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ExtremeBDS [4]
3 years ago
14

The longer the lever, the greater the

Physics
1 answer:
Simora [160]3 years ago
4 0

Answer: The longer the lever, the greater the force on the load will be.

Explanation:

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Select the correct answer.
elixir [45]

A) Energy is absorbed in the reaction.

Such reactions are called endothermic reactions.

5 0
3 years ago
Read 2 more answers
(I think it's D but idontknow)
sukhopar [10]

During an exothermic reaction; light and heat are released into the environment.

An exothermic reaction is one in which heat is released to the environment. This heat can be physically observed sometimes like in an a combustion reaction.

In an exothermic reaction, the enthalpy of the reactants is greater than the enthalpy of the products.

This heat lost is sometimes felt as the hotness of the vessel in which the reaction has taken place.

In conclusion, light and heat are released into the environment in an exothermic reaction.

Learn more: brainly.com/question/4345448

3 0
2 years ago
That is a difference between a law and a hypothesis?
telo118 [61]

Answer:

<em>A hypothesis</em> is a limited explanation of a phenomenon; a scientific theory is an in-depth explanation of the observed phenomenon.

<em> A law</em> is a statement about an observed phenomenon or a unifying concept, according to Kennesaw State University. ... However, Newton's law doesn't explain what gravity is, or how it works.

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2 years ago
How must the height of A relate to the height of B for the skateboarder to have just enough energy to complete the loop?
KiRa [710]

Answer:

The answer is D

Explanation:

8 0
3 years ago
Read 2 more answers
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
3 years ago
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