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Elanso [62]
3 years ago
10

5. List five materials that we use in compost?​

Physics
1 answer:
kramer3 years ago
6 0

Leaves.

Grass clippings.

Brush trimmings.

Manure (preferably organic)

Any non-animal food scraps: fruits, vegetables, peelings, bread, cereal, coffee grounds and filters, tea leaves and tea bags (preferably minus the staples)

Old wine.

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A solenoid 1.85 m long and 2.20 cm in diameter carries a current of 21.0 A. The magnetic field inside the solenoid is 25.0 mT. F
kobusy [5.1K]

Answer:

A wire carrying a 30.0-A current passes between the poles of a strong magnet that is perpendicular to its field and experiences a 2.16-N force on the 4.00 cm of wire in the field. What is the average field strength?

Explanation:

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8 0
2 years ago
A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati
Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

  • If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

       \tau = I * \alpha  (1)

  • Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
  • Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
  • τ = F*r (2)
  • For a solid uniform disk, the rotational inertia regarding an axle passing through its center  is just I = m*r²/2 (3).
  • Replacing (2) and (3) in (1), we can solve for α, as follows:

       \alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)

  • Since the angular acceleration is constant, we can use the following kinematic equation:

        \omega_{f}^{2}  - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)

  • Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

       0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)

  • Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

       \omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)

  • Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

        v = \omega * r (8)

  • where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of  the disk, r becomes the radius of the disk, 0.200 m.
  • Replacing this value and (7) in (8), we get:

       v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)

b)    

  • There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

       a_{t} = \alpha * r (9)

  • where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
  • Replacing this value and (4), in (9), we get:

       a_{t}  = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)

  • Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
  • The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

       a_{c} = \omega^{2} * r  (11)

  • Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
  • Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
  • Replacing this value and (7) in (11) we get:

       a_{c} = \omega^{2} * r   = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)

  • The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
  • Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

       a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)

6 0
3 years ago
Among the alkali earth metals, the tendency to react with other substances
padilas [110]
Answer D
In alkali earth metals reacrivity increases from top to bottom (opposite of b)
This is because as you go down, the electron shells increase by 1 shell. The farther away a shell is from the nucleus, the higher its tendency to react.
D is true because the more reactive an alkali metal is, the more vigorous the reaction will be with water.
4 0
3 years ago
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What is the meaning of critical angle in physics​
bonufazy [111]

Answer:

It's an Angle of incidence that provides a 90° angle but is also refracted at the same time. it's used to find the water-air boundary (which is 48.6 degrees). in addition, its an angle of incidence value.

6 0
3 years ago
You have a pick-up truck that weighed 4,000 pounds when it was new. you are modifying it to increase its ground clearance. when
jok3333 [9.3K]
U have to *modify it to increase its ground clearance*
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