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3 years ago

14

An electrician timed a box traveling over a conveyor and measured 25 sec to travel 30 feet. If the gearbox ratio is 20:1 and the

driver drum is 24", what is the RPM of the motor?

1 answer:

nataly862011 [7]3 years ago

4 0

Pppeejsidoocicjddkdkcocfoodoocoie

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How can tidal force from the moon affect our earth?

VARVARA [1.3K]

Answer:

sorry but I can understand the question

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2 years ago

Read 2 more answers

Zad. 2 W marszobiegu chłopiec przebiegł 900 m w ciągu 205 min, a następnie przeszedł 300m w tym danym czasie. Jaka była średnia

USPshnik [31]

Odpowiedź:

0,049 m / s

Wyjaśnienie:

Biorąc pod uwagę, że:

Dystans biegu = 900m

Czas trwania = 205 minut

Długość przejścia = 300 m

Zajęty czas = 205 minut

Średnia prędkość :

(Przebieg + pokonany dystans) / całkowity czas

Średnia prędkość :

(900 m +. 300 m) / 205 + 205

1200 m / 410 minut

Minuty do sekund

1200 / (410 * 60)

1200/24600

= 0,0487804

= 0,049 m / s

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3 years ago

a radio station broadcast a frequency 90500 Hz. these radio waves travel at speed of 30000 m/s. what is the wavelength of the ra

Feliz [49]

Answer:

0.331m

Explanation:

wave equation: v=f λ

v=30000m/s

f=90500 Hz

λ= $\frac{v}{f}$

λ= $\frac{30000}{90500}$

λ= 0.311m

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2 years ago

How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

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1 year ago

All galaxies follow the law of gravity. True or False

Juliette [100K]

The answer is true. All the galaxies in the universe follow the law of gravity.

<span>Based from the book, It's about Time: the Illusion of Einstein’s Time Dilation Explained, </span>

Einstein had explained that all the heavenly bodies in the universe follow the same scientific laws that are similar to our solar system. The stars and planets are held by the principles of inertia and gravity

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2 years ago