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butalik [34]
3 years ago
14

How does a second class lever make our work easier​

Physics
1 answer:
dmitriy555 [2]3 years ago
7 0

Answer:

In a second class lever, the load is located between the effort and the fulcrum. If the load is closer to the fulcrum than the effort, then less effort will be required to move the load. If the load is closer to the effort than the fulcrum, then more effort will be required to move the load.

Explanation:

from what i learned, if its farther away from the load, its easier to lift, like a wheel barrel

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Law of Conservation of Energy.
Tems11 [23]
Energy cannot be created nor be destroyed
3 0
3 years ago
A wire is used as a heating element that has a resistance that is fairly independent of its temperature within its operating ran
dedylja [7]

Answer:

Double the current

Explanation:

The energy delivered by the heater is related to the current by the following relation:

E= I^{2}R t

let R * t = k ( ∴ R and t both are constant)

so E= k I^{2}

Now let:

E2= k I₂^2

E2= 4E

⇒ k I₂^2= 4* k I^{2}

Cancel same terms on both sides.

I₂^2= 4* I^{2}

taking square-root on both sides.

√I₂^2 = √4* I^2

⇒I₂= 2I

If we double the current the energy delivered each minute be 4E.

3 0
3 years ago
On a cloudless day, the sunlight that reaches the surface of the earth has an intensity of about W/m². What is the electromagnet
Otrada [13]

Answer:

1.907 x 10⁻⁵ J.

Explanation:

Given,

Volume of space, V = 5.20 m³

Assuming the intensity of sunlight(S) be equal to 1.1 x 10³ W/m².

Electromagnetic energy = ?

E = \mu V

E = (\dfrac{S}{c})\times V

where c is the speed of light.

E = (\dfrac{1.1\times 10^3}{3\times 10^8})\times 5.20

E = 1.907\times 10^{-5}\ J

Hence, Electromagnetic energy is equal to 1.907 x 10⁻⁵ J.

4 0
4 years ago
Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 3
Scilla [17]

Explanation:

Given that,

Angle by the normal to the slip α= 60°

Angle by the slip direction with the tensile axis β= 35°

Shear stress = 6.2 MPa

Applied stress = 12 MPa

We need to calculate the shear stress applied at the slip plane

Using formula of shear stress

\tau=\sigma\cos\alpha\cos\beta

Put the value into the formula

\tau=12\cos60\times\cos35

\tau=4.91\ MPa

Since, the shear stress applied at the slip plane is less than the critical resolved shear stress

So, The crystal will not yield.

Now, We need to calculate the applied stress necessary for the crystal to yield

Using formula of stress

\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}

Put the value into the formula

\sigma=\dfrac{6.2}{\cos60\cos35}

\sigma=15.13\ MPa

Hence, This is the required solution.

3 0
3 years ago
For an object to be classified as a ____, it must meet certain definite criteria: It must be massive enough to pull itself into
Orlov [11]

Answer:

a planet

Explanation:

a planet is one which exerts these properties and therefore is the answer

5 0
3 years ago
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