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kirill [66]
4 years ago
11

The three forces shown act on a particle. what is the direction of the resultant of these three forces?

Physics
1 answer:
melisa1 [442]4 years ago
7 0
Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png

Solution:
We need to find the magnitude of the resultant on both x- and y-axis.

x-axis) The resultant on the x-axis is
F_x = 65 N\cdot cos 30^{\circ} - 30 N - 20 N\cdot sin 20^{\circ} = 19.45 N
in the positive direction.

y-axis) The resultant on the y-axis is
F_y = 65 N \cdot sin 30^{\circ} - 20 N \cdot cos 20^{\circ} = 13.70 N
in the positive direction.

Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
\tan \alpha =  \frac{F_y}{F_x} = \frac{13.70 N}{19.45 N}=0.7
from which we find 
\alpha=35^{\circ}
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7 0
3 years ago
What is the focal length of 2.30 D reading glasses found on the rack in a drugstore?
irakobra [83]

Answer:

Focal length, f = 0.43 meters

Explanation:

It is given that,

Power of the reading glasses, P = 2.3 D

We need to calculate the focal length of the reading glasses. The relationship between the power and the focal length inverse i.e.

P=\dfrac{1}{f}

f=\dfrac{1}{P}

f=\dfrac{1}{2.3\ D}

f = 0.43 m

So, the focal length of the reading glasses found on the rack in a drugstore is 0.43 meters.

8 0
4 years ago
A World-class sprinter can reach a top speed of about 11.5 m/s in the first 18.0 m of a race. What is the average acceleration o
irina [24]

Answer

a = 3.674 m / s ^ 2


t = 3.13 s

Using the kinematic equations for the movement we have:


h(t) = P_{0} + Vot + \frac{1}{2}at ^ 2 (1)


V_{f} = V_{0} + at (2)


Where:


P_{0} = initial position


V_{0}} = initial velocity


a = acceleration


t = time in seconds


V_{f} = final speed


We know:


P_{0}=0


V_{0}= 0

h = 18 m


V_{f} = 11.5\frac{m}{s}

  So:

 From (2) we have that: t =\frac{V_{f}}{a}


t =\frac{11.5}{a}

From (1) we have to:


h (t) = 0.5at ^ 2\\h = 18 = 0.5at ^ 2

Then we clear "a" to find the acceleration.


\frac{36}{t^2} = a\\a = \frac{36}{(\frac{11.5}{a})^2} \\\\a =\frac{11.5^2}{36}\\a = 3.674 m / s ^2

Then, the time it takes to reach this speed is:


t =\frac{V_{f}}{a}\\t =\frac{11.5}{3.674}\\t = 3.13 s

4 0
4 years ago
Read 2 more answers
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