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aleksandr82 [10.1K]
2 years ago
7

The arm of a crane at a construction site is 17.0 m long, and it makes an angle of 11.6 ◦ with the horizontal. Assume that the m

aximum load the crane can handle is limited by the amount of torque the load produces around the base of the arm. What maximum torque can the crane withstand if the maximum load the crane can handle is 643 N? Answer in units of N · m.
Physics
1 answer:
riadik2000 [5.3K]2 years ago
8 0

Answer:

τ_max ≈ 10,708 N.m

Explanation:

To find max torque, we need the component of force that is perpendicular to the direction of vector r.

Now, Torque is the perpendicular component of the force times the distance.

Thus;

τ_max = F x r

τ_max = (643 cos 11.6°) (17 )

τ_max ≈ 10,708 N.m

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Deer ticks can carry both Lyme disease and human granulocytic ehrlichiosis (HGE). In a study of ticks in the Midwest, it was fou
tiny-mole [99]

Answer:

a)0.024

b)0.148

Explanation:

Let 's represent the set of deer ticks Carrying Lyme disease with L and the set of deer ticks carrying Human Granulocytic Ehrlichiosis with H

Given:

P(L) = 0.16

P(H) = 0.10

P(L n H) = 0.1 ·P( L u H )

Hence, P( L u H) = 10 ·P( L nH)

(a)

Hence. using the equation. P(L U H) = P(L) + P(H) - P(L n H)

Hence, 10 · P(L n H ) = 0.16 + 0.1 - P(L n H )

Hence, 11 · P(L n H) = 0.16 + 0.1 = 0.26

Hence, P(L n H) = 0.26/11=0.024

(b)

We know that condition probability P(H ║ L) = p(L n H)/P(L)

hence, P(H ║ L) =(0.26/11)/0.16 =0.148

3 0
2 years ago
Energy conservation
balandron [24]

Answer:

7.328m/s

Explanation:

Given parameters:

height of table = 0.68m

final velocity of the ball = 6m/s

Unknown:

Initial velocity of ball = ?

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            V = U + 2gH

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 V is the final velocity

 U is the initial velocity

 g is the acceleration due to gravity

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    Since U is the unknown, let us make it the subject of the expression;

           U = V - 2gH

        U = 6 - (2 x 9.8 x 0.68)  = 7.328m/s(deceleration)

7 0
3 years ago
The two major parts of the optical telescope
nydimaria [60]
The "objective" (lens or mirror) is the major major major part of
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You can put a piece of film or a CCD right at the focal point of
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If you want to use the telescope for looking through and seeing stuff
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5 0
2 years ago
Easy defimation of newtons second law of motion
shtirl [24]
Hi pupil here's your answer ::

_______________________________

Newton's Second Law  of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
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8 0
3 years ago
a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
finlep [7]

Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

4 0
3 years ago
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