What caused day and night on a planet

What is the difference between reflection and refraction? What changes and what does not change.

wolverine [178]

Answer:

Reflection involves a change in direction of waves when they bounce off a barrier. Refraction of waves involves a change in the direction of waves as they pass from one medium to another.

Explanation:

8 0

1 year ago

Write a rule for the sequence. 3, -3, -9, -15. A. Start with 3 and add -6 repeatedly B. Start with -6 and add 3 repeatedly C. St

faust18 [17]

Substract two consecutive terms of the sequence to see if there is a common difference:

$\begin{gathered} (-3)-(3)=-3-3=-6 \\ (-9)-(-3)=-9+3=-6 \\ (-15)-(-9)=-15+9=-6 \end{gathered}$

As we can see, there is a common difference of -6.

Then, if a number of the sequence is given, the next one can be found by adding -6 (which is the same as subtracting 6).

Notice that the first term of the sequence is 3.

Then, the rule for the sequence is to start with 3 and add -6 repeatedly.

Therefore, the correct choice is option A) Start with 3 and add -6 repeatedly.

7 0

1 year ago

An electron with a speed of 1.2 × 107 m/s moves horizontally into a region where a constant vertical force of 5.2 × 10-16 N acts

Aliun [14]

Answer: 0.642mm

Explanation: F= force = 5.2×10^-16 N,

v = velocity of electron = 1.2×10^7 m/s,

m = mass of electron = 9.11×10^-31 kg.

We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.

Recall that f = ma.

Where a = acceleration

This acceleration of vertical because it occurred when the object deflected.

5.2×10^-16 = 9.11×10^-31 (ay)

ay = 5.2×10^-16 / 9.11×10^-31

ay = 5.71×10^14 m/s²

For the horizontal motion, x = vt

Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,

By substituting the parameters, we have that

0.019 = 1.27×10^7 × t

t = 0.019 / 1.27 × 10^7

t = 1.5×10^-9 s

The vertical distance (y) is gotten by using the formulae below

y = ut + at²/2

but u = 0

y = at²/2

y = 5.71×10^14 × (1.5×10^-9)²/2

y = 0.00128475/2

y = 0.000642m = 0.642mm

7 0

2 years ago

Read 2 more answers

Three small children of mas 20.0 kg, 24.0 kg, and 16.0 kg, respectively, hold hands, as

olya-2409 [2.1K]

Answer:

hope this helps youuuuu

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7 0

3 years ago

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 m /

Sidana [21]

Answer:

The maximum height is 2881.2 m.

Explanation:

Given that,

Acceleration = 29.4 m/s²

Time = 7.00 s

We need to calculate the distance

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times29.4\times7^2$

$s=720.3\ m$

We need to calculate the velocity

Using formula of velocity

$v=a\times t$

Put the value into the formula

$v=29.4\times7$

$v=205.8\ m/s$

We need to calculate the height

Using formula of height

$H=\dfrac{v^2}{2g}$

Put the value into the formula

$H=\dfrac{(205.8)^2}{2\times9.8}$

$H=2160.9\ m$

We need to calculate the maximum height

Using formula for maximum height

$H'=H+s$

Put the value into the formula

$H'=2160.9+720.3$

$H'=2881.2\ m$

Hence, The maximum height is 2881.2 m.

4 0

3 years ago