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Musya8 [376]
4 years ago
13

If legos are plastic can they be recycled?

Physics
2 answers:
DaniilM [7]4 years ago
8 0

Answer:

yes they can

If your bricks are worn and no longer suitable to play with, they can be recycled with the rest of your household plastic

Explanation:

raketka [301]4 years ago
5 0
Unfortunately, since Legos are made of plastic, they will stick around for millennia if they end up in landfills. Recycling Legos is hard, since they are made with an unusual plastic that is not accepted at many recycling centers. However, Legos are highly reusable.
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By applying a force of 55 N, a pulley system can lift a box with a mass of 20.0 kg. What
Gnesinka [82]

The mechanical advantage of the pulley is 3.56

Explanation:

The Mechanical Advantage (MA) of a pulley system is given by

MA=\frac{Load}{Effort}

where

Load is the weight of the object lifted

Effort is the force applied in input

For the pulley in this problem, we have:

Effort = 55 N

While the load is the weight of the box of mass m = 20.0 kg:

Load = mg = (20.0 kg)(9.8 m/s^2)=196 N

Substittuing, we find the MA of the pulley:

MA=\frac{196}{55}=3.56

#LearnwithBrainly

7 0
4 years ago
In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
Calculate the average force with which the Sun pulls the Earth. It will help to know
algol13

Answer:

7.9 x 10^21 pound-force

Explanation:

The average distance between the Earth and sun is 150 trillion meters, or 1.5 x 10^11 meters. The mass of the sun is 1.99 x 10^30 kilograms, while the Earth weighs in at 6.0 x 10^24 kilograms. The gravitational constant is 6.67 x 10^-11 meter^3 / (kilogram - second^2). So the Earth and sun pull on each other with a force equal to 3.52 x 10^22 newtons. The newton is a unit of force equal to a kilogram-meter/second^2. One newton is equal to 0.22 of the rarely used English unit called pound-force, so 3.52 x 10^22 newtons is 7.9 x 10^21 pound-force.

4 0
3 years ago
A student writes the following script for a scene in a futuristic movie: A rocket ship is landing on the Moon where many astrona
Sav [38]

Answer: No crash landing on the surface of the moon and there is no vibration due to low acceleration due to gravity.

Explanation:

The rocket ship can't land on the surface of the moon because it depends on winged flight for its controlled descent to the surface of the Earth.  Since the moon has no real atmosphere, the wings would not be able to glide the craft smoothly to the moon's surface.

The acceleration due to gravity in the moon is very low compared to the earth. The astronauts cannot experience vibration ripple through the Moon’s surface beneath their feet because the low level acceleration due to gravity will drastically affect the weight of the ship.

6 0
3 years ago
An electric current of 200 A is passed through a stainless steel wire having a radius R of 0.001268 m. The wire is L = 0.91 m lo
denis23 [38]

Answer:

The value is   T_m  =  435.2 \  K

Explanation:

From the question we are told that  

   The  current is  I  =  200 \ A

   The radius is  R =  0.001268 \  m

   The  length of the wire is  L  =  0.91 \  m\

    The  resistance is  R  =  0.126 \  \Omega

    The  outer surface temperature is  T _o  =  422.1 \  K

    The average thermal conductivity is  \sigma  =  22.5 W/mK

   

Generally the heat generated in the stainless steel wire is mathematically represented as  

    Q =  \frac{Power}{ \pi r^2L}

     Q =  \frac{I^2 R}{ \pi r^2L}

=>   Q =  \frac{200^2 * 0.126}{3.142 *  (0.001268)^2 * 0.91}

=>   Q =  1.096*10^{9}\  W/m^3

Generally the middle temperature is mathematically represented as

      T_m  =  T_o  + \frac{Q * r^2 }{ 6  * \sigma }

       T_m  =  422.1  +  \frac{1.096*10^{-9} * 0.001268^2}{6 * 22.5}

       T_m  =  435.2 \  K

4 0
3 years ago
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