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Svetlanka [38]
2 years ago
9

Exercises

Physics
1 answer:
Crank2 years ago
5 0

\\ \rm\Rrightarrow \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{1}{-10}+\dfrac{1}{38}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-19+5}{190}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-14}{190}

\\ \rm\Rrightarrow u=\dfrac{190}{-14}

\\ \rm\Rrightarrow u=13.6cm

Real

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An element is neutral when the charged is equal to zero. This means that the number of electrons and protons are equal. When an object has a NEGATIVE charge, its electrons are more than its protons. Also, the object has a POSITIVE charge when the number of its electrons are fewer compared to the number of the protons. 
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Red light has a wavelength of (7. 21x10^7)m. What is the frequency of this light?
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If the Earth absorbs more insolation from the sun than it radiates back, the Earth _____________.
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8 0
2 years ago
How much potential energy does a 50-N box have when lifted at a height of 1.5M?
nikitadnepr [17]

The correct answer is: Option (A) 75 J

Explanation:

First, be careful with the units here. As you can see it is mentioned that there is a 50N box. It means that the weight (<em>mg</em>) of the box is given as the unit is <em>Newton</em>, not its mass (which is in kg).

As,

Potential-energy = mass * acceleration-due-to-gravity * height

PE = m*g*h --- (A)


In equation (A), mg is actually the weight of the box, which is given.

mg = 50N

h = height = 1.5m

Plug the values in equation (A):

PE = 50 * 1.5  = <em>75 J (Option A)</em>

3 0
3 years ago
If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int
vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work done by the system on the surrounding


In this problem, the work done by the system is

W=-225 kcal=-941.4 kJ

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

Q=-5 \cdot 10^2 kJ=-500 kJ

with a negative sign as well because it is released by the system.


Therefore, by using the initial equation, we find

\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ

8 0
3 years ago
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