Answer:
0.5 m/s north
Explanation:
Take east to be +x, west to be -x, north to be +y, and south to be -y.
His displacement in the x direction is:
x = 20 m − 20 m = 0 m
His displacement in the y direction is:
y = 10 m
His total displacement is therefore 10 m north.
His velocity is equal to displacement divided by time.
v = 10 m north / 20 s
v = 0.5 m/s north
Answer:
D. Dylan is incorrect because a 90-degree launch angle results in the largest vertical range
Explanation:
Projectile is the motion of an object thrown into space. When an object is thrown into space, the only force which acts on it is the acceleration due to gravity.
An object thrown into space would reach maximum height (vertical range) if it is launched at an angle of 90 degrees. For maximum horizontal range, the object needs to be launched at an angle of 45 degrees.
Therefore Dylan is incorrect because a 90-degree launch angle results in the largest vertical range
Answer:
6.29 meters.
Explanation:
, where v is the speed of wave and f is the frequency of wave.
We are given that ,
The speed of sound is 346 m/s.
i..e v=346 m/s
A sound wave travels at a frequency of 55 H.
i..e f=55
the wavelength would be 6.29 meters.
This is based on another brainly answer
Link: brainly.com/question/12538018
Answer:
a) The current density ,J = 2.05×10^-5
b) The drift velocity Vd= 1.51×10^-15
Explanation:
The equation for the current density and drift velocity is given by:
J = i/A = (ne)×Vd
Where i= current
A = Are
Vd = drift velocity
e = charge ,q= 1.602 ×10^-19C
n = volume
Given: i = 5.8×10^-10A
Raduis,r = 3mm= 3.0×10^-3m
n = 8.49×10^28m^3
a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]
J = (5.8×10^-10) /(2.83×10^-5)
J = 2.05 ×10^-5
b) Drift velocity, Vd = J/ (ne)
Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)
Vd = (2.05×10^-5)/(1.36 ×10^10)
Vd = 1.51× 10^-5
