What is the IMA of the following pulley system?

Physics

2 answers:

Elis [28]2 years ago

7 0

Answer:

Explanation:

Mrac [35]2 years ago

4 0

The IMA of pulley system can be defined as the ratio of output force to input force.

And IMA stands for Ideal Mechanical advantage

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5. The force of gravity acting on the object is object’s ______________.

Dima020 [189]

I think it's mass...

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2 years ago

How many grams of calcium metal is required to react with 7.75 g water to produce calcium hydroxide and hydrogen gas?

Gekata [30.6K]

87.69 grams of calcium metal is required to react with 7.75 g water to produce calcium hydroxide and hydrogen gas.

Ca (OH)2+2H2 = CaH2 + 2H2O

Convert grams of hydrogen to moles of hydrogen.

H2 has a molar mass of 2.016 g/mol.

1 mol H2 x 8.400g H2

2.016g H2 equals 4.1667 mol H2

Calculate the moles of CaH2 using the molar ratio.

The molar ratio is 1 mol CaH2: 2 mol H2, according to the equation that is balanced.

2.0833 mol CaH2 = 4.1667 mol H2 1 mol CaH2 2 mol H2

grams of CaH2 to CaH2

CaH2 has a molar mass of 42.09 g/mol.

1.mol CaH2=87.69 g CaH2 2.0833mol CaH2=42.09 g CaH2

87.69 g CaH2 are required for the reaction.

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7 0

1 year ago

A crow drops a 0.11kg clam onto a rocky beach from a height of 9.8m. What is the kinetic energy of the clam when it is 5.0m abov

svp [43]

Answer:

The kinetic energy of the clam at a height of 5.0 m is 5.19 J and the speed of the clam at that height is 9.71 m/s.

Explanation: 

Mechanical energy is constant throughout the travel, we know that mechanical energy is calculated by adding potential energy and kinetic energy. Potential energy = $m \times g \times h$ , Kinetic energy = $\frac{1}{2} \times m \times v^{2}$ and Mechanical energy = $m \times g \times h+\frac{1}{2} \times m \times v^{2}$ Kinetic energy is zero at initial point. Now mechanical energy of clam with m=0.11kg,g=9.81 $\frac{m}{s^{2}}$ ,h=9.8 m is = 0.11×9.81×9.8 = 10.58 J.

Mechanical energy of clam at a height of 5.0 m = $0.11 \times 9.81 \times 5+\frac{1}{2} \times m \times v^{2}$ = $5.39+\frac{1}{2} \times m \times v^{2}$ . We know that mechanical energy is constant hence, mechanical energy of clam at height 9.8 m is equal to mechanical energy at height 5.0 m. This is represented as following

10.58 = $5.39+\frac{1}{2} \times m \times v^{2}$ 10.58 – 5.39 = $\frac{1}{2} \times m \times v^{2}$ 5.19 = $\frac{1}{2} \times m \times v^{2}$ kinetic energy of the clam is 5.19 J.

Now speed of the clam at height 5.0 m is 5.19 = $\frac{1}{2} \times 0.11 \times v^{2} \frac{5.19 \times 2}{0.11}=v^{2}$ 94.36 = $v^{2} \sqrt{94.36}=v \quad v$ = 9.71 m/s. The speed of the clam is 9.71 m/s.