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Galina-37 [17]
3 years ago
6

Find the tension in the two groups that are holding the 2.9 kg object in Pl., One makes an angle of 35.6° with respect to the ve

rtical group 2 is pulling horizontally
Physics
1 answer:
Stella [2.4K]3 years ago
4 0

Answer:

≈ 20.35 N [newton's of tension]

Explanation:

( (2.9 × 9.8) ÷ cos(35.6°) ) × sin (35.6°) =

( (28.42) ÷ (≈0.813) ) × (≈0.582) =

(≈34.96) × (≈0.582) = 20.3449446.... ≈ 20.35

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Answer:

helium-4 (90%) or tritium (7%).

Explanation:

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Which of the following can be an effect of a volcanic winter? select the two correct answers
Marysya12 [62]

it is B. Famine and C. Climate cooling

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A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A
ArbitrLikvidat [17]

Answer:

The mass of the solid cylinder is m  =  1612.5  \  kg

Explanation:

From the question we are told that

   The radius of the grinding wheel is R =  0.330 \ m

   The  tangential force is F_t =  250 \ N

    The angular acceleration is  \alpha  =  0.940 \ rad/s^2

The torque experienced by the wheel is mathematically represented as

     \tau  =  I  *  \alpha

Where  I  is the moment of inertia

The torque experienced by the wheel can also be  mathematically represented as

       \tau  =  F_t  * r

substituting values

       \tau  =  250 * 0.330

      \tau  = 82.5  \ N\cdot m

So

   82.5  =  I *  \alpha

    82.5  =  I *  0.940

So

   I  =  87.8 \ kg \cdot m^2

This moment of inertia can be mathematically evaluated as

     I  =  \frac{1}{2} * m* r^2

substituting values

  87.8  =  \frac{1}{2} * m* (0.330)^2

=>   m  =  1612.5  \  kg

5 0
3 years ago
8. Which element will form a 2-ion?<br> sulfur<br> Calcium<br> potassium<br> helium
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3 years ago
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The kinetic friction coefficient between a cabinet and the floor is 0.3. Mass of the cabinet is 300kg. A man pushes the cabinet
podryga [215]

Answer:

<h2>0.39m/s^2</h2>

Explanation:

Step one:

given data

mass m= 300kg

applied force F= 1000N

coefficient of friction μ= 0.3

Step two:

The net force Fn= applied force-friction force  

Fn=F-F1

F1= limiting force

F1=μ*m*g

F1=0.3*300*9.81

F1=882.9N

the Net force= 1000-882.9

Fn=117.1N

Step three:

we know that

F=ma

Fnet=ma

a= Fnet/m

a=117.1/300

a=0.39m/s^2

7 0
3 years ago
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