Which of the following statements is true of electromagnetic waves but NOT mechanical waves.

Which example has the least kinetic energy

Stells [14]

Answer:

D

Explanation:

Let’s calculate the kinetic energy for all of the choices.

a. (1/2)(100)(100)^2 = 50(10000)=500,000

b. (1/2)(100)(1)^2 = 50

c. (1/2)(10)(100)^2 = 5(10000) = 50,000

d. (1/2)(1)(1)^2 = 0.5

We can see that (d) has the least kinetic energy.

5 0

2 years ago

PLEASE HELP 100 POINTS! Please fill in the scale distance from sun and diversion factor, listing off the numbers works

steposvetlana [31]

Solved your another question same like this with scaling to Cm this time we go with metre(m)

Scale factor

1km=10³m
1m=10^{-3}km

Mercury

58000m

Ven us

108000m

Earth

150000m

Mars

228000m

Jupiter

778000m

Saturn

1430000m

Uranus

2870000m

Neptune

4500000m

7 0

2 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

If a charge at 60c flow in a conductor for 30 second then the current that flow in a conductor is

saw5 [17]

Explanation:

<h3>Given</h3>

- Charge = 60c

time = 30 sec

current

<h3>Solution </h3>

Current = Charge/time

I = V/T

I = 60/30

I = 2 ampere

More to know -

I = Current

V = Charge

T = Time

3 0

3 years ago

An airplane pilot flies due west at a speed of 216 km/hr with respect to the air. After flying for a half an hour, the pilot fin

Yuki888 [10]

Answer:

speed wind Vw = 54.04 km / h θ = 87.9º

Explanation:

We have a speed vector composition exercise

In the half hour the airplane has traveled X = 108 km to the west, but is located at coordinated 119 km west and 27 km south

Let's add the vectors in each coordinate axis

X axis (East-West)

-Xvion - Xw = -119

Xw = -Xavion + 119

Xw = 119 -108

Xwi = 1 km

Calculate the speed for time of t = 0.5 h

Vwx = Xw / t

Vwx= 1 /0.5

Vwx = - 2 km / h

Y Axis (North-South)

Y plane - Yi = -27

Y plane = 0

Yw = 27 km

Vwy = 27 /0.5

Vwy = 54 km / h

Let's use the Pythagorean theorem and trigonometry to compose the answer

Vw = √ (Vwx² + Vwy²)

Vw = R 2² + 54²

Vw = 54.04 km / h

tan θ = Vwy / Vwx

tan θ = 54/2 = 27

θ = Tan⁻¹ 1 27

θ = 87.9º

The speed direction is 87. 9th measure In the third quadrant of the X axis in the direction 90-87.9 = 2.1º west from the south

5 0

3 years ago