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Morgarella [4.7K]
2 years ago
5

A woman who weighs 500 N stands on an 8.0-m-long board that weighs 100 N. The board is supported at each end. The support force

at the right end is 3 times the support force at the left end. How far from the right end is the woman standing?

Physics
1 answer:
aniked [119]2 years ago
8 0

Answer:

The woman's distance from the right end is 1.6m = (8-6.4)m.

The principles of moments about a point or axis running through a point and summation of forces have been used to calculate the required variable.

Principle of moments: the sun of clockwise moments must be equal to the sun of anticlockwise moments.

Also the sun of upward forces must be equal to the sun of downward forces.

Theses are the conditions for static equilibrium.

Explanation:

The step by step solution can be found in the attachment below.

Thank you for reading this solution and I hope it is helpful to you.

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Answer:

(C) T

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Note: The figure of question is added in the attachment

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Consider a 2100-kg car cruising at constant speed of 70 km/h. Now the car starts to pass another car by accelerating to 110 km/h
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Explanation: Using the kinematic equation for the final velocity for a constant acceleration we have:

Vf=Vi+a*t

replacing the values the results is

a=(Vf-Vi)/t= (30.55 m/s-19.44 m/s)/5s= 2.22 m/s^2

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Finally to calculate the increment of power get the final velocity we have to use Neton second law to determine the Force applied to the car.

F=m* a=2100 Kg* 2.22 m/s^2= 4666.2 N

Then increment  power to accelerate is given by:

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What are two parts that make up a vector
mojhsa [17]

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3 0
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A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15° below the
Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, (\theta) = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ F_x=Fcos(\theta)          and             ⇒ F_y=Fsin (\theta)

⇒ F_x=15cos(15)                           ⇒ F_y=15sin (15)

⇒ Applying concept of  forces.

⇒ \sum F_x=F_n_e_t =F-f

⇒ F_n_e_t =F-f

⇒ ma =F-f       <em>  ...Newtons second law Fnet = ma</em>

⇒ a =\frac{F-f}{m}              

⇒ Plugging the values.

⇒ a =\frac{15cos(15)-0}{40}     <em>...f is the friction which is zero here.</em>

⇒ a =\frac{14.48}{40}

⇒ a=0.362\ ms^-^2

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0
2 years ago
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