The chicken's eggs are fertilized interally
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!
Answer:
mu = 0.56
Explanation:
The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v: final speed = 0m/s (the car stops)
v_o: initial speed in the interval of interest = 60km/h
= 60(1000m)/(3600s) = 16.66m/s
x: distance = 25m
BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

Furthermore, you use the relation between the friction force and the friction coefficient:

hence, the friction coefficient is 0.56
Answer:9.34 A/s
Explanation:
Given
radius of solenoid 
Emf induced 
no of turns per meter n=450
we know Induced EMF is given by

Magnetic Field is given by

thus 
Area of cross-section
where
solving integration we get

where r=distance from axis
R=radius of Solenoid



Answer:
9.43 m/s
Explanation:
First of all, we calculate the final kinetic energy of the car.
According to the work-energy theorem, the work done on the car is equal to its change in kinetic energy:

where
W = -36.733 J is the work done on the car (negative because the car is slowing down, so the work is done in the direction opposite to the motion of the car)
is the final kinetic energy
is the initial kinetic energy
Solving,

Now we can find the final speed of the car by using the formula for kinetic energy

where
m = 661 kg is the mass of the car
v is its final speed
Solving for v, we find
