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Rina8888 [55]
3 years ago
14

Determine for which class of lever the output force is always greater than the input force. for which class is the output force

always less than the input force?

Physics
2 answers:
umka2103 [35]3 years ago
5 0
The 3rd class lever is the <span>output force always less than the input force, becuase its mechanical advantage is always less than 1. this also due that in a 3rd class lever the effort arm is shorter than the load arm, that is why the output is lower than the input force. but 3rd class lever is a speed multiplier lever</span>
Ad libitum [116K]3 years ago
4 0

Answer:

  • <u>Second Class of lever:</u> The second class of the lever is considered the most efficient or have the most mechanical advantage in conveying more productivity or output while investing less input into the system.

Explanation:

<u>The second Class of lever:</u>

The second type of lever has the load,L in between the effort,E and the fulcrum,F. As the effort moves a larger distance to move the load a small distance. As there is more productivity or mechanical output in response to less input to the system.

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as you go above earths surface, the acceleration due to gravity will decrease. find the height, i meters, above the earths surfa
aleksklad [387]

Answer:

1.23×10⁸ m

Explanation:

Acceleration due to gravity is:

a = GM / r²

where G is the universal gravitational constant,

M is the mass of the planet,

and r is the distance from the center of the planet to the object.

When the object is on the surface of the Earth, a = g and r = R.

g = GM / R²

When the object is at height i above the surface, a = 1/410 g and r = i + R.

1/410 g = GM / (i + R)²

Divide the first equation by the second:

g / (1/410 g) = (GM / R²) / (GM / (i + R)²)

410 = (i + R)² / R²

410 R² = (i + R)²

410 R² = i² + 2iR + R²

0 = i² + 2iR − 409R²

Solve with quadratic formula:

i = [ -2R ± √((2R)² − 4(1)(-409R²)) ] / 2(1)

i = [ -2R ± √(1640R²) ] / 2

i = (-2R ± 2R√410) / 2

i = -R ± R√410

i = (-1 ± √410) R

Since i > 0:

i = (-1 + √410) R

R = 6.37×10⁶ m:

i ≈ 1.23×10⁸ m

8 0
3 years ago
When the saw slices wood, the wood exerts a 104-N force on the blade, 0.128 m from the blade’s axis of rotation. If that force i
Soloha48 [4]

Answer:

Explanation:

When saw slices wood by exerting a force on the wood , wood also exerts a reaction force on the saw in opposite direction which is equal to the force of action that is 104 N.

So torque exerted by wood on the blade

= force x perpendicular distance from the axis of rotation

= 104 x .128

=13.312 Nm.

Since this torque opposes the movement of blade , it turns the blade slower.

5 0
2 years ago
John goes grocery shopping with his mother. His job is to push the cart. The cart is
lora16 [44]

Answer:

Beacause he has more grocceries and food heavy

Explanation:

7 0
2 years ago
Answer and I will give you brainiliest <br><br><br>Please heeeelp​
Lostsunrise [7]

Answer:

T = 4.905[N]

Explanation:

In order to solve this problem we must perform a sum of forces on the vertical axis.

∑Fy = 0

We have two forces acting only, the weight of the body down and the tension force T up, as the body does not move we can say that it is system is in static equilibrium, therefore the sum of forces is equal to zero.

T-m*g=0\\T=0.5*9.81\\T=4.905[N]

5 0
3 years ago
The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: =E−Ryn2 In this equation Ry stands for
skelet666 [1.2K]

Answer:

  λ = 162 10⁻⁷ m

Explanation:

Bohr's model for the hydrogen atom gives energy by the equation

         E_{n} = - k²e² / 2m (1 / n²)

Where k is the Coulomb constant, e and m the charge and mass of the electron respectively and n is an integer

The Planck equation

           E = h f

The speed of light is

          c = λ f

          E = h c /λ

For a transition between two states we have

          E_{n} - E_{m} = - k²e² / 2m (1 / n_{f}² -1 / n_{i}²)

           h c / λ = -k² e² / 2m (1 / n_{f}² - 1/ n_{i}²)

           1 / λ = (- k² e² / 2m h c) (1 / n_{f}² - 1/n_{i}²)

The Rydberg constant with a value of 1,097 107 m-1 is the result of the constant in parentheses

Let's calculate the emission of the transition

            1 /λ = 1.097 10⁷ (1/10² - 1/8²)

            1 / λ = 1.097 10⁷ (0.01 - 0.015625)

            1 /λ = 0.006170625 10⁷

            λ = 162 10⁻⁷ m

3 0
3 years ago
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