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Alenkinab [10]
3 years ago
10

Two asteroids exert a gravitational force, F, on each other. Some time later, the asteroids are now three times as far from each

other as before. Which of the following represents the gravitational force at this distance? Select one:
a. F/9
b. F/6
c. F/3
d. F/2
Physics
1 answer:
dolphi86 [110]3 years ago
7 0

Answer:

a.F/9

Explanation:

Hello, I can help you wit this

the gravitational Force is described by the next equation

F=G*\frac{m_{1}*m_{2} }{r^{2} } \\

Where, G  is   the universal gravitational constant, M1 and M2 are the masses of the objects, and r is the distance between then.

Step 1

time 1

Let

m1= mass of the asteroid 1

m2= mass of the asteroid 2

distance 1=r1

put the values into the equation to find the force

F_{1}=G*\frac{m_{1}*m_{2} }{r^{2} }\\F_{1}=G*\frac{m_{1}*m_{2} }{r_{1}^{2} }

Step 2

time 2

r2=3r1

F_{2} =G*\frac{m_{1}*m_{2} }{(3*r_{1} )^{2} }\\

Step 3

compare the forces

F_{1}=G*\frac{m_{1}*m_{2} }{r^{2} }\\F_{2} =G*\frac{m_{1}*m_{2} }{9(r_{1}^{2}) }\\F_{2}=\frac{F_{1}}{9}

a.F/9

here it is observed that the force is inversely proportional to the square of the distance, at double distance, it is divided into 4, triple in 9, etc.

Have a great day.

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Diano4ka-milaya [45]

Answer:

the acceleration of the elevator is increasing

Explanation:

For this exercise we propose the solution using Newton's second law

         F -W = m a

         F = m (g + a)

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3 years ago
Two small balls, each of mass 5.0 g, are attached to silk threads 50 cm long, which are in turn tied to the same point on the ce
Rama09 [41]

In triangle ABC

Sin5 = BC/AC

BC = (AC) Sin5

BC = (50) Sin5 = 4.36 cm

r = distance between the two balls = 2 BC = 2 x 4.36 = 8.72 cm = 0.0872 m

q = charge on each ball

m = mass of each ball = 5 g = 0.005 kg

electric force between the two balls is given as

F = \frac{kq^{2}}{r^{2}}

using equilibrium of force in vertical direction

T Cos5 = mg eq-2

Using equilibrium of force in horizontal direction

T Sin5 = F eq-3

dividing eq-3 by eq-2

T Sin5 /(T Cos5) = F/mg

F = mg tan5

using eq-1

\frac{kq^{2}}{r^{2}} = mg tan5

inserting the values

\frac{(9 \times 10^{9})q^{2}}{(0.0872)^{2}} = 0.005 x 9.8

q = 2.03 x 10⁻⁷ C

sign of charge is same on both the balls. either it is negative or positive

4 0
3 years ago
A 1056-hertz tuning fork is sounded at the same time a piano note is struck. You hear three beats per second. What is the freque
REY [17]

Answer:

f_2 = 1053 Hz

or

f_2 = 1059 Hz

Explanation:

As we know that the beat frequency is given as

|f_1 - f_2| = f_{beat}

here we know that

f_1 = 1056 Hz

f_{beat} = 3 Hz

now we have

|1056 - f_2| = 3

now we have

1056 - f_2 = \pm 3

so we have

f_2 = 1053 Hz

or

f_2 = 1059 Hz

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Consider a sound wave moving through the air modeled with the equation s(x, t) = 5.00 nm cos(60.00 m−1x − 18.00 ✕ 103 s−1t). Wha
GaryK [48]

Answer:

Shortest time = 58.18 × 10^(-6) s

Explanation:

We are given;

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t))

Let us set x = 0 as origin.

Now, for us to find the time difference, we need to solve 2 equations which are;

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t1))

And

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t2))

Now, since the wave starts from maxima at time at t = 0, the required time would be the difference (t2 - t1)

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And

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Angle of the cos function is in radians, thus;

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So,

Required time = t2 - t1 = (116.36 × 10^(-6) s) - (58.18 × 10^(-6) s) = 58.18 × 10^(-6) s

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Answer:

Fa= 47.2N

Explanation:

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