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vazorg [7]
3 years ago
14

Which of the following is not a requirement of a practical fuel? A. It must occur in abundance in nature or be easy to produce.

B. It must be made up of elements that combine easily with oxygen. C. It must contain a large amount of stored energy. D. It must be in the solid state to be used.
Physics
1 answer:
Masteriza [31]3 years ago
7 0

Answer: Option D: It must be in the solid state to be used.

A material that is used to provide energy is for work is known as fuel. For practical use, it is important that the fuel must occur in abundance or can be easily produced. It should be made of elements that can easily combine with oxygen. This is required for burning. It must produce large amount of energy. So, it must contain large amount of stored energy. It is useful only when a small quantity can produce large amount of energy. It can be in any state, solid liquid or gas.

Therefore, it is not important that it must be in the solid state to be used.

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An object weighing 49 N is pushed across a floor by a force of 12 N. What is the acceleration of the object?
NISA [10]

Answer:

Explanation:

Given parameters:

Weight of object  = 49N

Force applied = 12N

Unknown:

Acceleration of object  = ?

Solution:

The acceleration of the object is found by dividing the force by the weight;

 Acceleration  = \frac{12}{49}   = 0.25m/s²

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2 years ago
A scooter travelling at 10 m/s speed up to 20 m/s in 4 sec. Find the acceleration of scooter.
Aleksandr-060686 [28]

Explanation:

The acceleration of the scooter is 2.5 m/s

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3 years ago
What is a plane mirror ? ​
g100num [7]

Answer:

A plain mirrior is a mirrior with flat reflective surface.

hope it is helpful for you.

6 0
3 years ago
Which statement is correct? When a positively charged atom looses an electron to a positively charged atom, two neutral atoms ar
vazorg [7]
I believe the answer is "When a neutral atom looses an electron to another neutral atom, two charged atoms are created."
3 0
3 years ago
Read 2 more answers
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
3 years ago
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