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insens350 [35]
4 years ago
5

How to classify an inner planet vs. outer planet

Physics
1 answer:
zvonat [6]4 years ago
7 0
Outer planets are farther away and made up of gases. Inner planets closer. It's pretty much self explanatory. Hope this helps.
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List the three components of effective communication and give an example of each
Marrrta [24]

Answer:

The three components of effective communication are:

  1. Sending Communication
  2. Receiving Communication
  3. Feedback

Explanation:

<h3>1) Sending Communication</h3>

One should think about what he is going to say, structure his message and focus clearly on a the purpose of sending a message. For example if an employer wants to communicate an employee to improve his performance, he should focus hid communication on the results rather than on his failures.

<h3 /><h3>2) Receiving Communication</h3>

The person on the other end of the communication should be a good listener. For example in the example given, even if the employer effectively communicates his side of the communication, but the employee is not paying his full attention to what his boss is saying, he wouldn't be able to make much of a difference.

<h3>3) Feedback</h3>

After sending and receiving communication, feedback should be given to complete the communication effectively. For example the employer may tell his boss about what he understood from his boss's speech, or what will he do to improve his performance.

5 0
3 years ago
Yellow-green light has a wavelength of 560 nm. What is its frequency?
Natasha2012 [34]
5.4 x 1014Hz
wavelength x frequency = the speed of light
7 0
3 years ago
Read 2 more answers
Can someone help me please??<br> This question is confusing to me:(
puteri [66]
Its d. the water isn't moving the fish, and that's how they remain stationary. they don't have any currents pushing on them. if this is wrong, its c, which is practically the same thing. but its most likely d.
7 0
4 years ago
A balloon is rising vertically upwards at a velocity of 10m/s. When it is at a height of 45m from the ground, a parachute bails
harina [27]

(a) 30.9 m

Let's analyze the motion of the parachutist. Its vertical position above the ground is given by

y=h+ut+\frac{1}{2}gt^2

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t=3 s , we find the height of the parachutist when it opens the parachute:

y=45 m+(10 m/s)(3 s)+\frac{1}{2}(-9.8 m/s^2)(3 s)^2=30.9 m

(b) 44.1 m

Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by

y = h + ut

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

Substituting t = 3 s, we find

y = 45 m + (10 m/s)(3 s) = 75 m

So the distance between the balloon and the parachutist after 3 s is

d = 75 m - 30.9 m = 44.1 m

(c) 8.2 m/s downward

The velocity of the parachutist at the moment he opens the parachute is:

v = u +gt

where

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t = 3 s,

v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s

where the negative sign means it is downward

After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of

a =+5 m/s^2

where we put a positive sign since this time the acceleration is upward.

The total distance he still has to cover till the ground is

d = 30.9 m

So we can find the final velocity by using

v^2-u^2 = 2ad

where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:

a = -5 m/s^2

Solving for v,

v=\sqrt{u^2 +2ad}=\sqrt{(19.4 m/s)^2+2(-5 m/s^2)(30.9 m)}=8.2 m/s

(d) 5.24 s

We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using

a=\frac{v-u}{t}

where

a = -5 m/s^2 is the deceleration

v = 8.2 m/s is the final velocity

u = 19.4 m/s is the initial velocity

t is the time

Solving for t, we find

t=\frac{v-u}{a}=\frac{8.2 m/s-19.4 m/s}{-5 m/s^2}=2.24 s

And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

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