Answer:

Explanation:
From the second law of Newton movement laws, we have:
, and we know that a is the acceleration, which definition is:
, so:

The next step is separate variables and integrate (the limits are at this way because at t=0 the block was at rest (v=0):

(This is the indefinite integral), the definite one is:

Answer:
Resistivity ρ=1.12 x 10^-4 Ωm
Explanation:
ρ= RA/l, where R is resistance, A is cross sectional area and l is length
A=πr^2
Note Current is given R is proportion to temperature and inversely proportional to Current R=(20+273)/14*10^-2 =2000Ω
⇒ρ=R*πr^2/l all length in metre.
fraction equation is<span>
F =µR
F=friction,µ=coefficient , R=reaction = mg
use same equation for b part, but the reaction is no longer mg because the plain is now inclined. Draw a forces diagram and you will see that the reaction force can be calculated from the weight of the object and inclination of the plain using trigonometry.</span>
Answer:
The downwind side of an obstacle such as a ridge. The addition of weight on top of a snowpack, usually from precipitation, wind drifting, or a person. An avalanche that releases from a point and spreads downhill collecting more snow - different from a slab avalanche. Also called a point-release or sluff.
Explanation:
Answer:

Explanation:
Given that
Height = h
Radius = R
From energy conservation

At point B
The minimum speed to complete the the circle

So the kinetic energy at point B




Without falling off at the top (point B)



