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klasskru [66]
3 years ago
10

A weight lifter picks up a barbell and 1. lifts it chest high 2. holds it for 30 seconds 3. puts it down slowly (but does not dr

op it). rank the work w that the weight lifter does during each of these three operations. label the quantities as w1, w2, and w3. (hint: think about how work is defined in terms of who is applying forces and who is doing work.) w3 = w2 = w1 w3 = w1 > w2 w2 > w1 > w3 none of the above w1 > w2 > w3 w3 > w2 > w1 w2 > w3 > w1 justify your ranking order.
Physics
1 answer:
Anna11 [10]3 years ago
7 0

1. lifts it chest high

The force opposing to this action is the force due to gravity. Therefore the work done is:

W1 = m g d

where m is mass of the barbell, g is gravity and d is displacement

 

2. holds it for 30 seconds

Work is a product of force and displacement, since there is no displacement, therefore work done is zero.

W2 = 0

 

3. puts it down slowly

If the barbell was dropped, then it would simply be a free fall. But since it was not, so the work done here is also equal to the weight of the barbell times displacement:

W3 = m g d

 

We can see that W1 = W3, and since W2 = 0, therefore the answer is:

<span>w3 = w1 > w2</span>

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3 years ago
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47. the beam is supported by two rods ab and cd that have cross-sectional areas of 12mm^2 and 8mm^2, respectively. determine the
Ugo [173]

Let the beam is of length L

Now the stress on both the end is same

now we can say that torque on the beam due to two forces must be zero

N_1* x =  N_2* (L - x)

also we know that stress at both ends are same

\frac{N_1}{12} = \frac{N_2}{8}

2*N_1 = 3*N_2

Now from two equations we have

\frac{3}{2}N_2*x = N_2* (L - x)

solving above equation we have

x = \frac{2}{5}L

<em>so the load is placed at distance 0.4L from the end of 12 mm^2 area</em>

8 0
3 years ago
The company where you work has obtained and stored five lasers in a supply room. You have been asked to determine the intensity
VikaD [51]

Answer:

a)  I = 5.79 10⁵ W/m² , b)  I = 2.58 10² W / m², c)   I = 8.03 10³ W / m² , d)     I = 5.3 10⁶ W / m², e)  I = 9 10¹ W / m² , f)  D> A> C> B> E

Explanation:

The intensity is defined as the power per unit area

       I = P / A

The area of ​​a circle is

      A = π r²

Laser A

Power P = 2.2 W

Diameter d = 2.9 mm = 2.9 10⁻³ m

Let's calculate

Area

      A =  π d² / 4

     A =  π (2.2 10⁻³)²/4

     A = 3.80 10⁻⁶ m²

Let's calculate the intensity

     I = 2.2 / 3.80 10⁻⁶

     I = 0.579 10⁶ W / m²

     I = 5.79 10⁵ W/m²

Laser B

The electric field is E = 440 V / m

Intensity average is

      I = E B / 2 μ₀

The relationship of the fields with the speed of light

      E / B = c

The intensity  

       I = EE / 2 μ₀ c

       I = 440² / (2 4π 10⁻⁷ 3 10⁸)

      I = 1.936 105/750

      I = 2.58 10² W / m²

Laser C

The magnetic field amplitude B = 8.2 10⁻⁶ T

      I = c / 2μ₀  B²

      I = 3 10⁸/2 4π 10⁻⁷ (8.2 10⁻⁶)²

      I = 8.03 10³ W / m²

Part D

Diameter d = 1.8 mm = 1.8 10⁻³ m

The radius is r = d / 2 = 0.9 10⁻³ m

The force is F = 9.0 10⁻⁸ N

The radiation pressure is on a reflective surface is

         P = 2S / c

         I = S =P c / 2

The definition of pressure is

         P = F / A

          I = F c / 2 A

          I = 9.0 10⁻⁸ 3 10⁸ / (2π (0.9 10⁻³)²)

          I = 5.3 10⁶ W / m²

Part E

Average energy density

         u = 3.0 10⁻⁷ J / m³

          I = S = c u

          I = 3 10⁸ 3.0 10⁻⁷

          I = 9 10¹ W / m²

Part F

Sort in descending order

The order is

  D> A> C> B> E

7 0
3 years ago
A line in the paschen series of hydrogen has a wavelength of 1880 nm. From what state did the electron originate?
NARA [144]

Answer:

n=4

Explanation:

Paschen series of hydrogen, is the series of transitions resulting from the hydrogen atom when electron jumps from a state of n\geq 4 to n_f=3.

Emission lines for hydrogen are given by:

\frac{1}{\lambda}=R_H(\frac{1}{n_f^2}-\frac{1}{n^2})

\lambda is the wavelength of the line emitted,R_H is the Rydberg constant for hydrogen, n_f is the final energy state of the electron and n is the energy state where the electron transition originated.

We have \lambda=1880nm=1880*10^{-9}m and n_f=3. Solving for n:

\frac{1}{\lambda R_H}=\frac{1}{n_f^2}-\frac{1}{n^2}\\\frac{1}{n^2}=\frac{1}{n_f^2}-\frac{1}{\lambda R_H}\\n^2=\frac{1}{\frac{1}{n_f^2}-\frac{1}{\lambda R_H}}\\n^2=\frac{1}{\frac{1}{3^2}-\frac{1}{1880*10^{-9}m(1.09737*10^7m^{-1})}}\\n^2=15.96\\n=4

6 0
4 years ago
CORRECT ANSWER WILL BE REWARDED BRAINLIEST
Mandarinka [93]

Answer:

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