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SSSSS [86.1K]
3 years ago
12

Two particles have a mass of 7.8 kg and 11.4 kg , respectively. A. If they are 800 mm apart, determine the force of gravity acti

ng between them. B. Compare this result with the weight of each particle. Find weight of the first particle. C. Find weight of the second particle.
Engineering
1 answer:
aleksley [76]3 years ago
3 0

Answer:

A) About 9.273 \times 10^{-9} newtons

B) 76.518 newtons

C) 111.834 newtons

Explanation:

A) F_g=\dfrac{GM_1M_2}{r^2} , where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:

F_g=\dfrac{(6.67408 \times 10^{-11})(7.8)(11.4)}{(0.8)^2}\approx 9.273 \times 10^{-9}

B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.

C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.

Hope this helps!

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Horitzontal and Vertical Lines

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You can use this command to generate horizontal and vertical dimensions. Creating a linear dimension is easy. All you have to do is start the command, specify the two points between which you want the dimension to be drawn and pick a point to fix the position of the dimension line.

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Which phrase best describes a safety-critical system?
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a poorly tighten terminal is often the cause of a/an ? a) open circuit b) circuit breaker interrupt c)short circuit d) ground fa
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Explanation:

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A 400 kg machine is placed at the mid-span of a 3.2-m simply supported steel (E = 200 x 10^9 N/m^2) beam. The machine is observe
Alenkinab [10]

Answer:

moment of inertia = 4.662 * 10^6 mm^4

Explanation:

Given data :

Mass of machine = 400 kg = 400 * 9.81 = 3924 N

length of span = 3.2 m

E = 200 * 10^9 N/m^2

frequency = 9.3 Hz

Wm ( angular frequency ) = 2 \pi f = 58.434 rad/secs

also Wm = \sqrt{\frac{g}{t} }  ------- EQUATION 1

g = 9.81

deflection of simply supported beam

t = \frac{wl^3}{48EI}

insert the value of t into equation 1

Wm^2 = \frac{g*48*E*I}{WL^3}   make I the subject of the equation

I ( Moment of inertia about the neutral axis ) = \frac{WL^3* Wn^2}{48*g*E}

I = \frac{3924*3.2^3*58.434^2}{48*9.81*200*10^9}  = 4.662 * 10^6 mm^4

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