Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is
X=0m/s
From the question we are told that
mosquito flying 2 m/s
against a 2 m/s headwind
Generally
The speed over the ground is the Flight Speed minus resistance speed
Generally the equation for the speed over the ground is mathematically given as
X=Flight Speed-resistance speed
Therefore
X=2-2
X=0m/s
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Answer:Visible light is a small part of the electromagnetic spectrum. The spectrum covers everything from gamma rays, x-rays, visible light, infrared, microwave and radio waves. Each part of the spectrum, including the different colors of visible light, have different wavelengths (the space between each wave).
Explanation:
Answer:For example, standard atmospheric pressure (or 1 atm) is defined as 101.325 kPa. The millibar, a unit of air pressure often used in meteorology, is equal to 100 Pa. (For comparison, one pound per square inch equals 6.895 kPa.)
Explanation:A pascal is a pressure of one newton per square metre, or, in SI base units, one kilogram per metre per second squared.
I hope this helps.... I'm sorry if it doesn't
Answer:
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Explanation:
From the exercise we know two information. The real speed and the experimental measured by the speedometer
Since the speedometer is only accurate to within 0.1km/h the experimental speed is
Knowing that we can calculate Kinetic energy for the real and experimental speed
Now, the potential error in her calculated kinetic energy is:
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