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mel-nik [20]
3 years ago
5

You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If

60-kg Dan sits on the left end of the board and 50-kg Tahreen on the right end of the board, where should 54-kg Komila sit to keep the board stable? Ignore the mass of the board and treat each student as point-like objects The positive r-direction is to the right, and the origin is at the center of the chair
Part A Determine the coordinate of the leftmost point on the board where Komila can sit to keep the board stable
Part B Determine the coordinate of the rightmost point on the board where Komila can sit to keep the board stable

Physics
1 answer:
kakasveta [241]3 years ago
5 0

Answer:

A) for leftmost point the coordinate is -0.28m that means it should be 0.28m towards the right.

B) for rightmost case the coordinate is 0.28m which is where komila should sit.

Explanation:

Detailed calculation and explanation is shown in the image below

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Answer: To answer this question, we will need the following equation: SPEED = DISTANCE/TIME (A multiplication and division triangle will be shown)i) The speed of the car is calculated by doing 100 metres/ 20 seconds which gives us 5 metres per second. ii) Rearranging the equation earlier, we can make the distance the subject of the equation so that we get SPEED x TIME = DISTANCE. We worked out the speed and the time was given as 1 minute 40 seconds but we cannot plug in the numbers yet as the time has to be converted to units of seconds (because our speed is in meters per second). 1 minute 40 seconds = 60 seconds + 40 seconds = 100 secondsWe then plug in the numbers to get the distance travelled = 5 metres per second x 100 seconds = 500 metres.

Explanation:

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You and your lab partner, Mel, are engrossed in a chemistry activity using beans to compute the average atomic mass of an elemen
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Answer:

C:

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Read 2 more answers
A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav
Vinvika [58]
<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

4 0
3 years ago
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