Answer:
20.96 h
Explanation:
The perimeter of the track is 2*pi*r = 20pi miles
In 10 hours, car B would have moved 20miles. So, when Car A leaves from point X, car B is 20pi - 20 miles from point X counter-clockwise and car A.
From here, we can express the distance of A from X like this:
xa = 3t
And the distance of B would be:
xb = 20pi - 20 - 2t
The time t where they would passed each other and put 12 miles between them would be the one where xa - xb is equal to 12:
xa - xb = 12
3t - (20pi - 20 - 2t) = 12
5t = 20 pi - 8
t = (20pi - 8)/5 = 10.96 h
Remember to add this value to the 10 hours car B had already been racing:
t = 20.96h
Answer:
the last one
Explanation:
Because it is a magnifying glass, it magnifies the object and makes it bigger than it appears
Answer:
(a) Wavelength is 0.436 m
(b) Length is 0.872 m
(c) 11.518 m/s
Solution:
As per the question:
The eqn of the displacement is given by:
![y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]](https://tex.z-dn.net/?f=y%28x%2C%20t%29%20%3D%20%281.22%20cm%29sin%5B14.4%20m%5E%7B-%201%7Dx%5Dcos%5B%28166%5C%20rad%2Fs%29t%5D)
(1)
n = 4
Now,
We know the standard eqn is given by:
(2)
Now, on comparing eqn (1) and (2):
A = 1.22 cm
K = 
where
A = Amplitude
K = Propagation constant
= angular velocity
Now, to calculate the string's wavelength,
(a) 
where
K = propagation vector
(b) The length of the string is given by:
(c) Now, we first find the frequency of the wave:
Now,
Speed of the wave is given by:
Answer:
the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Explanation:
Given the data in the question;
Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J
Mass of proton = 1.673 × 10⁻²⁷ kg
Charge of proton = 1.602 × 10⁻¹⁹ C
distance d = 2 m
we know that
Kinetic Energy = Charge of proton × Potential difference ΔV
so
Potential difference ΔV = Kinetic Energy / Charge of proton
we substitute
Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )
Potential difference ΔV = 20287.14 V
Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;
E = Potential difference ΔV / distance d
we substitute
E = 20287.14 V / 2 m
E = 10143.57 V/m or 1.01 × 10⁴ V/m
Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
