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2 years ago

A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task. 1.84 x 103 J

1 answer:

4 0

work will be equal to the potential energy gained by the person in climbing the stairs. work= potential energy gained = mgh W= 75kg*9.8m/s2*2.50m= 1837.5 J

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You have a light spring which obeys Hooke's law. This spring stretches 2.92 cm vertically when a 2.70 kg object is suspended fro

ehidna [41]

(a) 907.5 N/m

The force applied to the spring is equal to the weight of the object suspended on it, so:

$F=mg=(2.70 kg)(9.8 m/s^2)=26.5 N$

The spring obeys Hook's law:

$F=k\Delta x$

where k is the spring constant and $\Delta x$ is the stretching of the spring. Since we know $\Delta x=2.92 cm=0.0292 m$ , we can re-arrange the equation to find the spring constant:

$k=\frac{F}{\Delta x}=\frac{26.5 N}{0.0292 m}=907.5 N/m$

(b) 1.45 cm

In this second case, the force applied to the spring will be different, since the weight of the new object is different:

$F=mg=(1.35 kg)(9.8 m/s^2)=13.2 N$

So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):

$\Delta x=\frac{F}{k}=\frac{13.2 N}{907.5 N/m}=0.0145 m=1.45 cm$

(c) 3.5 J

The amount of work that must be done to stretch the string by a distance $\Delta x$ is equal to the elastic potential energy stored by the spring, given by:

$W=U=\frac{1}{2}k\Delta x^2$

Substituting k=907.5 N/m and $\Delta x=8.80 cm=0.088 m$ , we find the amount of work that must be done:

$W=\frac{1}{2}(907.5 N/m)(0.088 m)^2=3.5 J$

5 0

2 years ago

You apply the brakes of your car abruptly and your book starts sliding off the front seat. Three observers sitting in the car ex

krek1111 [17]

Answer:

All the observers are correct.

Explanation:

This is simply a problem of reference frames from which the motion of the book is being viewed by the various observers.

From their various reference frames, they are all correct.

Observer A must be in the inertial reference frame.

Observers who can explain the behavior of the book and the car by using the relationship between the sum of the forces and changing velocity are said to be observers in inertial reference frames.

This is clearly shown by what observer A noticed. There was a relative motion between the book and the car as she pointed out, making her to be in an inertial reference frame.

Similarly, observers in inertial reference frames can also explain the changes in velocity of objects by considering the forces exerted on them by other objects.

This is shown by observer B as he is able to notice how the force of the car affects the velocity of the book.

Observer C is actually in a non-inertial reference frame, as newtons law of force motion relationship are no longer observed. This occurs in the non inertial reference frame.

7 0

2 years ago

Anna pushes a box with a force of 8.00 newtons. She generates a power of 3.00 watts. How much time does it take for Anna to move

QveST [7]

Power is the energy in a system per time. It will have units of Watts which is equal to joules per second. It can be expressed as:

P = E / t

where E = Force x distance

P = Fd / t
t = Fd / P
t = 8 (9.72) / 3.0
t = 25.92 s

8 0

2 years ago

Read 2 more answers

1. A particular lever is 90.0% efficient. If 50.0 J of work are done on the lever, then how much work does the lever do on its l

laila [671]

Answer:

Explanation:

Using the efficiency formula;

Efficiency = Work done by the machine (output)/work done on the machine (input) ×100%

Efficiency =w/50 ×100

90 = 100w/50

Cross multiply

90×50 = 100W

4500 = 100W

W = 4500/100

W = 45Joules

Hence the lever does 45Joules of work on its load

2) Mechanical Advantage= Load/Effort

Given

MA = 4

Load = 500N

4 = 500/Effort

Effort = 500/4

Effort =125N

Hence the effort required to lift the load is 125N

8 0

1 year ago

Sasha is ordered Ampicillin 50mg/kg/day x 48 hours, to be given every 6 hours in 100mls of N/S run over 30 minutes. The tubing h

Anna007 [38]

Answer:

The correct dose = 1454.54 mg

and The jnfusion rate = 41.67 gitt/hr

Explanation: the correct dose will be 50mg/kg × kg/2.2 × 64lb

= 1454.54 mg

infusion rate will be

10 gtts/ml × 50mg/6 × 30/60

Infusion rate = 15000/360

= 41.67 gitt/hr

5 0

2 years ago