To solve this problem it is necessary to apply the rules and concepts related to logarithmic operations.
From the definition of logarithm we know that,
In this way for the given example we have that a logarithm with base 10 expressed in the problem can be represented as,
We can express this also as,
By properties of the logarithms we know that the logarithm of a power of a number is equal to the product between the exponent of the power and the logarithm of the number.
So this can be expressed as
Since the definition of the base logarithm 10 of 10 is equal to 1 then
The value of the given logarithm is equal to 6
Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor
Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.
<h3>What is the apparent weight of a body in a lift?</h3>
- Consider a body of mass m kept on a weighing machine in a lift.
- The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
- The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
<h3>How to solve the question?</h3>
- Here we have given with the actual weight of the body as 100lbs.
- This 100lb child is standing on the scale or the weighing machine, when it is riding .
- During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
- There is also<em> mg </em>downwards and a normal reaction in the upward direction.
- when we equate both the upward force and downward force, we get,
i.e. during riding the scale reads a weight less than that of actual weight.
- When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.
Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.
Learn more about the apparent weight of the body in a lift here:
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Answer:
(a) 1.16 s
(b)0.861 Hz
Explanation:
(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.
From the question,
If 1550 cycles is completed in (30×60) seconds,
1 cycle is completed in x seconds
x = 30×60/1550
x = 1.16 s
Hence the period is 1.16 seconds.
(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).
Mathematically, Frequency is given as
F = 1/T ........................... Equation 1
Where F = frequency, T = period.
Given: T = 1.16 s.
Substitute into equation 1
F = 1/1.16
F = 0.862 Hz
Hence thee frequency = 0.862 Hz
Answer:
3.75 × 10⁻⁸ N
Explanation:
Given:
Intensity of the electromagnetic wave, I = 150 W/m²
Sides of the board = 25 cm (= 0.25 m) and 30 cm (= 0.30 m)
therefore,
the area of the rectangular box, A = 0.25 × 0.30 = 0.075 m²
Now,
force exerted on the card by the radiation, F = 
here,
C is the speed of the light = 3 × 10⁸ m/s
on substituting the respective values, we get
F = 
or
F = 3.75 × 10⁻⁸ N
