Answer:
<h3> b. 1.18</h3>
Explanation:
The fundamental frequency in string is expressed as;
F1 = 1/2L√T/m .... 1
L is the length of the string
T is the tension
m is the mass per unit length
If the tension is increased by 40%, the new tension will be;
T2 = T + 40%T
T2 = T + 0.4T
T2 = 1.4T
The new fundamental frequency will be;
F2 = 1/2L√1.4T/m ..... 2
Divide 1 by 2;
F2/F = (1/2L√1.4T/m)/1/2L√T/m)+
F2/F = √1.4T/m ÷ √T/m
F2/F = √1.4T/√m ×√m/√T
F2/F = √1.4T/√T
F2/F = 1.18√T/√T
F2/F = 1.18
F2 = 1.18F
Hence the fundamental frequency of vibration changes by a factor of 1.18
Answer:
what is it on? like name one of the questions
Explanation:
Answer:
splashing
Explanation:
if you put in the cereal after the milk it will splash everywhere, causing a waste of milk, and a loss of time.
Answer:
Examples of man-made objects that spread an impulse over a large amount of time are "airbags" in vehicles and "arrestor beds" (for trucks).
Explanation:
The question above is highly related to the topic about "Impulse" in Physics.
"Impulse"<em> refers to an object's change in momentum (the amount of motion in an object) when a force acts upon it for an interval time.</em> When it comes to providing safety to people when it comes to vehicular crashes, impulse plays a vital role.
Let's take the example of airbags in vehicles. Once a vehicle collides with another object, the driver is carried by a forward motion. Without airbags, the time is normally shorter for the driver to be stopped by the windshield. This results to a greater force. However, with the presence of air-bags, the driver will hit the airbag, instead of the windshield. <u>This will lengthen the time of the impact, thus reducing the force.</u>
Another example are the arrestor beds for trucks. Arrestor beds have been designed in order for trucks to stop, since it's hard to maneuver them. <u>With the help of arrestor beds, trucks are able to come to a stop with a longer time interval, but decreased force.</u>
The magnitude of the induced emf is given by:
ℰ = |Δφ/Δt|
ℰ = emf, Δφ = change in magnetic flux, Δt = elapsed time
The magnetic field is perpendicular to the loop, so the magnetic flux φ is given by:
φ = BA
B = magnetic field strength, A = loop area
The area of the loop A is given by:
A = πr²
r = loop radius
Make a substitution:
φ = B2πr²
Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:
Δφ = ΔB2πr²
ΔB = change in magnetic field strength
Make another substitution:
ℰ = |ΔB2πr²/Δt|
Given values:
ΔB = 0.20T - 0.40T = -0.20T, r = 0.50m, Δt = 2.5s
Plug in and solve for ℰ:
ℰ = |(-0.20)(2π)(0.50)²/2.5|
ℰ = 0.13V