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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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The 45-g arrow is launched so that it hits and embeds in a 1.40 kg block. The block hangs from strings. After the arrow joins th
worty [1.4K]

Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg

where m_a is the mass of the arrow, m_b is the mass of the block, \Delta H of the change in height of the block after the collision, and v is the velocity of the arrow before it hit the block.

Solving for the velocity v, we get:

$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} } $

and we put in the numerical values

m_a = 0.045kg,

m_b = 1.40kg,

\Delta H = 0.4m,

g= 9.8m/s^2

and simplify to get:

\boxed{ v= 15.9m/s}

The arrow was moving at 15.9 m/s

6 0
3 years ago
Draw a simple circuit that lights up a bulb. ​
coldgirl [10]

Hope this helps

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5 0
2 years ago
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A top-fuel dragster accelerates from rest to a velocity of 100 m/s in 8 s. What is the acceleration?
seropon [69]

Answer:

100 m/s ÷ 8 = 12.5 m/s

Explanation:

You must put multiply (÷)

4 0
3 years ago
A motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a displacement x and attains a
iren [92.7K]

Answer:

√(6ax)

Explanation:

Hi!

The question states that during a time t the motorcyle underwent a displacement x at constant acceleration a starting from rest, mathematically we can express it as:

x=(1/2)at^2

Then the we need to find the time t' for which the displacement is 3x

3x=(1/2)a(t')^2

Solving for t':

t'=√(6x/a)

Now, the velocity of the motorcycle as a function of time is:

v(t)=a*t

Evaluating at t=t'

v(t')=a*√(6x/a)=√(6*x*a)

Which is the final velocity

Have a nice day!

3 0
3 years ago
Two identical small metal spheres with q1 &gt; 0 and |q1| &gt; |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
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