Answer:
d) -4.0
Explanation:
The magnification of a lens is given by

where
M is the magnification
q is the distance of the image from the lens
p is the distance of the object from the lens
In this problem, we have
p = 50 cm is the distance of the object from the lens
q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct
Also, q is positive since the image is real
So, the magnification is

From conservation of energy, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately
The given weight of Elliot is 600 N
From conservation of energy, the total mechanical energy of Elliot must have been converted to elastic potential energy. Then, the elastic potential energy from the spring was later converted to maximum potential energy P.E of Elliot.
P.E = mgh
where mg = Weight = 600
To find the height Elliot will reach, substitute all necessary parameters into the equation above.
250 = 600h
Make h the subject of the formula
h = 250/600
h = 0.4167 meters
Therefore, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately
Learn more about energy here: brainly.com/question/24116470
Answer:
b) N = 560 N, c) fr = 138.56 N, d) μ = 0.247
Explanation:
a) In the attachment we can see the free body diagram of the system
b) Let's write Newton's second law on the y-axis
N + T_y -W = 0
N = W -T_y
let's use trigonometry for tension
sin θ = T_y / T
cos θ = Tₓ / T
T_y = T sin θ
Tₓ = T cos θ
we substitute
N = W - T sin 30
we calculate
N = 640 - 160 sin 30
N = 560 N
c) as the system goes at constant speed the acceleration is zero
X axis
Tₓ - fr = 0
Tₓ = fr
we substitute and calculate
fr = 160 cos 30
fr = 138.56 N
d) the friction force has the formula
fr = μ N
μ = fr / N
we calculate
μ = 138.56 / 560
μ = 0.247
Answer: -0.84 rad/sec (clockwise)
Explanation:
Assuming no external torques act on the system (man + turntable), total angular momentum must be conserved:
L1 = L2
L1 = It ω + mm. v . r = 81.0 kg . m2 .21 rad/s – 56.0 kg. 3.1m/s . 3.1 m
L1 = -521.15 kg.m2/sec (1)
(Considering to the man as a particle that is moving opposite to the rotation of the turntable, so the sign is negative).
Once at rest, the runner is only a point mass with a given rotational inertia respect from the axis of rotation, that can be expressed as follows:
Im = m. r2 = 56.0 kg. (3.1m)2 = 538.16 kg.m2
The total angular momentum, once the runner has come to an stop, can be written as follows:
L2= (It + Im) ωf = -521.15 kg.m2/sec
L2= (81.0 kg.m2 + 538.16 kg.m2) ωf = -521.15 kg.m2/sec
Solving for ωf, we get:
ωf = -0.84 rad/sec (clockwise)