How is gravity "best" described?
2 answers:
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Answer:
F = 63N
Explanation:
M= 1.5kg , t= 2s, r = (2t + 10)m and
Θ = (1.5t² - 6t).
magnitude of the resultant force acting on 1.5kg = ?
Force acting on the mass =
∑Fr =MAr
Fr = m(∇r² - rθ²) ..........equation (i)
∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)
The horizontal path is defined as
r = (2t + 10)
dr/dt = 2, d²r/dt² = 0
Angle Θ is defined by
θ = (1.5t² - 6t)
dθ/dt = 3t, d²θ/dt² = 3
at t = 2
r = (2t + 10) = (2*(2) +10) = 14
but dr/dt = 2m/s and d²r/dt² = 0m/s
θ = (1.5(2)² - 6(2) ) = -6rads
dθ/dt =3(2) - 6 = 0rads
d²θ/dt = 3rad/s²
substituting equation i into equation ii,
Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)
∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]
∑F = 1.5(14*3+0) = 63N
F = √(Fr² +FΘ²) = √(0² + 63²) = 63N
a) 50.8 mm
b) The whole image (1:1)
c) It seems reasonable
Explanation:
a)
To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:
where
f is the focal length of the lens
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem:
f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)
p = 3.00 m is the distance of the person from the lens
Therefore, we can find q:
b)
Here we need to find the height of the image first.
This can be done by using the magnification equation:
where:
y' is the height of the image
y = 1.75 m is the height of the real person
q = 50.8 mm = 0.0508 m is the distance of the image from the lens
p = 3.00 m is the distance of the person from the lens
Solving for y', we find:
(the negative sign means the image is inverted)
Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.
c)
This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.
The time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.
The given parameters;
<em>Mass of the first object, m1 = 1 kg</em>
<em>Mass of the second object, m2 = 5 kg</em>
The final velocity of the objects during the downward motion is calculated as follows;
The time of motion of the object from the given height is calculated as;
The time of motion of each object is independent of mass of the object.
Thus, the time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.
Learn more about time of motion here: brainly.com/question/2364404