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3 years ago

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An astronaut's training centrifuge has a radius of 4.0m. If it goes round once every 2.5s, calculate the velocity of the end of

the centrifuge arm (4.0m from the pivot).

1 answer:

alexandr1967 [171]3 years ago

6 0

Answer: 10.048m/s

Explanation:

We know that the radius is r = 4.0m

And for rotating things, the tangential velocity (or the velocity of the end of the centrifuge arm) can be calculated as:

v = r*w

The period is T = 2.5s,

This means that if sin(w*t) describes this situation, we have that:

sin(w*t) = sin(w*(t + T))

and we know that:

sin(w*0) = 0

sin(w*(0 + T)) = Sin(w*T) = 0

this means that w*T = 2*pi

w = 2*pi/T = 2*pi/2.5s = (2*3.14)/2.5 s = 2.512 hz

Then the velocity can be calculated as

v = r*w = 4.0m*2.512hz = 10.048m/s

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What connects the upper motor neurons to lower motor neurons?

skad [1K]

Answer:

The upper motor neurons synapse in the spinal cord connect with anterior horn cells of lower motor neurons, usually via interneurons. The anterior horn cells are the cell bodies of the lower motor neurons and are located in the grey matter of the spinal cord.

Explanation:

Interneurons are the central nodes of neural circuits, enabling communication between the upper motor neurons, sensory or motor neurons located in the brain and spinal cord and they send signals to lower motor neurons or central nervous system (CNS) in the brain stem and spinal cord . When they get a signal from the upper motor neurons, they send another signal to your muscles to make them contract. They play vital roles in reflexes, neuronal oscillations, and neurogenesis in the adult mammalian brain.

Renshaw cells are among the very first identified interneurons. They are excited by the axon collaterals of the motor neurons. In addition, Renshaw cells make inhibitory connections to several groups of motor neurons.

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3 years ago

Assume that the home construction industry is perfectly competitive and in long-run competitive equilibrium. It follows that: A.

olga nikolaevna [1]

Answer:

B. Marginal cost equals long-run average total cost.

Explanation:

The zero profit condition implies that entry continues until all firms are producing at minimum long run average total cost. Since the marginal cost curve cuts the long run average total cost curve at its minimum point, marginal cost and long run average total cost must be equal in long run equilibrium.

4 0

3 years ago

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

elena55 [62]

<u>Answer:</u>

<em>Thunderbird is 995.157 meters behind the Mercedes</em>

<u>Explanation:</u>

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird decided to take a pit stop and slows down for 250 m. She spent 5 seconds in the pit stop.

Here final velocity $v=0 \ m/s$

initial velocity $u= 71 m/s$ distance

Distance covered in the slowing down phase = $250 m$

$v^2-u^2=2as$

$a= \frac {(v^2-u^2)}{2s}$

$a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2$

$v=u+at$

$t= \frac {(v-u)}{a}$

$= \frac {(0-71)}{(-10.082)}=7.042 s$

$t_1=7.042 s$

The car is in the pit stop for 5s $t_2=5 s$

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

$a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2$

$v=u+at$

$t= \frac{(v-u)}{a}$

$t= \frac{(71-0)}{6.81}= 10.425 s$

$t_3=10.425 s$

total time= $t_1 +t_2+t_3=7.042+5+10.425=22.467 s$

Distance covered by the Mercedes Benz during this time is given by $s=vt=71 \times 22.467= 1595.157 m$

Distance covered by the Thunderbird during this time= $250+350=600 m$

Difference between distance covered by the Mercedes and Thunderbird

= $1595.157-600=995.157 m$

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

6 0

3 years ago

R S ( M ) = 2 G M c 2 , where G is the gravitational constant and c is the speed of light. It is okay if you do not follow the d

padilas [110]

The provided question's answer is "Schwarzschild radius".

The conversion factor between mass and energy is the speed of light squared.

GM/r stands for gravitational potential energy, also known as energy per unit mass.

GM/rc² then has "mass per unit mass" units. In other words, as mass/mass splits out in a dimensional analysis, "dimensionless per unit."

The derivation yields a formula for time or space coordinate ratios requiring sqrt(1 - 2GM/rc²). This number becomes 0 when r=2GM/c2, or the formula becomes infinite if in the denominator. However, there is no justification for using c² as a conversion factor there. Consider the initial expression sqrt(1 - 2GM/rc²).

Assume that m is used as the test particle's mass instead of 1. Then you have sqrt(m - 2GMm/rc² and mass units. This expression denotes that the rest energy of the test mass m you introduced into the gravitational field is "gone" at that radius.

The 2 would be absent if the gravitational field were Newtonian. However, at the event horizon, Einstein gravity is slightly stronger than Newton gravity, resulting in the factor 2 in qualitative terms.

So, the given equation is of Schwarzschild radius.

Learn more about Schwarzschild radius here:

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2 years ago

A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w

Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, $a=3\ m/s^2$

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

$m=\dfrac{W}{g}$

$m=\dfrac{150}{9.8}$

$m=15.3\ kg$

Frictional force is given by :

$F=ma$

$F=15.3\times 3$

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0

4 years ago