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stepladder [879]
3 years ago
5

to lift a load to roof, you apply 21 n of force to a pulley. the pulley applies 21 n of force to the load. what is the mechanica

l advantage of the pulley?
Physics
1 answer:
MariettaO [177]3 years ago
7 0

Answer:

The mechanical advantage of the pulley is 1.

Explanation:

The mechanical advantage of the pulley is given as

MA=\frac{F}{f}

Here, MA is represent for the mechanical advantage of the pulley, F is load for lift and f applied force.

Given F =21 N and f= 21 N.

Substitute the given values, we get

MA=\frac{21N}{21N}

MA = 1.

Thus, the mechanical advantage of the pulley is 1.

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In this reaction diagram which part represents the doffrence in energy between the reactants and the products?
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Answer:

The correct answer is - option C. G.

Explanation:

In this reaction diagram, there is a representation of the reaction profile. The reaction profile shows the change that takes place during a reaction in the energy of reactants or substrate and products. In this profile, activation energy looks like a hump in the line, and the minimum energy required to initiate the reaction.

The overall energy of the reaction, including or excluding activation energy depends on the nature of the reaction if it is exothermic or endothermic. and products are represented by the G which shows the difference between the energy of the reactants and products.

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2 years ago
In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere
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Answer:

2.1\cdot 10^{21} electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

E=\frac{kQ}{r^2}

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a  charged sphere, so we have:

E_b=3.00\cdot 10^6 N/C is the maximum electric field strength tolerated by the air before breakdown occurs

r=1.00 km = 1000 m is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C

Assuming that the cloud is negatively charged, then

Q=-333.3 C

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e=-1.6\cdot 10^{-19}C

The number of excess electrons on the cloud is

N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}

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