Answer:
0.5 s
Explanation:
From the question given above, the following data were obtained:
Number of circle (n) = 2
Time (t) = 1 s
Period =?
Period of a wave is simply defined as the time taken to make one complete oscillation. Mathematically, it can be expressed as:
T = t / n
Whereb
T => is the period
t => is the space time
n => is the number of circle or oscillation.
With the above formula, we can obtain the period of the wave as follow:
Number of circle (n) = 2
Time (t) = 1 s
Period =?
T = t / n
T = 1 / 2
T = 0.5 s
Thus, the period of the wave is 0.5 s
Answer:
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
Explanation:
The bulk modulus is represented by the following differential equation:
Where:
- Bulk module, measured in pascals.
- Sample volume, measured in cubic meters.
- Local pressure, measured in pascals.
Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:
This resultant expression is solved by definite integration and algebraic handling:
The final volume is predicted by:
If , and , then:
Change in volume due to increasure on pressure is:
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
4 water has a low heat capacity and a high vaporization temperate and coasts have low temps
Answer:
1456 N
Explanation:
Given that
Frequency of the piano, f = 27.5 Hz
Entire length of the string, l = 2 m
Mass of the piano, m = 400 g
Length of the vibrating section of the string, L = 1.9 m
Tension needed, T = ?
The formula for the tension is represented as
T = 4mL²f²/ l, where
T = tension
m = mass
L = length of vibrating part
F = frequency
l = length of the whole part
If we substitute and apply the values we have Fri. The question, we would have
T = (4 * 0.4 * 1.9² * 27.5²) / 2
T = 4368.1 / 2
T = 1456 N
Thus, we could conclude that the tension needed to tune the string properly is 1456 N
Here when an object is placed on the level floor then in that case there are two forces on the object
1). Weight of object downwards (mg)
2). Normal force due to floor which will counterbalance the weight (N)
so when no force is applied on the box at that time normal force is counter balanced by weight.
Now here it is given that A person tried to lift the box upwards
So now there are two forces on the box
1). Applied force of person
2). Normal force due to ground
So now these two forces will counter balance the weight of the crate
So we can write an equation for force balance like
given that
here
m = 30 kg and
g = acceleration due to gravity = 10 m/s^2
now from above equation
So force applied by the person must be 150 N