Answer:
C
Explanation:
When an object has a greater mass than the drag force pushing it back then it is moving forward therefore making car C faster than car D. If car D was much smaller it wouldn't have enough mass to push past the drag force.
Answer:
Make the surfaces little more smoother. ...
Lubrication is another way to make a surface smoother. ...
Make the object more streamlined. ...
Reduce the Normal force acting between the surfaces in contact. ...
Reduce the contact between the surfaces, so that less number of bonds will be formed.
Answer:
the net force applied to the car is zero.
Explanation:
According to Newton's second law, the acceleration of an object (a) is directly proportional to the net force applied (F):
where m is the object's mass.
In this problem, the car is moving with constant velocity: this means that the acceleration is zero, a = 0. Therefore, according to the previous equation, the net force must also be zero: F = 0. So, the correct answer is
the net force applied to the car is zero.
The answer is no moons<span> at all. That's right, </span>Venus<span> (and the planet Mercury) are the only two planets that don't </span>have<span>a single natural </span>moon<span> orbiting them. Figuring out why is one question keeping astronomers busy as they study the Solar System.</span>
Answer:
a) the distance between her and the wall is 13 m
b) the period of her up-and-down motion is 6.5 s
Explanation:
Given the data in the question;
wavelength λ = 26 m
velocity v = 4.0 m/s
a) How far from the wall is she?
Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;
x = λ/2
we substitute
x = 26 m / 2
x = 13 m
Therefore, the distance between her and the wall is 13 m
b) What is the period of her up-and-down motion?
we know that the relationship between frequency, wavelength and wave speed is;
v = fλ
hence, f = v/λ
we also know that frequency is expressed as the reciprocal of the time period;
f = 1/T
Hence
1/T = v/λ
solve for T
Tv = λ
T = λ/v
we substitute
T = 26 m / 4 m/s
T = 6.5 s
Therefore, the period of her up-and-down motion is 6.5 s
