Answer:
displacement at 45 s = 30
65 s = 50
So the average speed over the interval from 45 s to 65 s is
(50 - 30) cm / 20 s = 1 cm / sec
As a check an average speed of 1 cm / sec for 20 sec will produce a
displacement of 1 cm / sec * 20 sec = 20 cm or from 30 to 50 cm
Answer
given,
mass of glider = 0.23 Kg
spring constant = k = 4.50 N/m
spring stretched to 0.130 m
The springs potential energy =


U = 0.038 J
at x = 0,the only energy will be kinetic .


v² = 0.3304
v = 0.575 m/s
displacement of the glider
using conservation of energy



x = 0.678 m
Answer:
depth of well is 163.30 m
Explanation:
Given data
speed of sound = 343 m/s
timer = 6.25 s
to find out
depth of well
solution
let us consider depth d
so equation will be
depth = 1/2 ×g ×t² ..............1
and
depth = velocity of sound × time .................2
here we have given time 6.25 that is sum of 2 time
when stone reach at bottom that time
another is sound reach us after stone strike on bottom
so time 1 + time 2 = 6.25 s
so from equation 1 and 2 we get
1/2 ×g ×t² = velocity of sound × time
1/2 ×9.8 × t1² = 343 × (6.25 - t1 )
t1 = 5.77376 sec
so height = 1/2 ×g ×t²
height = 1/2 ×9.8 × (5.773)²
height = 163.30 m
Answer:
marblebrainiest plz\c cmarble
Explanation:
Answer:
Explanation:
General equation of the electromagnetic wave:
![E(x, t)= E_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]](https://tex.z-dn.net/?f=E%28x%2C%20t%29%3D%20E_0sin%5B%5Cfrac%7B2%5Cpi%7D%7B%5Clambda%7D%28x-ct%29%2B%5Cphi%20%5D)
where
Phase angle, 0
c = speed of the electromagnetic wave, 3 × 10⁸
wavelength of electromagnetic wave, 698 × 10⁻⁹m
E₀ = 3.5V/m
Electric field equation
![E(x, t)= 3.5sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6x-2.7\times 10^{15}t)]](https://tex.z-dn.net/?f=E%28x%2C%20t%29%3D%203.5sin%5B%5Cfrac%7B2%5Cpi%7D%7B6.98%5Ctimes10%5E%7B-7%7D%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CE%28x%2C%20t%29%3D%203.5sin%5B%7B9%20%5Ctimes%2010%5E6%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CE%28x%2C%20t%29%3D%203.5sin%5B%7B9%20%5Ctimes%2010%5E6x-2.7%5Ctimes%2010%5E%7B15%7Dt%29%5D)
Magnetic field Equation
![B(x, t)= B_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]](https://tex.z-dn.net/?f=B%28x%2C%20t%29%3D%20B_0sin%5B%5Cfrac%7B2%5Cpi%7D%7B%5Clambda%7D%28x-ct%29%2B%5Cphi%20%5D)
Where B₀= E₀/c

![B(x, t)= 1.2\times10^{-8}sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6x-2.7\times 10^{15}t)]](https://tex.z-dn.net/?f=B%28x%2C%20t%29%3D%201.2%5Ctimes10%5E%7B-8%7Dsin%5B%5Cfrac%7B2%5Cpi%7D%7B6.98%5Ctimes10%5E%7B-7%7D%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CB%28x%2C%20t%29%3D%201.2%5Ctimes10%5E%7B-8%7Dsin%5B%7B9%20%5Ctimes%2010%5E6%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CB%28x%2C%20t%29%3D%201.2%5Ctimes10%5E%7B-8%7Dsin%5B%7B9%20%5Ctimes%2010%5E6x-2.7%5Ctimes%2010%5E%7B15%7Dt%29%5D)