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gayaneshka [121]
4 years ago
15

When did they start using the word "Scrubbed" for launches?

Physics
1 answer:
VladimirAG [237]4 years ago
6 0

Answer:

Many speculation exists for why the term 'scrubbed' is used in relation to launches. <em>One of these is that during world war II, when the Air force planned bombing raids</em>. These  Air force missions had their briefings done with illustration of the details of the launch and and flight drawn on chalkboards. <em>When some of mission flight takeoffs were canceled (due to bad weather or a malfunction or other reasons) the details were then scrubbed off of the chalkboard</em>. This idea of the mission <em>being scrubbed meaning that it is no longer to be followed through with</em>, proceeded to this modern day rocket and aviation launches.

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If the magnets are brought close together, they will
murzikaleks [220]

It depends on which side. Opposites attract, so north and south would attract to each other and collide, while north and north or south and south would go away from eachother.

7 0
4 years ago
Read 2 more answers
What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is
ss7ja [257]

Answer:

g' = g/9 = 1.09 m/s²

Explanation:

The magnitude of free fall acceleration at the surface of earth is given by the following formula:

g = GM/R²   ----- equation 1

where,

g = free fall acceleration

G = Universal Gravitational Constant

M = Mass of Earth

R = Distance between the center of earth and the object

So, in our case,

R = R + 2 R = 3 R

Therefore,

g' = GM/(3R)²

g' = (1/9) GM/R²

using equation 1:

g' = g/9

g' = (9.8 m/s)/9

<u>g' = 1.09 m/s²</u>

3 0
3 years ago
g A 150-kg object and a 450-kg object are separated by 4.90 m. (a) Find the magnitude of the net gravitational force exerted by
klemol [59]

Answer:

BBbbb

Explanation:

B is the correct answer:))))

4 0
3 years ago
3 A picture is supported by two vertical strings; if the weighi
andrezito [222]

Answer:

<em>The force exerted by each string is 25 N</em>

Explanation:

<u>Net Force</u>

The net force is the vector sum of forces acting on a body. The net force is a single force that represents the effect of the original forces on the body's motion. It gives the particle the same acceleration as all those actual forces together as described by Newton's second law of motion.

The picture described in the problem is hanging at rest supported by two vertical strings. This means that the net force acting on it is zero.

Assume the magnitude of each of these equal forces is F, and the picture has a weight of W=50 N, thus the net force is:

F + F - W

The positive signs indicate an upwards direction and the negative sign means a downwards direction. Since the net force is zero:

F + F - W = 0

2F = W

F = W/2 = 50 N/2

F = 25 N

The force exerted by each string is 25 N

7 0
3 years ago
To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(
NeX [460]

Answer:

The transverse displacement is   y(1.51 , 0.150) = 0.055 m    

Explanation:

 From the question we are told that

     The generally equation for the mechanical wave is

                    y(x,t) = Acos (kx -wt)

     The speed of the transverse wave is v = 8.25 \ m/s

     The amplitude of the transverse wave is A = 5.50 *10^{-2} m

     The wavelength of the transverse wave is \lambda = 0540 m

      At t= 0.150s , x = 1.51 m

 The angular frequency of the wave is mathematically represented as

          w = vk

Substituting values  

         w = 8.25 * 11.64

        w = 96.03 \ rad/s

The propagation constant k is mathematically represented as

                  k = \frac{2 \pi}{\lambda}

Substituting values

                  k = \frac{2 * 3.142}{0. 540}

                   k =11.64 m^{-1}

Substituting values into the equation for mechanical waves

      y(1.51 , 0.150) = (5.50*10^{-2} ) cos ((11.64 * 1.151 ) - (96.03  * 0.150))

       y(1.51 , 0.150) = 0.055 m    

4 0
4 years ago
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