Answer:
The final velocity of the runner at the end of the given time is 2.7 m/s.
Explanation:
Given;
initial velocity of the runner, u = 1.1 m/s
constant acceleration, a = 0.8 m/s²
time of motion, t = 2.0 s
The velocity of the runner at the end of the given time is calculate as;

where;
v is the final velocity of the runner at the end of the given time;
v = 1.1 + (0.8)(2)
v = 2.7 m/s
Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.
Radio waves have a low frequency and a long wavelength. (A. low; long)
Explanation:
Electromagnetic waves are periodic oscillations of the the electric and the magnetic field in a plane perpendicular to the direction of motion the wave.
All electromagnetic waves travel in a vacuum at the same speed, which is known as the speed of light:
Electromagnetic waves are also divided into 7 different types, according to their wavelength and frequency. From shortest to longest wavelength (and therefore, from highest to lowest frequency), we have:
Gamma rays
X rays
Ultraviolet
Visible light
Infrared radiation
Microwaves
Radio waves
As we can see from the list, radio waves are the waves with lowest frequency and longest wavelength. They are used in telecommunications, satellite communications, and in general in long-distance transmission of signals. So, the correct answer is
Radio waves have a low frequency and a long wavelength.
Learn more about electromagnetic waves:
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I don't understand the question
Explanation:
Please give it clearly
Answer:
The distance from the top of the stick would be 2l/3
Explanation:
Let the impulse 'FΔt' acts as a distance 'x' from the hinge 'H'. Assume no impulsive reaction is generated at 'H'. Let the angular velocity of the rod about 'H' just after the applied impulse be 'W'. Also consider that the center of percussion is the point on a bean attached to a pivot where a perpendicular impact will produce no reactive shock at the pivot.
Applying impulse momentum theorem for linear momentum.
FΔt = m(Wl/2), since velocity of center of mass of rod = Wl/2
Similarly applying impulse momentum theorem per angular momentum about H
FΔt * x = I * W
Where FΔt * x represents the impulsive torque and I is the moment of inertia
F Δt.x = (ml² . W)/3
Substituting FΔt
M(Wl/2) * x = (ml². W)/3
1/x = 3/2l
x = 2l/3