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3 years ago

What is a continuous range of a single feature such as a wave lengt

Physics

1 answer:

Evgen [1.6K]3 years ago

7 0

I think you're looking for "spectrum".

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Is work being done on a barbell when a weight lifter is holding the barbell over his head?

AVprozaik [17]

I don’t see any picture sorry I wish I can help tho

7 0

2 years ago

The tension in cable da has a magnitude of tda=6.27 lb. find the cartesian components of tension tda, which is directed from d t

Oliga [24]

Complete Question

The Complete Question is attached below

We have that the Cartesian components of tension $T_{da}$ is

$T_{DA}=-4.433i-3.49j+2.735k$

From the Question we are told that

$M_{da}=6.27 lb\\\\w=9.50ft\\\\d=6.60ft\\\\h=4.50ft$

$\vec {DA}=-4.7i-3.7j+2.9k)ft$

$\vec {DB}=-1.9i-3.7j+1.9k)ft\\\\\vec {DC}=-1.9i+5.8j-1.6k)ft$

Generally the equation for $T_{DA}$ is mathematically given as

$T_{DA}=\phi_{DA}* M_{da}$

Where

$\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{(-4.7)^2+(-3.7)^2+(2.9)^2}\\\\\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}$

Therefore

$T_{DA}=\phi_{DA}* M_{da}$

$T_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}* 6.27$

$T_{DA}=-4.433i-3.49j+2.735k$

For more information on this visit

brainly.com/question/20746649?referrer=searchResults

5 0

3 years ago

A bicyclist, initially at rest, begins pedaling and gaining speed steadily for 4.90s during which she covers 25.0m.

emmasim [6.3K]

The bicyclist accelerates with magnitude a such that

25.0 m = 1/2 a (4.90 s)²

Solve for a :

a = (25.0 m) / (1/2 (4.90 s)²) ≈ 2.08 m/s²

Then her final speed is v such that

v ² - 0² = 2a (25.0 m)

Solve for v :

v = √(2 (2.08 m/s²) / (25.0 m)) ≈ 10.2 m/s

Convert to mph. If you know that 1 m ≈ 3.28 ft, then

(10.2 m/s) • (3.28 ft/m) • (1/5280 mi/ft) • (3600 s/h) ≈ 22.8 mi/h

8 0

2 years ago

Some superconductors are capable of carrying a very large quantity of current. If the measured current is 1.00 ´ 105 A, how many

Zielflug [23.3K]

Answer:

The $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

Explanation:

Given :

Current $I = 1 \times 10^{5}$ A

Charge of electron $e = 1.6 \times 10^{-19}$ C

Time $t = 1$ sec

From the formula of current,

Current is the number of charges flowing per unit time.

$I = \frac{ne}{t}$

Where $n =$ number of charges means in our case number of electrons

$n = \frac{It}{e}$

$n = \frac{1 \times 10^{5} }{1.6 \times 10^{-19} }$

$n = 6.25 \times 10^{23}$

Therefore, $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

5 0

3 years ago

What is the independent, and dependent variable? what should songs bobs conclusion be?

Fudgin [204]

The dependent variable is the slime on Gary's shell, because it's depending on other factors (independent factors).

6 0

3 years ago