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valentina_108 [34]

2 years ago

6

A 5.00-m-long uniform ladder, weighing 200 N, rests against a smooth vertical wall with its base on a horizontal rough floor, a

distance of 1.20 m away from the wall. The coefficient of static friction between the ladder and the floor is 0.200. How far up the ladder, measured along the ladder, can a 600-N person climb before the ladder begins to slip

1 answer:

Morgarella [4.7K]2 years ago

5 0

Answer:

0.488 m

Explanation:

If θ be the angle ladder makes with the plane

cos θ = 1.2 / 5

Tan θ = 4.04

Let the height a person of weight 600 N can climb be h from the ground .

Distance from the base point where ladder touches the floor = h / tanθ

= h / 4.04

Total reaction force = total downward force

R = 200 + 600

800 N

Frictional force = μ R

= .2 x 800

= 160 N

Taking moment of force about the point on the ladder where it touches the floor and balancing them

200 x 1.2 x .5 + 600 x h / tanθ = μ R x 1.2 / tanθ ( reaction at the top point of ladder where it touches the wall is R₁ and

R₁ =μ R )

= 200 x 1.2 x .5 + 600 x h / tanθ = 160 x 1.2 / tanθ

120 - 600 h / 4.04 = 47.52

120 - 47.52 = 600 h / 4.04

72.48= 148.51 h

h = 0.488 m

=

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Answer:

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Explanation:

We are given that

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We have to find the direction of electric field at z=a on the positive z-axis if $\lambda_1$ and $\lambda_2$ are positive.

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2 years ago

A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external forceof 3.0 N, always tangent to the t

kodGreya [7K]

The work done by the force is 47.1 J

Explanation:

The work done by a force in moving an object is given by

$W=Fd cos \theta$ (1)

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$\theta$ is the angle between the direction of the force and the motion of the object

In this problem, the force applied to the object is

F = 3.0 N

This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

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where r = 2.5 m is the radius.

Now we can substitute into eq.(1) to find the work done:

$W=(3.0)(15.7)(cos 0)=47.1 J$

Learn more about work:

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2 years ago

(a) If two sound waves, one in a gas medium and one in a liquid medium, are equal in intensity, what is the ratio of the pressur

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Answer:

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Explanation:

Given;

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density of liquid, $\rho _l$ = 972 kg/m³

speed of sound in gas, $C_g$ = 376 m/s

speed of sound in liquid, $C_l$ = 1640 m/s

The of the sound wave is given by;

$I = \frac{P_o^2}{2 \rho C} \\\\P_o^2 = 2 \rho C I\\\\p_o = \sqrt{2 \rho CI}$

Where;

$P_o$ is the pressure amplitude

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(b) when the pressure amplitudes are equal, the ratio of the intensities is given as;

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2 years ago