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EleoNora [17]
3 years ago
11

Jeff is a landscaping contractor and lifts a rock weighing 600 pounds by wedging a board under the rock. Jeff weighs 150 pounds

and puts all of his weight on the lever. How much mechanical advantage did the lever provide to Jeff in lifting the rock? A) 0 B) 2 C) 4 D) 8
Physics
2 answers:
jonny [76]3 years ago
6 0

Answer: 4

The mechanical advantage is the ratio of the force exerted  by the object to the force applied to do work on it.

Here, Jeff tried to lift a rock weighing 600 pounds by wedging board under the rock. Jeff who weighs 150 pounds uses all his weight to exert force on lever and lift rock.

Mechanical advantage, M.A.=\frac{weight\hspace{1mm}of\hspace{1 mm}rock}{weight\hspace{1mm}of\hspace{1 mm}Jeff}=\frac{600 pounds}{150 pounds}=4.

Therefore, the mechanical advantage that lever provided to Jeff in lifting rock is 4.

attashe74 [19]3 years ago
6 0
Answer: 4

The lever enabled Jeff to lift 600 pounds with merely 150 pounds.
By definition, the mechanical advantage is 600/150 = 4.
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mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
An experimental tungsten light bulb filament has a length of 5 cm and a diameter of 0.074 cm. The filament is basically just a w
adell [148]

Answer:

power emitted is 1.75 W

Explanation:

given data

length l = 5 cm = 5 ×10^{-2} m

diameter d = 0.074 cm = 74 ×10^{-5} m

total filament emissivity = 0.300

temperature = 3068 K

to find out

power emitted

solution

we find first area that is π×d×L

area = π×d×L

area = π×74 ×10^{-5}×5 ×10^{-2}

area = 1162.3892  ×10^{-5} m²

so here power emitted  is express as

power emitted  = E × σ × area × (temperature)^4

put here all value

power emitted  = 0.300× 5.67 × 1162.3892  ×10^{-5}  × (3068)^4

power emitted = 1.75 W

5 0
3 years ago
A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 9.25 mm into four pipes, each
irinina [24]

Answer:

1.24 m/s

Explanation:

Metric unit conversion:

9.25 mm = 0.00925 m

5 mm = 0.005 m

The volume rate that flow through the single pipe is

\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s

This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of

\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74\times10^{-5} m^3/s

So the flow speed of each of the narrower pipe is:

v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}

v_n = \frac{9.74\times10^{-5}}{\pi 0.005^2} = 1.24 m/s

8 0
3 years ago
Which of the following statements best describes energy conservation in heat engines?
Lorico [155]
B is the correct answer
5 0
2 years ago
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sergejj [24]

Answer:\sqrt{2}T

Explanation:

Given

object of mass m is suspended from spring and set in oscillation with time Period T

We know Time period of a mass in oscillation is given by

T=2\pi \sqrt{\frac{m}{k}}

where k=spring constant

When mass m is replaced by a mass of 2 m time period is given by

T'=2\pi \sqrt{\frac{2m}{k}}

T'=\sqrt{2}\times 2\pi \sqrt{\frac{m}{k}}

T'=\sqrt{2}T

i.e. New time period becomes \sqrt{2} times of previous one

                         

7 0
3 years ago
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