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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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Two routes connect an origin and a destination. Routes 1 and 2 have performance functions t1 = 2 + X1 and t2 = 1 + X2, where the
Musya8 [376]

Solution :

Given

$t_1=2+x_1$

$t_2=1+x_2$

Now,

$P(h

$0.4=1-P(h \geq5)$

$0.6=P(h \geq5)$

$0.6= e^{\frac{-x_1 5}{3600}}$

Therefore,   $x_1=368 \ veh/h$

                        $=\frac{368}{1000} = 0.368$

Given,   $t_1=2+x_1$

                 = 2 + 0.368

                 = 2.368 min

At user equilibrium, $t_2=t_1$

∴  $t_2$ = 2.368 min

$t_2=1+x_2$

$2.368=1+x_2$

$x_2 = 1.368$

$x_2 = 1.368 \times 1000$

    = 1368 veh/h

7 0
3 years ago
Determine the initial void ratio, the relative density and the unit weight (in pounds per cubic foot) of the specimens for each
Elina [12.6K]

The initial void ratio is the <em>parameter </em>which is used to show the structural foundations for each <em>specimen of sand </em>so that the method and speed of compression would be <em>measured</em>.

Relative density is the mass per unit volume of each specimen of sand which is <em>measured </em>and it has to do with the<em> relative ratio</em> of the density of the sand.

Unit weight is the the exact weight per cubic foot of the sand which is measured.

Please note that your question is incomplete so I gave you a general overview to help you better understand the concept

Read more here:

brainly.com/question/15220801

5 0
3 years ago
List and describe three classifications of burns to the body.
DiKsa [7]

AnswerWhat Are the Classifications of Burns? Burns are classified as first-, second-, or third-degree, depending on how deep and severe they penetrate the skin's surface. First-degree burns affect only the epidermis, or outer layer of skin. The burn site is red, painful, dry, and with no blisters.

Explanation:

8 0
3 years ago
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Briefly explain how each of the following influences the tensile modulus of a semicrystalline polymer and why:(a) molecular weig
marin [14]

Answer:

(a) Increases

(b) Increases

(c) Increases

(d) Increases

(e) Decreases

Explanation:

The tensile modulus of a semi-crystalline polymer depends on the given factors as:

(a) Molecular Weight:

It increases with the increase in the molecular weight of the polymer.

(b) Degree of crystallinity:

Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.

(c) Deformation by drawing:

The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.

(d) Annealing of an undeformed material:

This also results in an increase in the tensile strength of the material.

(e) Annealing of  a drawn material:

A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.

5 0
3 years ago
5) Calculate the LMC wal thickness of a pipe and tubing with OD as 35 + .05 and ID as 25 + .05 A) 4.95 B) 5.05 C) 10 D) 15.025
padilas [110]

Answer:

B) 5.05

Explanation:

The wall thickness of a pipe is the difference between the diameter of outer wall and the diameter of inner wall divided by 2. It is given by:

Thickness of pipe = (Outer wall diameter - Inner wall diameter) / 2

Given that:

Inner diameter = ID = 25 ± 0.05, Outer diameter = OD = 35 ± 0.05

Maximum outer diameter = 35 + 0.05 = 35.05

Minimum inner diameter = 25 - 0.05 = 24.95

Thickness of pipe = (maximum outer wall diameter - minimum inner wall diameter) / 2 = (35.05 - 24.95) / 2 = 5.05

or

Thickness = (35 - 25) / 2 + 0.05 = 10/2 + 0.05 = 5 + 0.05 = 5.05

Therefore the LMC wall thickness is 5.05

6 0
3 years ago
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