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3 years ago

if the ima of a machine is 4 and the effort force is 6.0 newtons, then the force exerted on the resistance by the machine is

Chemistry

1 answer:

Nata [24]3 years ago

6 0

Answer:

24 N

Explanation:

Mechanical advantage is the rate of load overcome to the effort applied to overcome the load or resistance. Also, it is the quotient of the resistance force to the effort force.

i.e MA = $\frac{L}{E}$

Given that, MA = 4, and L = 6.0 N, then;

4 = $\frac{L}{6}$

cross multiply to have

L = 4 x 6

= 24 N

L = 24 N

Thus, the force exerted on the resistance by the machine is 24 N.

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drek231 [11]

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4 years ago

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Andru [333]

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3 years ago

A chemist titrates 110.0 mL of a 0.7684 M methylamine (CH3NH2) sotion with 0.4469 M HNO3 solution at 25 °C. Calculate the pH at

Firdavs [7]

Answer: The pH at equivalence point for the given solution is 5.59.

Explanation:

At the equivalence point,

$n_{HNO_{3}} = n_{CH_{3}NH_{2}}$

So, first we will calculate the moles of $CH_{3}NH_{2}$ as follows.

$n_{CH_{3}NH_{2}} = 0.764 M \times \frac{110 ml}{1000 ml/L}$

= 0.0845 mol

Now, volume of $HNO_{3}$ present will be calculated as follows.

Volume = $\frac{\text{no. of moles}}{\text{Molarity}}$

= $\frac{0.0845}{0.4469 M}$

= 0.1891 L

Therefore, the total volume will be the sum of the given volumes as follows.

110 ml + 189.1 ml

= 299.13 ml

or, = 0.2991 L

Now, $[CH_{3}NH_{3}^{+}] = \frac{0.0845 mol}{0.2991 L}$

= 0.283 M

Chemical equation for this reaction is as follows.

$CH_{3}NH_{3}^{+} + H_{2}O \rightleftharpoons CH_{3}NH_{2} + H_{3}O^{+}$

As, $k_{a} = \frac{k_{w}}{k_{b}}$

= $\frac{10^{-14}}{10^{-3.36}}$

= $2.29 \times 10^{-11}$

Now, $[HNO_{3}] = \sqrt{k_{a}[CH_{3}NH_{3}^{+}]}$

= $\sqrt{2.29 \times 10^{-11} \times 0.283}$

= $2.546 \times 10^{-6}$

Now, pH will be calculated as follows.

pH = $-log [H_{3}O^{+}]$

= $-log (2.546 \times 10^{-6})$

= 5.59

Thus, we can conclude that pH at equivalence point for the given solution is 5.59.

6 0

3 years ago

If this same quantity of energy were transferred to 2.5 kg of water at it's boiling point what fraction of the water would be va

garik1379 [7]

Answer:

Fraction of water that can be vaporized = 0.0961 or 9.61%

Note : The question is incomplete. The complete question is given below:

A serving of Cheez-Its releases 130 kcal (1 kcal = 4.18 kJ) when digested by your body.

If this same quantity of energy were transferred to 2.5 kg of water at its boiling point, what fraction of the water would be vaporized?

Explanation:

Energy released by a serving of Cheeze-Its = 130 kcal

Since 1 kcal = 4.18 kJ, 130 kcal = 130 * 4.18 kJ = 543.4 kJ

Heat of vaporization (evaporating or condensing) Hv, of water = 2260 J/g

From formula, quantity of heat, Q = mass * heat of vaporization

mass of water = 2.5 kg = 2500 g

Q = 2500 g * 2260 J/g

Q = 5650000 J = 5650 kJ

Therefore 2.5 kg of water requires 5650 kJ of heat for complete vaporization

Fraction of water that can be vaporized by 543.4 kJ energy = 543.4/5650

Fraction of water that can be vaporized = 0.0961 or 9.61%

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3 years ago

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Lapatulllka [165]

Answer:

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Explanation:

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2 years ago