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masha68 [24]
2 years ago
5

A standard bathroom scale is placed on an elevator. A 28 kg boy enters the elevator on the first floor and steps on the scale. W

hat will the scale read (in newtons) when the elevator begins to accelerate upward at 0.5 m/s2
Physics
1 answer:
kramer2 years ago
4 0

Answer:

Explanation:

Newton's Second Law is pretty much the standard for all motion that involves a force. It applies to gravitational force and torque and friction and weight on an elevator. The main formula for force is

F = ma. We have to adjust that to take into account that when the elevator is moving up, that "surge" of acceleration weighs down a bit on the scale, causing it to read higher than the actual weight until the acceleration evens out and there is no acceleration at all (no acceleration simply means that the velocity is constant; acceleration by definition is a change in velocity, and if there is no change in velocity, there is 0 acceleration). The force equation then becomes

F_n-w=ma  where F_n is normal force. This is what the scale will read, which is what we are looking for in this problem (our unknown). Since we are looking for F_n, that is what we will solve this literal equation for:

F_n=ma+w .  m is the mass of the boy, a is the acceleration of the elevator (which is going up so we will call that acceleration positive), and w is weight. We have everything but the unknown and the weight of the boy. We find the weight:

w = mg so

w = 28(9.8) and

w = 274.4 N BUT rounding to the correct number of significance we have that the weight is actually

w = 270 N.

Filling in the elevator equation:

F_n=28(.50)+270 and according to the rules of significant digits, we have to multiply the 28(.50) {notice that I did add a 0 there for greater significance; if not that added 0 we are only looking at 1 significant digit which is pretty much useless}, round that to 2 sig fig's, and then add to 170:

F_n=14+270 and adding, by the rules, requires that we round to the tens place to get, finally:

F_n=280N  So you see that the surge in acceleration did in fact add a tiny bit to the weight read by the scale; conversely, if he were to have moved down at that same rate, the scale would have read a bit less than his actual weight). Isn't physics like the coolest thing ever!?

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Spiderman, whose mass is 74.0 kg, is dangling on the free end of a 11.0-m-long rope, the other end of which is fixed to a tree l
Anestetic [448]

Answer:

W = -1844.513 J

Explanation:

GIVEN DATA:

mass of spider man is m  74 kg

vertical displacement if spider is 11 m

final displacement  =  11 cos 60.6 =  - 6.753 m

change in displacement is  = -6.753 - (-11) = 4.25 m

gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N

work done by gravity is W = F \delta r cos\theta

W = 725.2 \times 4.25 \times cos 180

where 180 is the angle between spiderman weight and displacement

W = -1844.513 J

7 0
3 years ago
A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia
aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

2) The potential energy at <u>point A</u> is 19620 J

3) The kinetic energy at <u>point B</u> is 10010 J

4) The potential energy at <u>point C</u> is zero

5) The kinetic energy at <u>point C</u> is 19820 J

6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

E_{t} = KE_{i} + PE_{i}

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The<em> total energy</em> is:

E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

   

2) The <em>potential energy</em> at point A is:

PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

KE_{A} + PE_{A} = KE_{B} + PE_{B}

KE_{B} = KE_{A} + PE_{A} - PE_{B}

Since

KE_{A} + PE_{A} = KE_{i} + PE_{i}

we have:

KE_{B} = KE_{i} + PE_{i} - PE_{B} =  19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J

5) The <em>kinetic energy</em> of the roller coaster at point C is:

KE_{i} + PE_{i} = KE_{C} + PE_{C}            

KE_{C} = KE_{i} + PE_{i} = 19820 J      

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

KE_{C} = \frac{1}{2}mv_{C}^{2}

v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

Read more here:

brainly.com/question/21288807?referrer=searchResults

I hope it helps you!

3 0
3 years ago
What force is acting as the centripetal force for the International Space Station as it orbits the Earth?
laila [671]

Answer:

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Explanation:

6 0
2 years ago
A wet bar of soap slides down a ramp 9.2 m long inclined at 8.0∘ .
xenn [34]

Answer:

The time taken by the bar to reach the bottom t=4.886s

Given:

Displacement of the bar S=9.2m

Angle of inclination \theta=8.0^{\circ}

Coefficient of friction factor \mu k=0.056

To find:

How long it takes to reach the bottom ‘t’

<u>Step by Step Explanation:</u>

Solution:

We know that the formula for weight of the soap bar is given as

F_{g}=m g \sin \theta

The frictional force acting on this soap bar is determined by

F_{f}=\mu m g \cos \theta

To determine the constant acceleration of the bar, we derive as

F=m a

Here F=F_{g}-F_{f} and thus

F_{g}-F_{f}=m am g \sin \theta-\mu m g \cos \theta=m a

a=g \sin \theta-\mu g \cos \theta

WhereF_{g}=Force imparted due to weight

F_{f}=Frictional Force

m=Mass of the bar

g=Acceleration due to gravity

a=Acceleration of the bar

\sin \theta and \cos \theta are the angles involved in the system

If the bar starts from the rest

Equations of motion involved in calculating the displacement of the bar is given as

s=\frac{1}{2} a t^{2}, From this

 a t^{2}=2 s

t^{2}=\frac{2 s}{a}

t=\sqrt{\frac{2 s}{a}}

Where s= displacement or length moved by the bar

a=Acceleration of the bar

t=Time taken to reach bottom

Substitute all the known values in the above equation we get

t=\sqrt{\frac{2 \times 9.2}{a}} and we know that

a=g \sin \theta-\mu g \cos \theta

=9.8 \times \sin 8.0-0.056 \times 9.8 \times \cos 8.0

=1.364-0.543

a=0.821

t=\sqrt{\frac{2 \times 9.2}{0.821}}

t=\sqrt{\frac{19.6}{0.821}}

t=\sqrt{23.87332}

t=4.886s

Result:

Thus the time taken by the bar to reach the bottom is t=4.886s

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3 years ago
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7 0
3 years ago
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