The period of the wave would be halved
Answer:
a) the distance between her and the wall is 13 m
b) the period of her up-and-down motion is 6.5 s
Explanation:
Given the data in the question;
wavelength λ = 26 m
velocity v = 4.0 m/s
a) How far from the wall is she?
Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;
x = λ/2
we substitute
x = 26 m / 2
x = 13 m
Therefore, the distance between her and the wall is 13 m
b) What is the period of her up-and-down motion?
we know that the relationship between frequency, wavelength and wave speed is;
v = fλ
hence, f = v/λ
we also know that frequency is expressed as the reciprocal of the time period;
f = 1/T
Hence
1/T = v/λ
solve for T
Tv = λ
T = λ/v
we substitute
T = 26 m / 4 m/s
T = 6.5 s
Therefore, the period of her up-and-down motion is 6.5 s
Answer:
1224km/hr
Explanation:
To convert from m/s to km/hr
1000m = 1km
Divide both sides by 1000
1m = 1/1000 km................. (1)
60×60 seconds = 1 hr
3600s = 1hr
Divide both sides by 3600
1s = 1/3600 .............(2)
Divide (2) by (1)
1m/s = 1/1000 ÷ 1/3600 km/hr
1m/s = 1/1000 × 3600/1 km/hr
1m/s = 3600/1000 km/hr
1m/s = 3.6 km/hr .............(3)
To convert 340m/s to km/hr
Multiply (3) by 340
1× 340m/s = 3.6 × 340 km/hr
340m/s = 1224km/hr
I hope this was helpful, please mark as brainliest
Explanation:
Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.
→ The tangential component of acceleration is rate of increase in the speed of plane so,
→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).
dy/dx = d(0.4x²)/dx
= 0.8x
Take the derivative again,
d²y/dx² = d(0.8x)/dx
= 0.8
at x= 5 Km
dy/dx = 0.8(5)
= 4
![p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} } }{\frac{d^{2y} }{dx^{2} } }](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B%5B1%2B%20%28%5Cfrac%7Bdy%7D%7Bdx%7D%29%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%20%7D%7B%5Cfrac%7Bd%5E%7B2y%7D%20%7D%7Bdx%5E%7B2%7D%20%7D%20%7D)
now insert the values,
![p = \frac{[1+(4)^{2}]^{\frac{3}{2} } }{0.8} = 87.62 km](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B%5B1%2B%284%29%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%7D%7B0.8%7D%20%20%3D%2087.62%20km)
→ Now the normal component of acceleration is given by
= (200)²/(87.6×10³)
aₙ = 0.457 m/s²
→ Now the total acceleration is,
![a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}](https://tex.z-dn.net/?f=a%20%3D%20%5B%28a_%7Bt%7D%29%5E%7B2%7D%20%2B%28a_%7Bn%7D%20%29%5E%7B2%7D%20%5D%5E%7B0.5%7D)
![a = [(0.8)^{2} + (0.457)^{2}]^{0.5}](https://tex.z-dn.net/?f=a%20%3D%20%5B%280.8%29%5E%7B2%7D%20%2B%20%280.457%29%5E%7B2%7D%5D%5E%7B0.5%7D)
a = 0.921 m/s²
If all the mass of fuel and oxygen is burned to form gases of combustion, the downward velocity of these gases would be 12,505 m/s.
<h3>
Conservation of linear momentum</h3>
The principle of conservation of linear momentum states that, the total momentum of an isolated system is constant.
The downward velocity of thes gases is calculated as follows;
v1(m1 + m2) = v2(m2)
305(1000 + 25) = v2(25)
312,625 = 25v2
v2 = 312,625/25
v2 = 12,505 m/s
Thus, If all the mass of fuel and oxygen is burned to form gases of combustion, the downward velocity of these gases would be 12,505 m/s.
Learn more about linear momentum here: brainly.com/question/7538238
