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Darina [25.2K]
2 years ago
13

Identify and sketch all the external forces acting on the chair. Because the chair can be represented as a point particle of mas

s m, draw the forces with their tails centered on the black dot in the middle of the chair. Be certain to draw your forces so that they have the correct orientation. Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The length of the vectors will not be graded.
Physics
1 answer:
Monica [59]2 years ago
5 0

Answer:

y axis normal (N) and the weight (W)

x axis  pplied force (F) and friction force (fr)

Explanation:

If we have a chair on a horizontal surface, the normal (N) and the weight (W) of the body act on the vertical axis.

On the x axis, the applied force (F) acts in the direction of movement and the friction force (fr) in the opposite direction of movement.

In this exercise we assume that the body tends to move to the right, all the forces can be seen in the adjoint

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Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in 2.70 s, while
Vinvika [58]

Answer:

Explanation:

Given

Time taken to reach ground is t=2.7\ s

Malda initial velocity u=95\ cm/s

Let h be the height of Cliff

using h=ut+\frac{1}{2}at^2

where, u=initial velocity

t=time

In first case chirpy drop downward thus u=0

h=0+\frac{1}{2}(9.8)(2.7)^2

h=35.72\ m

For Milada there is horizontal velocity u=95 cm/s=0.95 m/s[/tex]

time taken to reach the ground will be same so distance traveled in this time with 0.95 m/s horizontal velocity is given by

R=u\times t

R=0.95\times 2.7=2.43\ m    

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3 years ago
A balloon filled with helium has a small hole on the right side of the balloon. How would Pascal explain why the entire balloon
Morgarella [4.7K]

Answer:

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4 0
2 years ago
A 65.0-kg woman steps off a 10.0-m diving platform and drops straight down into the water. 1) If she reaches a depth of 3.20 m,
timurjin [86]

Answer:

F=2627.6N

Explanation:

The work done by this resistive force while traveling a distance <em>d</em> underwater would be:

W=F.d=-Fd

where the minus sign appears because the force is upwards and the displacement downwards.

This work is equal to the change of mechanical energy. At the diving plataform and underwater, when she stops moving, the woman has no kinetic energy, so all can be written in terms of her total change of gravitational potential energy:

W=\Delta E=U_f-U_i=mgh_f-mgh_i=mg(h_f-h_i)

Putting all together:

F=-\frac{W}{d}=-\frac{mg(h_f-h_i)}{d}=-\frac{(65kg)(9.8m/s^2)(-3.2m-10m)}{3.2m}=2627.6N

7 0
2 years ago
Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa
tester [92]

Answer:

Part a)

t = 2.83 s

Part b)

Ball thrown downwards =v_f = 23.8 m/s

Ball thrown upwards =v_f = 23.8 m/s

Part c)

d = 22.24 m

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

\Delta y = 0 = v_y t + \frac{1}{2}at^2

0 = 13.9 t - \frac{1}{2}(9.81) t^2

t = 2.83 s

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

v_f^2 - v_i^2 = 2 a \Delta y

v_f^2 - (13.9)^2 = (2)(-9.81)(-19.1)

v_f^2 = 567.9

v_f = 23.8 m/s

Part c)

Relative speed of two balls is given as

v_{12} = v_1 - v_2

v_{12} = (13.9) - (-13.9) = 27.8 m/s

now the distance between two balls in 0.8 s is given as

d = v_{12} t

d = 27.8 \times 0.8

d = 22.24 m

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Do you mean in general or in a piece of paper?
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