Ask question

Ask question

All categories

soldi70 [24.7K]

3 years ago

12

A uniform beam resting on two pivots has a length L = 6.00 m and weight M = 220 lbs. The pivot under the left end exerts a norma

l force n1 on the beam, and the second pivot, located a distance l = 4.00 m from the end exerts a normal force n2. A woman with a weight of 130 lbs steps onto the left end of the beam and begins walking to the right. How far can the woman walk before the beam tips?

1 answer:

gulaghasi [49]3 years ago

4 0

Answer:

x = 4,138 m

Explanation:

For this exercise, let's use the rotational equilibrium equation.

Let's fix our frame of reference on the left side of the pivot, the positive direction for anti-clockwise rotation

∑ τ = 0

n₁ 0 - W L / 2 + n₂ 4 - W_woman x = 0

x = (- W L / 2 + 4n2) / W_woman

Let's reduce the magnitudes to the SI System

M = 6 lbs (1 kg / 2.2 lb) = 2.72 kg

M_woman = 130 lbs = 59.09 kg

Let's write the transnational equilibrium equation

n₁ + n₂ - W - W_woman = 0

n₁ + n₂ = W + W_woman

n₁ + n₂ = (2.72 + 59.09) 9.8

At the point where the system begins to rotate, pivot 1 has no force on it, so its relation must be zero (n₁ = 0)

n₂ = 605,738 N

Let's calculate

x = (-2.72 9.8 6/2 + 4 605.738) / 59.09 9.8

x = 4,138 m

You might be interested in

You want to move a heavy box with mass 30.0 across a carpeted floor. You pull hard on one of the edges of the box at an angle 30

ivolga24 [154]

Answer:

Explanation:

Check attachment for solution

6 0

3 years ago

A material that does not conduct electricity is called a... what?

xxTIMURxx [149]

A material that does not conduct electricity is known as an insulator

3 0

3 years ago

Charge is flowing through a conductor at the rate of 420 C/min. If 742 J. of electrical energy are converted to heat in 30 s., w

Y_Kistochka [10]

Answer:

3.53 V

Explanation:

Electric charge: The is the rate of flow of electric charge along a conductor.

The S.I unit of electric charge is C.

Mathematically it is expressed as,

Q = It ............................ Equation 1

Where Q = electric charge, I = current, t = time.

I = Q/t.......................... Equation 2

From the question, charge flows through the conductor at the rate of 420 C/mim

Which means in 1 min, 420 C of charge flows through the conductor.

Hence,

Q = 420 C, t = 1 min = 60 seconds

Substitute into equation 2

I = 420/60

I =7 A

Also

P = VI......................... Equation 3

Where P = power, V = potential drop, I = current.

V = P/I................... Equation 4

Note: Power = Energy/time

From the question, P = 742/30 = 24.733 W. and I = 7 A.

Substitute these values into equation 4

V = 24.733/7

V = 3.53 V

Hence the potential drop across the conductor = 3.53 V

4 0

3 years ago

Mark creates a graphic organizer to review his notes about electrical force. Which labels belong in the regions marked X and Y?

sveta [45]

Answer:

The correct answer is A

Explanation:

The question requires as well the attached image, so please see that below.

Coulomb's Law.

The electrical force can be understood by remembering Coulomb's Law, that describes the electrostatic force between two charged particles. If the particles have charges $q_1$ and $q_2$ , are separated by a distance r and are at rest relative to each other, then its electrostatic force magnitude on particle 1 due particle 2 is given by:

$|F|=k \cfrac{q_1 q_2}{r^2}$

Thus if we decrease the distance by half we have

$r_1 =\cfrac r2$

So we get

$|F|=k \cfrac{q_1 q_2}{r_1^2}$

Replacing we get

$|F|=k \cfrac{q_1 q_2}{(r/2)^2}\\|F|=k \cfrac{q_1 q_2}{r^2/4}$

We can then multiply both numerator and denominator by 4 to get

$|F|=k \cfrac{4q_1 q_2}{r^2}$

So we have

$|F|=4 \left(k \cfrac{q_1 q_2}{r^2}\right)$

Thus if we decrease the distance by half we get four times the force.

Then we can replace the second condition

$q_{2new} =2q_2$

So we get

$|F|=k \cfrac{q_1 q_{2new}}{r_1^2}$

which give us

$|F|=k \cfrac{q_1 2q_2}{r_1^2}\\|F|=2\left(k \cfrac{q_1 q_2}{r_1^2}\right)$

Thus doubling one of the charges doubles the force.

So the answer is A.

8 0

3 years ago

Read 2 more answers

An insect 5.25 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

Zigmanuir [339]

Answer:

(A) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

Explanation:

height of the insect (h) = 5.25 mm = 0.525 cm

distance of the insect (s) = 25 cm

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)

index of refraction (n) = 1.7

(A) we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{∞} -\frac{1}{-12.5} )$

$\frac{1}{f} =(0.7)(0 + \frac{1}{12.5} )$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) if the lens is reversed, the radius of curvatures would be interchanged

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = 12.5 cm

we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{12.5} -\frac{1}{∞} )$

$\frac{1}{f} =(0.7)( \frac{1}{12.5} - 0)$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

7 0

3 years ago