PLS, THIS IS URGENT!! Read the following passage that explains the image you examined at the beginning of the lesson in order to answer the following question. Why is it important to have a range of tools that can measure all parts of the electromagnetic spectrum?
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Answer:

IVE BEEN TRYNA FIND THE ANSWER FOR THE PAST HOUR T-T

Explanation:

Answer:

3.2 x10^11 bits of storage

Explanation:

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.

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

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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:

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:

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:

woh its really exist?

Explanation:

cakra

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:

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.

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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

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

The answer is c.velocity is speed and direction

Answer:

The change in momentum of the car is 16380.8 kg.m/s

Explanation:

Given;

mass of the car, m = 1600 kg

time of motion, t = 4.20s

force of friction on the car, F = 3900 N

final velocity of the car after the brakes were applied, v = 0

The initial velocity of the car during the motion is calculated as;

[tex]F = ma = \frac{mu}{t} \\\\mu = Ft\\\\u = \frac{Ft}{m} \\\\u = \frac{3900\times 4.2}{1600} \\\\u = 10.238 \ m/s[/tex]

The change in momentum of the car is calculated as;

ΔP = mu

ΔP = 1600 x 10.238

ΔP = 16380.8 kg.m/s

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

Explanation: the fourth is correct the fifth is incorrect

Explanation:

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

Answer:

the tensile stress of the steel wire is 2.5 x 10³ Nm⁻².

Explanation:

Given;

applied force, F = 5000 N

cross sectional area of the steel wire, A = 0.2 m²

The tensile stress of the steel wire, is calculated as;

[tex]\sigma = \frac{F}{A} \\\\\sigma = \frac{500}{0.2} \\\\\sigma = 2500 \ N/m^2\\\\\sigma = 2.5 \ \times \ 10^3 \ Nm^{-2}[/tex]

Therefore, the tensile stress of the steel wire is 2.5 x 10³ Nm⁻².

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:

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:

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:

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:

mass is 5 kg

Explanation:

F=m/a

M=f/a

F=5000

a=1000

5000/1000=5

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:

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: C

Explanation:

The model which represents best of transverse wave with very high crests and low troughs is the third graph. Hence, option C is correct.

Sound is a physical perturbation from an equilibrium position that travels through an elastic medium of material. It is also possible to define sound solely subjectively, as that which is regarded by the ear, but this definition lacks clarity and is overly constrictive because it is useful to discuss sounds that are produced by devices other than the human ear, such as dog whistles and sonar hardware, which cannot be done here by the auditory system.

The characteristics of sound waves should be examined first in any study of sound. Transverse and longitudinal waves are the two fundamental forms of waves, and they differ by the direction wherein they move.

The first graph is uniform trough and crest, but the third one has higher crest but lower trough.

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