Answer:
Mass of object in Earth = 1.58 kg.
Free-fall acceleration on planet X = 12.47 m/s²
Explanation:
We know weight is the product of mass and acceleration due to gravity.
Mass is a constant.
W = mg
On planet X, an object weighs 19.7 N.
On planet B where the magnitude of the free-fall acceleration is 1.92 g (where g = 9.8 m/s^2 is the gravitational acceleration on Earth), the object weighs 29.8 N.
Mass of object in Earth = 1.58 kg.
We need to find free-fall acceleration on planet X.
Free-fall acceleration on planet X = 12.47 m/s²
Answer:
62.5 %
Explanation:
Let the initial intensity of unpolarized light is Io.
After first polariser the intensity of light becomes I'.
So,
Now it passes through another polariser. The angle between the first polariser and the second polariser is given by Ф. The intensity is I''.
According to the law of Malus
Here, Ф = 30 degree
The percentage change in the intensity is given by
= 62.5 %
Answer:
The maximum velocity is
Explanation:
From the question we are told that
The coefficient of static friction is
Generally the maximum acceleration can be mathematically represented as
Where g is the acceleration due to gravity and the value is
substituting vlues
The question is incomplete. However, I have made an attempt to guess what the question is asking for.
Answer:
The original frequency of the guitar string is 507 Hz.
The frequency is increased after the tension is increased.
Its new frequency is 508 Hz.
Explanation:
The tuning fork has a frequency of 512 Hz.
With a beat frequency of 5 Hz, the guitar has a frequency of
The relationship between frequency, <em>f</em>, and tension, <em>T</em>, is given by
provided other factors are constant.
It follows that when tension is increased, frequency increases.
Therefore, when the tension in the string is increased, its frequency increases.
From the question, this increase in frequency caused a reduced beat frequency of 4 Hz.
This means the new frequency is
In the first choice, there is an increase in the frequency from 507 Hz to 508 Hz.
The second choice indicates a decrease from 517 Hz to 516 Hz.
Since we have established there is an increase in frequency, the first choice is correct. The original frequency of the guitar string is 507 Hz.