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Lina20 [59]
3 years ago
10

Which of the following statements best describes ergonomics?

Physics
2 answers:
Romashka-Z-Leto [24]3 years ago
6 0

A.

Ergonomics involves the study of the human body.

B.

Ergonomics is the science of adjusting environments, tasks, or procedures to fit you, the healthcare worker.

C.

Ergonomics is the study of human psychology.

D.

Ergonomics is the science of adjusting you, the healthcare worker, to fit the environment, tasks, or procedures.

emmainna [20.7K]3 years ago
3 0
Can you provide the multiple choices​ ?
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What is 3,278,000 as scientific notation?
jonny [76]
3.278*10^6 I think. Sorry if it’s wrong.
4 0
3 years ago
A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s^2. If the student use
polet [3.4K]
The net force of the object is equal to the force applied minus the force of friction. 
                         Fnet = ma = F - Ff
                           12 kg x 0.2 m/s² = 15 N - Ff
The value of Ff is 12.6 N. This force is equal to the product of the normal force which is equal to the weight in horizontal surface and the coefficient of friction.
                             Ff = 12.6 N = k(12 kg)(9.81 m/s²)
The value of k is equal to 0.107. 
7 0
3 years ago
In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point
Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ  X volume of the oil drop

volume of the oil drop (spherical) =  (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

=  (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0
3 years ago
A perfectly elastic collision occurs between a 15.0-kg mass traveling at 3.50 m/s and a 9.00-kg mass traveling at 2.35 m/s. if t
BaLLatris [955]
Momentum is conserved in a collision. Momentum is mass*velocity, so you can find your answer by calculating initial and final momentums and setting them equal to each other.

15kg * 3.50 m/s + 9kg * 2.35 m/s = 73.65 kg m/s

73.65 = 9 * 2.8 + 15x

solve for x
x= 3.23

The final velocity is 3.23 m/s
5 0
3 years ago
With what force will a car hit a tree if the car has a mass of 3,415 kg and it is accelerating at
weeeeeb [17]

Answer:

F= 17,075\ N

Explanation:

When the car is under an accelerating force and hits a tree, the instant force received by the tree is the same force that is accelerating the car.

The accelerating force can be calculated using Newton's second law:

F=m\cdot a

Where m is the mass of the car and a is the acceleration.

F=3,415\ kg\cdot 5\ m/s^2

\boxed{F= 17,075\ N}

3 0
3 years ago
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