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2 years ago

A typical adult human has a mass of about 70 kg.

1 answer:

3 0

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
a. FM GmMmr2
= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
= 2.40 x 10-3 N

b. FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%

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A balloon drifts 140m toward the west in 45s ; then the wind suddenly changes and the balloon flies 90m toward the east in the n

Bogdan [553]

Answer: 140 m

Explanation:

Let's begin by stating clear that motiont is the change of position of a body at a certain time. So, during this motion, the balloon will have a trajectory and a displacement, being both different:

The trajectory is the path followed by the body, the distance it travelled (is a scalar quantity).

The displacement is the distance in a straight line between the initial and final position (is a vector quantity).

So, according to this, the distance the balloon traveled during the first 45 s (its trajectory) is 140 m.

But, if we talk about displacement, we have to draw a straight line between the initial position of the balloon (point 0) to its final position (point 90 m). Being its displacement 95 m.

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2 years ago

An electric winch is used to raise a 40-kg package and a 10-kg package vertically up the side of a building as pictured in the d

just olya [345]

Answer:

Explanation:

40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.

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2 years ago

A 9V battery is directly connected to each of 3 LED bulbs. Select the statement that accurately describes this circuit. A) A dir

Likurg_2 [28]

First off, you can cross out alternating current because a 9V battery doesn't give out AC, it gives out solely DC. If the battery is connected to each battery individually, then they are in parallel. So, according to Kirchhoff's Voltage Law, in parallel, V total = V1 = V2= V3..
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2 years ago

Read 2 more answers

A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision

BigorU [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

$f= -75 \ cm = - 0.75 \ m$

$P = -1.33 \ diopters$

Explanation:

From the question we are told that

The image distance is $d_i = -75 cm$

The value of the image is negative because it is on the same side with the corrective glasses

The object distance is $d_o = \infty$

The reason object distance is because the object father than it being picture by the eye

General focal length is mathematically represented as

$\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}$

substituting values

$\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}$

=> $f= -75 \ cm = - 0.75 \ m$

Generally the power of the corrective lens is mathematically represented as

$P = \frac{1}{f}$

substituting values

$P = \frac{1}{-0.75}$

$P = -1.33 \ diopters$

7 0

2 years ago

The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th

Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

A freight train has a mass of $1.83\times 10^{7}\,kg$ . The wheels of the locomotive push back on the tracks with a constant net force of $7.50\times 10^{5}\,N$ , so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?

If locomotive have a constant net force ( $F$ ), measured in newtons, then acceleration ( $a$ ), measured in meters per square second, must be constant and can be found by the following expression:

$a = \frac{F}{m}$ (1)

Where $m$ is the mass of the freight train, measured in kilograms.

If we know that $F = 7.50\times 10^{5}\,N$ and $m = 1.83\times 10^{7}\,kg$ , then the acceleration experimented by the train is:

$a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}$

$a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$

Now, the time taken to accelerate the freight train from rest ( $t$ ), measured in seconds, is determined by the following formula:

$t = \frac{v-v_{o}}{a}$ (2)

Where:

$v$ - Final speed of the train, measured in meters per second.

$v_{o}$ - Initial speed of the train, measured in meters per second.

If we know that $a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$ , $v_{o} = 0\,\frac{m}{s}$ and $v = 22.222\,\frac{m}{s}$ , the time taken by the freight train is:

$t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s} }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }$

$t = 542.265\,s$

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

6 0

2 years ago