]m[/tex] and velocity [tex]v[/tex] would be [tex]p = m \cdot v[/tex]. Rewrite this equation to find an expression for mass [tex]m\![/tex] given velocity [tex]v\![/tex] and momentum [tex]p\![/tex]:

[tex]\displaystyle m = \frac{p}{v}[/tex].

Right before collision, the momentum of this ball is [tex]p = 8.85\; \rm kg \cdot m \cdot s^{-1}[/tex] while its velocity is [tex]v \approx 8.85\; \rm m \cdot s^{-1}[/tex]. Therefore, the mass of this ball would be:

[tex]\begin{aligned}m &= \frac{p(\text{right before landing})}{v(\text{right before landing})} \\ &\approx \frac{8.85\; \rm kg \cdot m \cdot s^{-1}}{8.85\; \rm m \cdot s^{-1}} \approx 1\; \rm kg\end{aligned}[/tex].

Answer:

gravity

Explanation:

there eould be a different gravitational strenght on both of the planets causing it to weigh more, or less

In mars, the effect of gravity on the object is weaker compared to while on earth. This weakness is what causes the reduction in the weight of the bag of sugar while on mars.

We must understand that the mass of an object may differ depending on its location on the planet. This is due to the effect of gravity on the object.

Gravity is simply defined as any force which attracts an object towards the centre of the earth.

From the question given, we are told that the weight of sugar is 10N on the earth and 4N on Mars. This difference in their mass is attributed to the effect of gravity on the object.

In mars, the effect of gravity on the object is weaker compared to while on earth. This weakness is what causes the reduction in the weight of the bag of sugar while on mars.

Learn more here: https://brainly.com/question/17205409

Explanation:

Hey there!

Given that 4kg ball struck by 12 N force.

So, we need acceleration.

Force = Mass × Acceleration

And, so

Acceleration = Force / Mass.

Acceleration = 12/4

Acceleration is equal to 3m/s² in SI units.

Hope it helps :)

Answer:

i. 15.6 m/s

ii. I = 1.44 KNs

Explanation:

The impulse, I, on a body is the product of force applied on it and the time it acts.

i.e I = F x t

Impulse is sometimes expressed as the change in momentum of a body. It is measured in Ns.

i. mass, m, of the player = 92 kg

initial velocity of the player, u = 9.4 m/s

final velocity of the player, v = 6.2 m/s

Since he bounces back on hitting the pole, then the sign of initial and final velocities are of opposite sign.

So that,

change in velocity of the player = final velocity - initial velocity

                                          = 6.2 - (-9.4)

                                         = 6.2 + 9.4

                                         = 15.6 m/s

change in velocity of the player is 15.6 m/s

ii. Impulse, I = m(v - u)

                    = 92 x 15.6

                    = 1435.2

Impulse on the player is 1.44 KNs.

Answer:

YAYA MAN

Explanation:

Answer:

Conduction, radiation and convection all play a role in moving heat between Earth's surface and the atmosphere. Since air is a poor conductor, most energy transfer by conduction occurs right near Earth's surface

Answer:

Gold is metallic, with a yellow color when in a mass, but when finely divided it may be black, ruby, or purple. It is the most malleable and ductile metal; 1 ounce (28 g) of gold can be beaten out to 300 square feet. It is a soft metal and is usually alloyed to give it more strength.

Formula  --

Molecular Weight (lb/m o l)  28.96

Critical Temp. (°F)  N/A

Critical Pressure (p s i a)  N/A

Boiling Point (°F)  -317.8

Melting Point (°F)  N/A

Psat  70°F (p s i a)  (note 1)

Explanation:

The final velocity of the bus  is 9.83 m/s

The given parameters;

initial velocity of the bus, = 30 km/h

constant increase in speed, = 1.5 m/s

time of motion, t = 6.8 s

The initial velocity of the bus in m/s;

[tex]v = 30 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 8.33 \ m/s[/tex]

The acceleration of the bus;

[tex]a = \frac{\Delta v}{\Delta t} = \frac{1.5 \ m/s}{6.8 \ s } = 0.22 \ m/s^2[/tex]

The final velocity of the bus at the given time of motion;

[tex]v= u + at\\\\v = 8.33 + (0.22 \times 6.8)\\\\v = 9.83 \ m/s[/tex]

Thus, the final velocity of the bus  is 9.83 m/s

Learn more here: https://brainly.com/question/17956487

b...radio waves answer

Answer:

350 J of energy is transferred by radiation.

Explanation:

Given that,

600 J of energy is transferred from the chemical  store in the battery to the thermal store of surroundings.

Heat energy = 250 J

Sound energy = x

According to the conservation of energy,

600 = 250 +x

x = 600-250

x = 350 J

So, 350 J of energy is transferred by radiation.

Answer:

177 coulomb

Explanation:

Given data

I=2.95A

T=60

The relationship between current I and quantity of charge Q.

Q=IT

Substitute

Q=2.95*60

Q=177 coulomb

Answer:

Speed = 106.67mph

Explanation:

Let the first motion be A.

Given the following data;

Speed A = 40mph

Time A = 4 hours

To find the distance covered in A;

Distance = speed * time

Distance = 40 * 4

Distance = 160 miles.

To find the speed required to cover the above distance in 1.5 hours;

Speed = distance/time

Speed = 160/1.5

Speed = 106.67mph

Answer: 5/17 seconds or roughly .3 seconds

Explanation:

We will hear our echo once it has traversed the canyon to reach the end, bounce off, then return back to our ears.

To find how long it travels in one direction:

[tex]\frac{50m}{340 \frac{m}{s}} = \frac{5}{34} s[/tex] (or about .14 seconds)

For it to return back to us we double that time.

The time for it to reach our ears would be 5/17 seconds or roughly .3 seconds.

Answer:

0.57 kg

Explanation:

We are given that

Force, F=20 N

Acceleration ,a =[tex]35m/s^2[/tex]

We have to find the mass of Aero-plane.

To find the value of mass we will divide the force by acceleration.

We know that

Force, F=ma

Using the formula

[tex]20=m\times 35[/tex]

[tex]\frac{20}{35}=m[/tex]

[tex]m=0.57 kg[/tex]

Hence, the mass of the aero-plane=0.57 kg

Answer:

the mass of object is 2000

Answer:

Newtons 3sd law

Force = mass × acceleration

3rd photo about momentum

Answer:

-606.73C

Explanation:

According to coulombs law

F = kq1q2/r^2

Substitute the given value

83,506 = 9*10^9(-0.198)q2/(3004)^2

83,506 =  -1.242*10^9q2/9,024,016

-1.242*10^9q2 = 7.54*10^11

q2 = -7.54*10^11/1.242*10^9

q2 = -606.73C

hence the magnitude of the other charge is -606.73C

Answer:

b. d. f.

Explanation:

Got it right on Edge

Answer:

2. Gina gathers information about a court case.

4. Hana interviews and advises a person who has been accused of a crime.

6. Dewayne creates the paperwork for a business contract.