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xxMikexx [17]
10 months ago
6

(HURRY 20 MINS)

Physics
1 answer:
leonid [27]10 months ago
4 0

Answer:

Roles and responsibilities

Explanation:

Definition of safety plan:

"A Safety Plan is a written document that describes the process for identifying the physical and health hazards that could harm workers, <em>procedures to prevent accidents</em>, and steps to take when accidents occur. Written safety plans can be comprehensive, such as an injury and illness prevention program, or they can be specific to a particular activity, hazard, or piece of equipment. The written safety plan is your blueprint for keeping workers safe."

Alternative definition

"What is an OSHA Safety Plan? An OSHA Safety Plan is a written plan that describes the potential hazards in the workplace, <u><em>and the company policies</em></u>, controls, and <u><em>work practices</em></u> used to minimize those hazards."

elements of a safety plan:

Basic Safety Plan Elements

Policy or goals statement

<u><em>List of responsible persons</em></u>

Hazard identification

<em>Hazard controls and safe practices</em>

<em>Emergency and accident response</em>

Employee training and communication

<em>Recordkeeping</em>

I say roles and responsibilities because it makes sense that if it's your responsibility and possibly something that could be dangerous -- hence a safety plan --  you would have to sign it before working. I hope this helps!

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All electric devices are required to have identifying plates that specify their electrical characteristics. The plate on a certa
masha68 [24]

Answer:

R= 20 ohm  

Explanation:

Given that

Current ,I = 6 A

Voltage difference ,ΔV = 120 V

Lets take resistance of the stem iron = R

We know that ,the relationship between current ,voltage difference and resistance is given as

ΔV = I R

R=\dfrac{\Delta V}{I}

Now by putting the values in the above equation we get

R=\dfrac{120}{6}\ ohm

R= 20 ohm

Therefore the resistance of the steam  will be 20 ohm.

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3 years ago
A car traveling at 60 mph has how much more energy than a car going at 15 mph?
VikaD [51]

Answer:

No, if a car is going faster. The RPM is obviously higher. If that is higher, you can burn through gas and energy much faster. A car going at 15mph would be cruising and wouldn't have to worry too much about burning our your vehicle.

Explanation:

7 0
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Why does increasing the number of trials increase confidence in the results of the experiment?
Paladinen [302]
It increases confidence because the more times you conduct the same experiment over and over should either prove your hypothesis right and wrong and eliminate any random occurrences that might affect your results.
8 0
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What type of material is wrapped around an object could cause the object to lose the most heat
Dmitrij [34]
<span>A sheet of copper could cause the object to lose the most amount of heat. Copper is an essential element and a good conductor of heat. Heat can transfer from one end of a piece of copper to the other end.</span>
7 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
2 years ago
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