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3 years ago

What is involved in a career exploration

Physics

2 answers:

tensa zangetsu [6.8K]3 years ago

8 0

Answer:

would be...Learning about the salary, education and job market of a particular career.

Explanation:

Dmitriy789 [7]3 years ago

4 0

Answer:

Learning about the salary, education and job market of a particular career.

so your answer is option E

Hope this helps

Have a great day (:

Explanation:

You might be interested in

How to find the velocity of an object in a circular path?

Luda [366]

First of all, let's just talk about the speed, and not get wound up
in the velocity. OK ?

If a fly is sitting on the rim of the wheel and the wheel is rotating, then for
each full revolution of the wheel, the fly travels the circumference of the
wheel, which is (2 π) x (radius of the wheel).

In 'N' revolutions, the fly travels (2 N π) x (the radius). and so on.

So if the wheel is going, let's say 71 revs per minute (RPM), a point
on the rim is moving at (2 π times 71) x (the radius) per minute.

Another way to say it:

Speed of a point on the circle = (2 π) x (rotation frequency) x (radius).

The 'rotation frequency' takes care of the unit of time, and the 'radius'
takes care of the unit of length, so the result is a speed.

7 0

3 years ago

Read 2 more answers

Suppose you put an ice cube into a cup of hot tea. In what direction does energy in the form of heat flow? What happens to the i

Semenov [28]

Answer:

The energy flows between the ice and the tea equally. The table below shows the temperatures of several different objects made of the same material.

6 0

3 years ago

You are trying to hear your friend give directions to new store in town. But from your distance (1 point) of 15 m you only hear

Marat540 [252]

Answer:

option D

Explanation:

given,

Intensity of sound = 20 dB

distance = 15 m

intensity of sound is increased to = 50 dB

distance between the sound level = ?

Using relation

$L_2 = L_1 - |20(log \dfrac{r_2}{r_1})|$

L₁ = 20 dB L₂ = 50 dB r₁ = 15 m r₂ = ?

$log (\dfrac{r_2}{r_1}) = \dfrac{L_1 -L_2}{20}$

$\dfrac{r_2}{r_1}= 10^{\dfrac{|L_1 -L_2|}{20}}$

$r_2 =r_1 10^{\dfrac{|L_1 -L_2|}{20}}$

$r_2 =15 \times 10^{\dfrac{|20-50|}{20}}$

$r_2 =15 \times 10^{-1.5}$

r₂ = 0.47 m

r₂ = 47 cm

hence, the correct answer is option D

7 0

3 years ago

Can someone please help me with this physics question? I'm desperate!

Lelu [443]

Answer:

a) 2·√10 seconds

b) Linda should be approximately 30.6 meters

c) Jenny's speed at the 100-m mark is approximately 6.325 m/s

Explanation:

The speed with which Linda is running = 8.6 m/s

The point Jenny starts = The 80-m mark

The acceleration of Jenny = 1.0 m/s²

a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows

Δs = u·t + (1/2)·a·t²

Δs = Distance = 100-m - 80-m = 20-m

u = The initial velocity of Jenny = 0

a = Jenny's acceleration = 1.0 m/s²

∴ 20 = 0×t + (1/2) × 1 × t² = t²/2

20 = t²/2

t = √(20 × 2) = 2·√10

The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds

b) The distance Linda runs in t = 2·√10 seconds, d = v × t

Given that Linda's velocity, v = 8.6 m/s, we have;

d = 8.0 × 2·√10 = 16·√10

The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters

Therefore, Linda should be approximately (50.6 - 20) meters = 30.6 meters behind Jenny when Jenny starts running

c) Jenny's speed at the 100 m mark is given as follows;

v = u + a·t

t = 2·√10 seconds, a = 1.0 m/s², u = 0

∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325

Jenny's speed at the 100-m mark ≈ 6.325 m/s

3 0

3 years ago

The transverse standing wave on a string fixed at both ends is vibrating at its fundamental frequency of 250 Hz. What would be t

hodyreva [135]

Answer:

Explanation:

fundamental frequency, f = 250 Hz

Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.

the formula for the frequency is given by

$f=\frac{1}{2l}\sqrt{\frac{Tl}{m}}$ .... (1)

Now the length is doubled ans the tension is four times but the mass remains same.

let the frequency is f'

$f'=\frac{1}{2\times 2l}\sqrt{\frac{4T\times 2l}{m}}$ .... (2)

Divide equation (2) by equation (1)

f' = √2 x f

f' = 1.414 x 250

f' = 353.5 Hz

7 0

3 years ago