Average speed = total distance / total time
total distance = 40 + 20 = 60km
total time taken = 10 + 5 = 15 minutes
Average speed = 60/15 = 4km/min
Answer:
same 0.81m
Explanation:
in this problem if we assume there no resistance of any sort. and we apply the energy conservation
change in Potential energy = change in kinetic energy
mgh = 0.5mv^2
gh = 0.5v^2
the above relation suggests that the speed at the bottom is only depending on the height it is released from not on the shape, mass or radius.
so at the bottom
put h = 0.81m
9.81 * 0.81 * 2 = v^2
v=3.99 m/s
both CYLINDER and SPHERE will have same velocity at the bottom if released from the same height irrespective of shape and size
Answer:
(a) ω = 1.57 rad/s
(b) ac = 4.92 m/s²
(c) μs = 0.5
Explanation:
(a)
The angular speed of the merry go-round can be found as follows:
ω = 2πf
where,
ω = angular speed = ?
f = frequency = 0.25 rev/s
Therefore,
ω = (2π)(0.25 rev/s)
<u>ω = 1.57 rad/s
</u>
(b)
The centripetal acceleration can be found as:
ac = v²/R
but,
v = Rω
Therefore,
ac = (Rω)²/R
ac = Rω²
therefore,
ac = (2 m)(1.57 rad/s)²
<u>ac = 4.92 m/s²
</u>
(c)
In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:
Centripetal Force = Frictional Force
m*ac = μs*R = μs*W
m*ac = μs*mg
ac = μs*g
μs = ac/g
μs = (4.92 m/s²)/(9.8 m/s²)
<u>μs = 0.5</u>
The difference between the two is, well for one
Spectrum: The entire range that the "<em>waves" </em>could be such, as visible light, x-ray's and so on.
Waves: These are different because they aren't telling you or showing the entire spectrum just which they length that they are.
<em>It may confuse you but it makes sense to me (Sorry)</em>
Answer:
The speed of q₂ is 
Explanation:
Given that,
Distance = 0.4 m apart
Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.
We need to calculate the speed of q₂
Using conservation of energy
Put the value into the formula
Hence, The speed of q₂ is
