The formula we can use in
this case would be:
v = sqrt (T / (m / l))
Where,
v = is the velocity of the
transverse wave = unknown (?)
T = is the tension on the
rope = 500 N
m = is the mass of the
rope = 60.0 g = 0.06 kg
l = is the
length of the rope = 2.00 m
Substituting the given values into the equation to search
for the speed v:
v = sqrt (500 N/(0.06 kg /2 m))
v = sqrt (500 * 2 / 0.06)
v = sqrt (16,666.67)
<span>v = 129.10 m/s</span>
Answer:
a) transparent
b) transparent
c) opaque
Explanation:
In the first one, the light rays go completely through, so it is transparent.
The second one I'm not too sure about. It is refraction so it's going through a different material, but the fact that it went through makes me say transparent.
Last one, the light rays are reflecting off the surface so it's opaque.
Please feel free to correct me if I'm wrong. This is just my understanding
Answer: The smallest effort = 300N
Explanation:
Using one of the condition for the attainment of equilibrium:
Clockwise moment = anticlockwise moments
900 × 1 = 3 × M
Where M = the weight of the strong man
3M = 900
M = 900/3 = 300N
Therefore, 300N is the smallest effort that the strongman can use to lift the goat
Answer:
20m/s
Explanation:
Given parameters:
Distance of throw = 50m
Time = 3s
Unknown:
Average speed = ?
Solution:
Average speed is distance divided by time;
Average speed =
Insert the parameters and solve;
Average speed =
= 16.66667m/s = 20m/s
Answer:
The answer to the question is
At an altitude of 1413 km or 1.222·R above the north pole the weight of an object reduced to 67% of its earth-surface value
Explanation:
We make use of the gravitational formula as follows
F = G
where
m₁ = mass of the object
m₂ = mass of the earth
d = distance between the two objects and
G = gravitational constant
if at the altitude the weight is reduced to 67 % of its weight on earth then
with all other variables remaining constant, we have
67% F = G
=0.67× G×
cancelleing like ternss from both sides we have
1/R₂² =0.67/R₁² or r₁²/R₂² =0.67 and R₁/R₂ = 0.8185
or R₂ = R₁/0.8185 or 1.222·R = (6.37×10⁶ m)/0.8185 = 7.783×10⁶ m
Hence at an altitude = 1.413×10⁶ m the weight of the object will reduce to 67% its earth-surface value