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omeli [17]
3 years ago
9

As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil

l cause the object to be in equilibrium?
3.8 N at 108° counterclockwise from
F1->

6.3 N at 162° counterclockwise from
F1->

3.8 N at 162° counterclockwise from
F1->

6.3 N at 108° counterclockwise from
F1->

Physics
1 answer:
galina1969 [7]3 years ago
7 0

Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

F_1 = 5.7\ N\\\\F_2 = 1.9\ N

We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

F1->

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A disk shaped grindstone of mass 3.0 kg and radius 8 cm is spinning at 600 rpm. After the power is shut off the frictional torqu
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Answer:

\theta=50\ revolution

Explanation:

It is given that,

Mass of the grindstone, m = 3 kg

Radius of the grindstone, r = 8 cm = 0.08 m

Initial speed of the grindstone, \omega_i=600\ rpm=62.83\ rad/s

Finally it shuts off, \omega_f=0

Time taken, t = 10 s

Let \alpha is the angular acceleration of the grindstone. Using the formula of rotational kinematics as :

\alpha =\dfrac{\omega_f-\omega_i}{t}

\alpha =\dfrac{0-62.83}{10}

\alpha =-6.283\ rad/s^2

Let \theta is the number of revolutions of the grindstone after the power is shut off. Now using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

\theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha }

\theta=\dfrac{-62.83^2}{2\times -6.283}

\theta=314.15\ radian

\theta=49.99\ revolution

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\theta=50\ revolution

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122397.7

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A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much work was done?
jeka57 [31]

Answer:

<h3>The answer is 45 J</h3>

Explanation:

The work done by an object can be found by using the formula

<h3>workdone = force × distance</h3>

From the question

distance = 3 meters

force = 15 newtons

We have

workdone = 15 × 3

We have the final answer as

<h3>45 J</h3>

Hope this helps you

7 0
3 years ago
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