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3 years ago

A 1200N load is to be lifted with 200N effort using a first class lever. At what distance

Physics

1 answer:

kipiarov [429]3 years ago

7 0

Explanation:

Hey there!!

Let's simply work with it.

Here,

load = 1200N

Effort = 200N

Load distance = 15cm

We have,

According to the principle of lever.

L×LD = E×ED.

1200×15 = 200× ED.

18000 = 200ED.

$ed = \frac{18000}{200}$

Therefore, Effort Distance = 90cm.

Hope it helps....

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What is a disadvantage of using moving water to produce electricity

klemol [59]

Most of the time the towns get flooded and many people die and lose their homes and their jobs, businesses by the floods and the natural environment is destroyed ,air pollution is produced, the ecosystem could be disrupted and dams are known to fall over and dams are really expensive to build, The dams also cause geological damage, example earthquakes and spills happen.

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3 years ago

An 800-g block of ice at 0.00°C is resting in a large bath of water at 0.00°C insulated from the environment. After an entropy c

Allisa [31]

Answer:

Unmeltedd ice = 308.109 g

Explanation:

Gibbs Free energy:

A systems Gibbs Free Energy is defined as the free energy of the product of the absolute temperature and the entropy change less than the enthalpy change.

Therefore, G = ΔH-TΔS

where G is Gibbs Free Energy

ΔH is enthalpy change

T is absolute temperature

ΔS is entropy change

Here since there is a phase change, therefore G will be 0.

∴ΔH = TΔS

Given: Temperature, T = 0°C = 273 K

Entropy change,ΔS = 600 J/K

Latent heat of fusion of water = 333 J/g

∴ΔH = TΔS

∴ΔH = 273 x 600

= 163800 J

So this is the amount of enthalpy that will be used into melting of ice.

∴ΔH = mass of ice melted x latent heat of fusion of water

Mass of ice melted = ΔH / latent heat of fusion of water

= 163800 / 333

= 491.891 g

This is the mass of ice melted.

And initial amount of ice is 800 g

Amount of ice left after melting = Initial amount of ice - amount of ice melted

= 800-491.891

= 308.109 g

Amount of ice remained after melting = 308.109 g

8 0

3 years ago

Ideal mechanical advantage is equal to the displacement of the effort force divided by the displacement of the load.

ELEN [110]

This would be false. hope this helps. good luck :)

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3 years ago

If the mass of each ball is 12 kg and they experience a gravitational force of magnitude 2.00

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Answer:

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I'm just in jss2 but I read physics. this is what I think

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3 years ago

Two identical bowling balls are rolling on a horizontal floor without slipping. The initial speed of both balls is v = 10 m/s. B

Lapatulllka [165]

Answer:

The difference between frictionless ramp and a regular ramp is that on a frictionless ramp the ball cannot roll it can only slide, but on a regular ramp the ball can roll without slipping.

We will use conversation of energy.

$K_A_1 + U_A_1 = K_A_2 + U_A_2\\\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_A$

Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.

For the ball B;

$K_B_1 + U_B_1 = K_B_2 + U_B_2$

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The initial velocities of the balls are equal. Their maximum climbing point will be proportional to their final potential energy. Since their initial kinetic energies are equal, their final potential energies must be equal as well.

Hence, both balls climb the same point.

Explanation:

4 0

3 years ago