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3 years ago

What must exist for the engineering process to begin?

Physics

2 answers:

posledela3 years ago

7 0

Answer:

a. a problem

Explanation:

The engineering process begins by defining a problem and completing background research on the problem.

Vlad1618 [11]3 years ago

5 0

Answer:

Explanation:

rzp-yyib-oiv

plèase jóin ón góógle mèèt

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On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s

Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

6 0

3 years ago

Which of the below statements are true?

andrezito [222]

Answer:

Explanation:

Displacement is the distance from the start point to the endpoint, displacement disregard the path taken or the amount traveled.

if you start at point A, then go to point B, and back to point A, the displacement is zero because you started and ended at the same point.

for this question, pretend you started at point A, went east 20 km to point B, and then west 8 km to point C, your displacement is 12 km. 12 km is the distance between point A and point C.

4 0

3 years ago

Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

3 years ago

(a) What is the fluid speed in a fire hose with a 9.00-cm diameter carrying 80.0 L of water per second? (b) What is the flow rat

son4ous [18]

Answer:

12.5752053801 m/s

$80\times 10^{-3}\ m^3/s$

No.

Explanation:

Q = Volume flow rate = $80\ L/s=80\times 10^{-3}\ m^3/s$

d = Diameter of pipe = 9 cm

A = Area = $\dfrac{\pi}{4}d^2$

Volume flow rate is given by

$Q=Av\\\Rightarrow v=\dfrac{Q}{A}\\\Rightarrow v=\dfrac{80\times 10^{-3}}{\dfrac{\pi}{4} (9\times 10^{-2})^2}\\\Rightarrow v=12.5752053801\ m/s$

Velocity of fluid is 12.5752053801 m/s

The volume flow rate in m³/s is $80\times 10^{-3}\ m^3/s$

The flow of fluid does not depend on the type of water used. Hence the answers would be same. If Q is constant v will be the same irrespective of the type of water used.

8 0

3 years ago

Use the data provided to calculate the amount of heat generated for each cylinder height. Round your answers to the nearest tent

luda_lava [24]

100m: 4.9 kJ

200m: 9.8 kJ

1000m: 49.1 kJ

8 0

3 years ago

Read 2 more answers