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3 years ago

1. A force acting on an object in the upward direction is 3 N. The force that would

2 answers:

5 0

Answer: ow consider a book sliding from left to right across a tabletop. Sometime in the prior history of the book, it may have been given a shove and set in motion from a rest position. Or perhaps it acquired its motion by sliding down an incline from an elevated position. Whatever the case, our focus is not upon the history of the book but rather upon the current situation of a book sliding to the right across a tabletop.

Explanation:

DENIUS [597]3 years ago

4 0

Answer:

See the explanation below

Explanation:

We must perform a sum of forces on the body, that sum of forces is equal to zero. That is, the body is in balance and does not move.

ΣF = 0

3 - 3 = 0

This force is negative and acts by pointing downwards.

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Gas

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The gas state of matter has the most energy because of how freely the molecules move

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Which type(s) of electromagnetic radiation emitted by the Sun are absorbed by Earth’s atmosphere and do not reach Earth’s surfac

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3 years ago

Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ

bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; $H_0 = 51km/s/Mly$

Age of the universe; $t = \ ?$

We know that, the reciprocal of the Hubble's constant ( $H_0$ ) gives an estimate of the age of the universe ( $t$ ). It is expressed as:

$Age\ of\ Universe; t = \frac{1}{H_0}$

Now,

Hubble's constant; $H_0 = 51km/s/Mly$

We know that;

$1\ light\ years = 9.46*10^{15}m$

$1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m$

Therefore;

$H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\$

Now, we input this Hubble's constant value into our equation;

$Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years$

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0

3 years ago

The kitchen in your house measures 10 ft by 10 ft. You need to have 50 footcandles of illumination in the room. How many 60 watt

tekilochka [14]

Answer:

<u>6 bulbs</u> are needed to illuminate the room.

Explanation:

Given:

Measurement of kitchen (A) = 10 ft by 10 ft = 100 sq. ft

Number of footcandles (n) = 50

Lumens emitted by 1 bulb = 834

Number of bulbs (N) = ?

We are also given,

1 foot candle = 1 lumen/sq. ft

So, 50 foot candles = 50 lumens/sq. ft

Now, for an area of 1 sq. ft 50 lumens are emitted.

So, for an area of 100 sq. ft, lumens emitted = 50 × 100 = 5000 lumens

Now, one bulb emits = 834 lumes

Therefore, number of bulbs required for emitting 5000 lumens is given as:

$N=\frac{Number\ of\ lumens}{Number\ of\ lumens\ by\ 1\ bulb}\\\\N=\frac{5000}{834}=5.995\approx6$

So, 6 bulbs are needed to illuminate the room.

7 0

3 years ago