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klasskru [66]
2 years ago
10

A weight lifter picks up a barbell and 1. lifts it chest high 2. holds it for 30 seconds 3. puts it down slowly (but does not dr

op it). rank the work w that the weight lifter does during each of these three operations. label the quantities as w1, w2, and w3. (hint: think about how work is defined in terms of who is applying forces and who is doing work.) w3 = w2 = w1 w3 = w1 > w2 w2 > w1 > w3 none of the above w1 > w2 > w3 w3 > w2 > w1 w2 > w3 > w1 justify your ranking order.
Physics
1 answer:
Anna11 [10]2 years ago
7 0

1. lifts it chest high

The force opposing to this action is the force due to gravity. Therefore the work done is:

W1 = m g d

where m is mass of the barbell, g is gravity and d is displacement

 

2. holds it for 30 seconds

Work is a product of force and displacement, since there is no displacement, therefore work done is zero.

W2 = 0

 

3. puts it down slowly

If the barbell was dropped, then it would simply be a free fall. But since it was not, so the work done here is also equal to the weight of the barbell times displacement:

W3 = m g d

 

We can see that W1 = W3, and since W2 = 0, therefore the answer is:

<span>w3 = w1 > w2</span>

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a loaded sack of total mass is 1000 gramme falls down from the floor of a lorry 200cm high, calculate the workdone by the gravit
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Answer:

W = 20 J

Explanation:

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The mass of a loaded sack, m = 1000 g = 1 kg

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W = mgh

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W = 1 × 10 × 2

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8 0
3 years ago
The drill used by most dentists today is powered by a small air-turbine that can operate at angular speeds of 350000 rpm. These
alexdok [17]

Answer:

θ  = 6.3 *10³ revolutions

Explanation:

Angular acceleration of the drill

We apply the equations of circular motion uniformly accelerated

ωf= ω₀ + α*t  Formula (1)

Where:  

α : Angular acceleration (rad/s²)  

ω₀ : Initial angular speed ( rad/s)  

ωf : Final angular speed ( rad

t :  time interval (s)

Data

ω₀ = 0

ωf = 350000 rpm = 350000 rev/min

1 rev = 2π rad

1 min= 60 s

ωf = 350000 rev/min =350000*(2π rad/60 s)

ωf = 36651.9 rad/s

t = 2.2 s

We replace data in the formula (2) :

ωf= ω₀ + α*t

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α = 36651.9 / (2.2)

α = 17000 rad/s²

Revolutions made by the drill

We apply the equations of circular motion uniformly accelerated

ωf²= ω₀ ²+ 2α*θ Formula (2)

Where:  

θ : Angle that the body has rotated in a given time interval (rad)

We replace data in the formula (2):  

(ωf)²= ω₀²+ 2α*θ

(36651.9)²= (0)²+ 2( 17000 )*θ

θ = (36651.9)²/ (34000 )

θ  = 39510.64 rad = 39510.64 rad* (1 rev/2πrad)

θ  = 6288.31 revolutions

θ  = 6.3 *10³ revolutions

3 0
3 years ago
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