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2 years ago

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A spinning satellite begins to unfold two solar panels as shown. How does a satellites ANGULAR MOMENTUM change as its panels ext

end?
A. It drops to zero.
B. It decreases.
C. It remains constant.
D. It increases.

1 answer:

Soloha48 [4]2 years ago

6 0

As the satellite panels extend, the angular velocity decreases due to drag force, and hence it will cause a decrease in the angular momentum of the satellite.

<h3>What is angular momentum?</h3>

Angular momentum is defined as the quantity of rotation of a body, which is the product of mass, velocity and radius.

L = mvr

L = mωr²

where;

m is mass of the object
v is velocity
r is radius
ω is angular velocity

As the satellite panels extend, the angular velocity decreases due to drag force, and hence it will cause a decrease in the angular momentum of the satellite.

Learn more about angular momentum here: brainly.com/question/4126751

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The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12. How many grains are there in the ball

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Given :

The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12.

To Find :

How many grains are there in the ball?

Solution :

Volume of ball of the ballpoint is :

$V = \dfrac{4 \pi r^3}{3}\\\\V = \dfrac{4\times 3.14 \times 0.5^3}{3}\ mm^3\\\\V = 0.523\ mm^3$

Now, grain size of 12 has about 520000 grains/mm³.

Therefore, number of grains are :

$n = 520000\times 0.523\ grains\\\\n = 271960\ grains$

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3 years ago

At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

$\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

$\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

so

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

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3 years ago

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Newton's second law allows calculating the response for the person's acceleration while leaving the trampoline is:

-4.8 m / s²

Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body

F = m a

Where the bold letters indicate vectors, F is the force, m the masses and the acceleration

The free body diagram is a diagram of the forces without the details of the body, in the attached we can see the free body diagram for this system

$F_t$ -W = m a

Whera $F_t$ is the trampoline force

Body weight is

W = mg

We substitute

$F_t$ - mg = ma

a = $\frac{F_t - m g}{m}$

Let's calculate

a = $\frac{375 - 75 \ 9.8 }{75}$

a = -4.8 m / s²

The negative sign indicates that the acceleration is directed downward.

In conclusion using Newton's second law we can calculate the acceleration of the person while leaving the trampoline is

-4.8 m / s²

Learn more here: brainly.com/question/19860811

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3 years ago

There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target. Part A A p

tamaranim1 [39]

Explanation:

A pilot flying a weight; it takes t = 3.4 s to hit the ground, during which it travels a horizontal distance d = 120 m

h = g/2*t^2

h= 4.903*3.4^2

h=56.701 m

V = d/t

V= 120/3.4

V=35.3 m/sec

Now the pilot does a run at the same height but twice the speed.

How much time (t1) does it take the weight to hit the ground?

t1 = t = 3.4 sec (depending just upon height)

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