Answer:
A mass of 4 Kg rest on the horizontal plane. The plane is gradually inclined until at an angle of θ=15
∘
with the horizontal,the mass just being to slide what is the coffeficient of static friction between the block & the surface.
Explanation:
Answer:
Angle with the +x axis is θ = 79.599degree
Then the velocity of owner = 1.235m/s
Explanation:
Given that the mass of dog is m1 =26.2 kg
velocity of dog is u1 = 3.02 m/s (north)
mass of cat is m2 = 5.3 kg
velocity is u2 = 2.74 m/s (east )
Mass of owner is M = 65.1 kg
Consider the east direction along +x axis andnorth along +y
momentum of dog is Py = m1 x u1
= 79.124 kg.m/s (j)
momentum of cat is Px = m2 x u2
= 14.522 kg.m/s (i)
Then the net magnitude of momentum is P = (Px2 + Py2)1/2
= 80.445
Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree
Then the velocity of owner is v = P / M = 1.235 m/s
Answer:
They become the same exact tempature
Explanation:
Since they got connected it should be the same.
Answer : The final temperature is, 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.


where,
= specific heat of ice = 
= specific heat of water = 
= mass of ice = 50 g
= mass of water = 200 g
= final temperature = ?
= initial temperature of ice = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:


Therefore, the final temperature is, 