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4 years ago

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. One of these is that during world war II, when the Air force planned bombing raids. These Air force missions had their briefings done with illustration of the details of the launch and and flight drawn on chalkboards. 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. This idea of the mission being scrubbed meaning that it is no longer to be followed through with, proceeded to this modern day rocket and aviation launches.

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Compare the time period of two simple pendulums of length 4m and 16m at a place.

Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L) is defined as:

$T=2\,\pi\,\sqrt{ \frac{L}{g} }$

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

5 0

3 years ago

Describe in terms of kinetic and potential energy what happens if an apple falls from a tree and comes to rest on the ground( wr

Lunna [17]

Answer:

An apple hanging at a branch has potential energy due its position. It can be written as PE= mgh where m is the mass of the apple h is the distance between the apple and the ground and g is the acceleration due to gravity.

as the apple falls from the tree it loses its potential energy and gains kinetic energy due to the movement of the apple. Its kinetic energy will be given by KE= 1/2mv² where m is the mass of the apple and v is the speed with which the apple falls.

As the apple falls the height or the distance reduces and PE becomes reduces. But it gains Kinetic energy due to its speed.

But when the apple falls to the ground and comes to rest its kinetic energy is converted to potential energy.

thus the total energy remains the same. it changes from one form to the other but remains unaltered.

6 0

3 years ago

The earth's magnetic field is associated with the:

ra1l [238]

The earth's magnetic field is associated with the core.

4 0

4 years ago

Read 2 more answers

which drawing below represents the electric field surrounding two objects that have equal magnitude changes of opposite polarity

Fantom [35]

Answer:

the 3rd one

Explanation:

j cause I think so

8 0

3 years ago

A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio

Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

$F_c = m\frac{v^2}{r}$

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

$v = r\omega$

with ω denoting the angular velocity, which we are given. With that, the above becomes:

$F_c = m\frac{v^2}{r}=m\omega^2 r$

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

$F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}$

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

$\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}$

and so 45 rev/min = 4.71 rad/s.

$\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27$

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.

3 0

3 years ago