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1.Cell phones convert sound waves into______waves.
-light waves
-sound waves
-radio waves

2.Electromagnetic waves used in cell phone communications are called?
-microwave
-radio wave
-infrared
-all of the above

3.To send out a radio signal far and wide...it is called?
-broadcast
-casting
-antenna

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.

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

Explanation: the fourth is correct the fifth is incorrect

Answer:

80 N

Explanation:

mass = 20 kg

acceleration = 4 m/s^2

Force = mass * acceleration

         = 20 *4  

         = 80 N

Answer:

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

Explanation:

Answer:

3.2 x10^11 bits of storage

Explanation:

Force of a crate : 300 N

Given

0.25 meters wide and 1.5 meters long

Pressure = 800 Pa

Required

Force

Solution

P = F/A

A = 0.25 x 1.5

A = 0.375 m²

P = 800 Pa = 800 N/m²

Input the value :

F = P x A

F = 800 N/m² x 0.375 m²

F = 300 N

v=u+at

v=0+8.0×30

v=240

Answer:

0.8A

Explanation:

because im very intelligent and beautiful so the answer is correct and its 0.8A

Answer:

See the explanation below.

Explanation:

Work in physics is defined as the product of force by the distance that the body travels in the direction of the force.

It can be represented by means of the following equation.

[tex]W=F*d[/tex]

where:

W = work [J]

F = force [N]

d = distance [m]

In the given example the work is zero since the tree does not move, therefore the bear exerts a force on the tree. But there is no talk of movement of the tree, therefore the work is zero.

Answer:

D. blinking ofeyea

Explanation:

not really sure

PE = 450 m

Given

The initial problem might be like this

A 1.0-kg ball is thrown into the air with an initial velocity of 30 m/s.

Required

Potential energy

Solution

h max = vo²sin²θ/2g

h max = 30²sin²90/2.10

h max = 45 m

PE = m.g.h

PE = 1 x 10 x 45

PE = 450 m

The magnitude of the pressure : P = 56,250 N/m

Given

Area of 0.004 m²

Force 225 N

Required

the magnitude of the pressure

Solution

Pressure = force exerted per unit of surface area(perpendicular)

Can be formulated :

P = F/A

Input the value :

P = 225 N / 0.004 m

P = 56,250 N/m

Answer:

C. Some waves

Explanation:

The answer is some waves because transverse waves are not the only types of waves we have on the earth. We also have longlitudal waves, which means only some waves can only be transverse. The answer to the question should be C. Some waves.

Answer:

to have courage and show the strength of your character. ... A person with true grit has passion and perseverance. Goals are set and followed through.

Explanation:

for example: i had the passion and the courage to finish my school work and chores for i can go out with my friends and to get out of the house for a bit and have fun so i had the courage to finish chores from the house and do some schoolwork for i can finish ad hang out with some friends. so i had the strength to finish everything because i didn't want to be doing school work and chores any more so i had the passion to finish up quickly  

Momentum = (mass) x (velocity)

Momentum m₁  =  (6 kg) x (13 m/s east)  =  78 kg-m/s east

Momentum m₂  =  (14 kg) x (7 m/s east)  =  98 kg-m/s east

Since they're both in exactly the same direction you can simply addum up to get their total momentum.  It's 176 kg-m/s east .

The  momentum of a system that consists of two objects should be 176 kg m/s on east hand side.

Since we know that

Momentum = (mass) x (velocity)

So,

Momentum m₁  =  (6 kg) x (13 m/s east)  =  78 kg-m/s east

Momentum m₂  =  (14 kg) x (7 m/s east)  =  98 kg-m/s east

Since they are on same direction so the total momentum should be

= 78 + 98

= 176 kg-m/s east

hence, The  momentum of a system that consists of two objects should be 176 kg m/s on east hand side.

Learn more about momentum here: https://brainly.com/question/24815698

Answer:

A device that is used to break an electric circuit is called electric switch

Answer:

A. Laser 1 has its first maximum closer to the central maximum.

B. 0.077 mm

Explanation:

A) Which laser has its first maximum closer to the central maximum?

Since for a double slit interference, sinθ = mλ/d where θ is the angle between the central axis and the maximum point, m = order of fringe, λ = wavelength of light and d = width of slit.

Also tanθ = y/D where θ is the angle between the central axis and the maximum point, y = distance between maximum point and central axis and D = distance between slit and screen = 4.60 mm.

Since θ is small,  sinθ ≅  θ and  tanθ ≅ θ

So, sinθ ≅ tanθ  

mλ/d = y/D

So, y = mλD/d for the first maximum, m = 1.

So,  y = (1)λD/d

y = λD/d

For the Laser 1, λ = d/20

So, y = λD/d

y = (d/20)D/d  

y = D/20

y = 4.60 mm/20

y = 0.23 mm

For the Laser 2, λ' = d/15

So, y' = λ'D/d

y' = (d/15)D/d  

y' = D/15

y' = 4.60 mm/15

y' = 0.307 mm

Since y = 0.23 mm < y' = 0.307 mm, Laser 1 has its first maximum closer to the central maximum.

B) What is the distance between the first maxima (on the same side of the central maximum) of the two patterns?

The distance between the first maxima (on the same side of the central maximum) of the two patterns is Δy = y' - y = 0.307 mm - 0.23 mm = 0.077 mm

Answer:

Explanation:

[tex]V^2=V^2_i+2a(x_f-x_i)[/tex]

[tex]0.76^2=0^2+2a(15.2)[/tex]

[tex]0.5776=30.4a[/tex]

[tex]a=0.019m/s^2[/tex]

Answer:

(10 L)(100° C) + (90 L)(10°) = (100 L)T

1,000 + 900 = 100T

1,900 = 100T

100T = 1,900

T = 1,900/100

Explanation:

Answer:

KE = 1/2 m v^2

Explanation:

Answer:

A 50 kg ball traveling at 20 m/s would have 4 times more kinetic energy.

A 50 kg ball traveling at 5 m/s would have 4 times less kinetic energy.

A 50 kg person falling at 10 m/s would have the same kinetic energy.

Explanation:

hope this helps:)

Answer: c

Explanation:

Sound waves cannot travel through a medium

Answer:

(a) The potential difference between the spheres is 750 kVA

(b) The charge on the smaller sphere is 6.[tex]\overline 6[/tex] μC

(c) The charge on the smaller sphere, Q₁ = 13.[tex]\overline 3[/tex] μC

Explanation:

(a) The given parameters are;

The charge on the inner sphere, q = 5.00 μC

The radius of the inner sphere, r = 3.00 cm = 0.03 m

The charge on the larger sphere, Q = 15.0 μμC

The radius of the larger sphere, R = 6.00 cm = 0.06 m

The potential difference between two concentric spheres is given according to the following equation;

[tex]V_r - V_R = k \times q \times \left ( \dfrac{1}{r} - \dfrac{1}{R} \right)[/tex]

Where;

R = The radius of the larger sphere = 0.06 m

r = The radius of the inner sphere = 0.03 m

q = The charge of the inner sphere = 5.00 × 10⁻⁶ C

Q = The charge of the outer sphere = 15.00 × 10⁻⁶ C

k = 9 × 10⁹ N·m²/C²

Therefore, by plugging in the value of the variables, we have;

[tex]V_r - V_R = 9 \times 10^9 \times 5.00 \times 10^{-6} \times \left ( \dfrac{1}{0.03} - \dfrac{1}{0.06} \right) = 750,000[/tex]

The potential difference between the spheres, [tex]V_r - V_R[/tex] = 750,000 N·m/C = 750 kVA

(b) When the spheres are connected with a wire, the charge, 'q', on the smaller sphere will be added to the charge, 'Q', on the larger sphere which as follows;

[tex]Q_f[/tex] = Q + q = (5 + 15) × 10⁻⁶ C = 20 × 10⁻⁶ C

[tex]Q_f[/tex] = 20 × 10⁻⁶ C

From which we have;

Q₁/Q₂ = R/r

Where;

Q₁ = The new charge on the on the larger sphere

Q₂ = The new charge on the on the smaller sphere

[tex]Q_f[/tex] = 20 × 10⁻⁶ C = Q₁ + Q₂

∴ Q₁ = 20 × 10⁻⁶ C - Q₂ = 20 μC - Q₂

∴ (20 μC - Q₂)/Q₂ = 0.06/(0.03) = 2

20 μC - Q₂ = 2·Q₂

20 μC = 3·Q₂

Q₂ = 20 μC/3

The charge on the smaller sphere, Q₂ = 20 μC/3 = 6.[tex]\overline 6[/tex] μC

(c) Q₁ = 20 μC - Q₂ = 20 μC - 20 μC/3 = 40 μC/3

The charge on the smaller sphere, Q₁ = 40 μC/3 = 13.[tex]\overline 3[/tex] μC.

The potential difference is 7.5*10^5V and the charge on the smaller sphere is 6.67uC while the charge on the larger sphere is 13.34uC

Data;

for the inner sphere;

[tex]v_i = \frac{kq}{r} = \frac{(9*10^9)(5*10^-^6)}{0.03} \\v_i = 1.5*10^6V[/tex]

for the outer sphere;

[tex]v_o = \frac{KQ}{R} = \frac{(9*10^9)(15*10^-^6)}{0.06} = 2.25 * 10^6V[/tex]

The difference in potential is

[tex]\delta V = v_o - v_i = 2.25*10^6 - 1.5*10^6 = 7.5*10^5V[/tex]

[tex]q + Q = 5 + 15 = 20 \mu C = 20*10^-^6C ..eq(i)[/tex]

The sharing of charge continues till they attain a point of equal potential

[tex]v_i = v_o \\\frac{kq}{r} = \frac{kQ}{R} \\\frac{q}{0.03} = \frac{Q}{0.06} \\Q= 2q ...eq(ii)\\[/tex]

let's solve for equation (i) and equation (ii)

[tex]q+2q = 20\mu C\\q = 6.67 \mu C[/tex]

The charge on the smaller sphere is 6.67uC

The charge on the larger sphere is

[tex]Q = 2q\\q = 6.67 \mu C\\Q = 2 * 6.67 = 13.34 \mu C[/tex]

The charge on the larger sphere is 13.34uC

Learn more on potential difference on a charged particle here;

https://brainly.com/question/9383604

Answer:

As the Sun ages it will enter another stage of stellar evolution where it's atmosphere begins to inflate.As the Sun heats up and expands, life on Earth will become increasingly difficult. Long before the Sun becomes a red giant some 4 or 5 billion years from now, our planet will be rendered uninhabitable.

~I HOPE IT HELPS♡

Answer:

it is 9m/s , hope it helps