Answer:
The wavelength of wave is 7.5 meter.
Given:
Speed of wave = 1500 
Frequency of wave = 200 Hz
To find:
Wavelength of wave = ?
Formula used:

Where
= wavelength of the wave
v = speed of wave
n = frequency of wave
Solution:
Wavelength of wave is given by,

Where
= wavelength of the wave
v = speed of wave
n = frequency of wave

= 7.5 m
The wavelength of wave is 7.5 meter.
Answer:
2.28
Explanation:
From mirror formula,
1/f = 1/u+1/v .......... Equation 1
Where f = focal length of the mirror, v = image distance, u = object distance.
Note: The focal length mirror is positive.
make v the subject of the equation,
v = fu/(u-f)............ Equation 2
Given: f = 2.5 cm, u = 1.4 cm
Substitute into equation 2
v = 2.5(1.4)/(1.4-2.5)
v = 3.5/-1.1
v = -3.2 cm.
Note: v is negative because it is a virtual image.
But,
Magnification = image distance/object distance
M = v/u
Where M = magnification.
Given: v = 3.2 cm, u = 1.4 cm
M = 3.2/1.4
M = 2.28.
Thus the magnification of the tooth = 2.28.
Answer:
Animals are classifield as two main groups in the animal kingdom : Vertebrates and invertebrates .
Explanation:
keep it up....
At speeds over 30 mph, you should maintain a following distance of at least <u>three full seconds</u> behind the vehicle ahead of you.
As a general rule and common sense at a speed of 30 mph you can leave three full seconds so that you can achieve a prudent distance between the car you are driving and the car in front in order to be able to perform some kind of maneuver if an accident or unforeseen event occurs.
To count the full three seconds you can use the technique of counting the Mississippis as follows: Mississippi one, Mississippi two, Mississippi three.
<h3>What is an accident?</h3>
An accident is an unexpected event that generally causes damage, injury or negative consequences.
Learn more about accident at: brainly.com/question/28070413
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Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2