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3 years ago

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determine the force of gravitational attraction between a 78kg boy sitting 2 meters away from a 65kg girl.

1 answer:

bixtya [17]3 years ago

8 0

Answer:

$F=8.45\times 10^{-8}\ N$

Explanation:

Given that,

Mass of a boy is 78 kg

Mass of a girl is 65 kg

We need to find the force of gravitational attraction between them if they are 2 m away.

The formula for the gravitational force is given by :

$F=\dfrac{Gm_1m_2}{r^2}\\\\F=\dfrac{6.67\times 10^{-11}\times 78\times 65}{(2)^2}\\\\=8.45\times 10^{-8}\ N$

So, the force between them is $8.45\times 10^{-8}\ N$ .

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Unlike a roller coaster, the seats in a Ferris wheel swivel so that the rider is always seated upright. An 80-ft-diameter Ferris

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Answer:

a

$F_A =425.42 \ N$

b

$F_A_H = 358.58 \ N$

Explanation:

From the question we are told that

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The apparent weight of the passenger at the lowest point is mathematically represented as

$F_A_L = F_c + W$

Where $F_c$ is the centripetal force on the passenger, which is mathematically represented as

$F_c =m * r * w^2$

Where $w$ is the angular velocity which is mathematically represented as

$w = \frac{2* \pi }{T}$

substituting values

$w = \frac{2* 3.142 }{24}$

$w = 0.2618 \ rad/s$

and r is the radius which is evaluated as $r = \frac{d}{2}$

substituting values

$r = \frac{24.383}{2}$

$r = 12.19 \ ft$

So

$F_c = 40 * 12.19* (0.2618)^2$

$F_c = 33.42 \ N$

W is the weight which is mathematically represented as

$W = 40 * 9.8$

$W = 392 \ N$

So

$F_A = 33.42 + 392$

$F_A =425.42 \ N$

The apparent weight of the passenger at the highest point is mathematically represented as

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substituting values

$F_A_H = 392 - 33.42$

$F_A_H = 358.58 \ N$

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3 years ago

A uniform 2.50m ladder of mass 7.30kg is leaning against a vertical wall while making an angle of 51.0degree with the floor. A w

Zigmanuir [339]

Answer:

19.95 J

Explanation:

The center of mass of the ladder is initially at a height of:

$h_1=\frac{L}{2}sin\theta$

The center of mass of the ladder ends at a height of:

$h_2= \frac{L}{2}sin90$ =L/2

So, the work done is equal to the change in potential energy which is:

W = PE = $mg(h_2-h_1)$

now $h_2-h_1= 1-sin\theta$

therefore

W = [mgL/2]×[1 - sin(theta)]

W = [(7.30)(9.81)(2.50)/2]×[1-sin(51°)]

solving this we get

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3 years ago

A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en

Virty [35]

Complete Question

The complete question is shown on the first uploaded image

Answer:

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Explanation:

The objective of this solution is to obtain the acceleration of the particle

Now looking at Newton law which is mathematically represented as

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$a = \frac{qvB}{m}$

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So substituting $(2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T$ for B

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$a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j \ + (3*4)i \ \cdot \ k)$

According to vector multiplication

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So

$a = 0.003* 6 =0.018\ m/s^2$

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3 years ago