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2 years ago

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Physics

1 answer:

wlad13 [49]2 years ago

5 0

Answer:

the correct answer is B

Newton/columb

I would appreciate if my answer is chosen as a brainliest answer

You might be interested in

If a chemical is completely consumed in a chemical reaction, what is the substance?

Dima020 [189]

Is a process in which One set of substances, called REACTANTS, is converted to a new set of substances is called PRODUCTS.

--In other words, a chemical reaction is the process by which a chemical change occurs.

5 0

3 years ago

IBM has a fast computer that it calls the Blue Gene/L that can do '136.8

dmitriy555 [2]

Answer:

138.6 megacalculations

Explanation:

This is a pretty straightforward one.

All it needs is to convert the degree of measurement.

Dimensions in physics are attributed names, which state the power to which they're are raised. Just as how

Kilo and Mega means the numbers are raised to the power of 3 and 6 respectively. There also exists the ones that indicates how small, such as milli and micro, which are to the powers of -3 & -6.

The question says the IBM computer calculates at an astonishing 136.8 teracalculations.

Tera in physics means it's raised to the power of 12. Thus, the IBM calculates at an astonishing rate of

136.8*10^12 calculations per second.

We're then asked how many calculations it does in 1 micro second. Like I had highlighted earlier, 1 micro second is 1 raised to the power of -6. Or succinctly put,

1 micro second = 1*10^-6.

If the IBM does

138.6*10^12 = 1 second,

Then it does

x = 1*10^-6 second.

When we cross multiply, we have

138.6*10^12 * 1*10^-6, and that is

138.6*10^6 calculations, or say, 138.6 megacalculations.

The IBM does 138.6 megacalculations in 1 micro second, which is still astonishing, by the way

6 0

3 years ago

Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec

Dmitrij [34]

Answer:

(a) $F_{net}$ = 4.19 x $10^{-5}$ N, and its direction is towards $m_{2}$ .

(b) It must be placed inside a hollow shell.

Explanation:

Let, $m_{1}$ = 165 kg, $m_{2}$ = 465 kg, $m_{3}$ = 60 kg, and the distance between $m_{1}$ and $m_{2}$ is 0.340 m.

(a) Since $m_{3}$ is placed midway between $m_{1}$ and $m_{2}$ , then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = $\frac{Gm_{1}m_{2} }{r^{2} }$

Where all variables have their usual meaning.

Then,

a. $F_{net}$ = $F_{23}$ - $F_{13}$

$F_{13}$ = $\frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }$

= 2.25 x $10^{-5}$ N

$F_{23}$ = $\frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }$

= 6.44 x $10^{-5}$ N

∴ $F_{net}$ = = 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

The net force exerted by the two masses on the 60 kg object is 4.19 x $10^{-5}$ N.

(ii) / $F_{net}$ / = / $F_{23}$ / - / $F_{13}$ /

= 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

(iii) The direction of the net force is to the right i.e towards $m_{2}$ .

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0

3 years ago

On sunday a strong gale blows near beth’s house. the best estimate for the wind speed is ____ miles/hour.

Rufina [12.5K]

Answer:

B. 47-54 miles/hour

Explanation:

Gale is a strong wind which is depicted by red warning flag. According to U.S. National Weather Service gale is a sustained surface wind. It is also used to refer winds from tropical coastal areas.

On the basis of force of wind gale is divided in four groups:

Near gale - 32-38 mph
Gale - 39-46 mph
Strong gale - 47-54 mph
Storm - 55-63 mph

8 0

3 years ago

Question 1 of 25

finlep [7]

Answer:

2.753*10^-11N

Explanation:

According to Newton's law of gravitation, the force between the masses is expressed as;

F = GMm/d²

M and m are the distances

d is the distance between the masses

Given

M = 3.71 x 10 kg

m = 1.88 x 10^4 kg

d = 1300m

G = 6.67 x 10-11 Nm²/kg

Substitute into the formula

F = 6.67 x 10-11* (3.71 x 10)*(1.88 x 10^4)/1300²

F = 46.52*10^(-6)/1.69 * 10^6

F = 27.53 * 10^{-6-6}

F = 27.53*10^{-12}

F = 2.753*10^-11

Hence the gravitational force between the asteroid is 2.753*10^-11N

6 0

2 years ago