In what state of matter does oxygen exist at room temperature
2 answers:
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Answer:
Explanation:
Applied force, F = 18 N
Coefficient of static friction, μs = 0.4
Coefficient of kinetic friction, μs = 0.3
θ = 27°
Let N be the normal reaction of the wall acting on the block and m be the mass of block.
Resolve the components of force F.
As the block is in the horizontal equilibrium, so
F Cos 27° = N
N = 18 Cos 27° = 16.04 N
As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .
The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N .... (1)
The vertically downward force acting on the block is mg - F Sin 27°
= mg - 18 Sin 27° = mg - 8.172 ... (2)
Now by equating the forces from equation (1) and (2), we get
mg - 8.172 = 6.42
mg = 14.592
m x 9.8 = 14.592
m = 1.49 kg
Thus, the mass of block is 1.5 kg.
Refer to the diagram shown below.
We want to find y in terms of d, φ and θ.
By definition,
Therefore
y = x tan(θ) (1)
y = (x - d) tan(φ) (2)
Equate (1) and (2).
![(x - d) \, tan(\phi) = x \, tan(\theta) \\ x[tan(\phi) - tan(\theta)] = d \, tan(\phi) \\ x= \frac{d tan(\phi)}{tan(\phi)-tan(\theta)}](https://tex.z-dn.net/?f=%28x%20-%20d%29%20%5C%2C%20tan%28%5Cphi%29%20%3D%20x%20%5C%2C%20tan%28%5Ctheta%29%20%5C%5C%20x%5Btan%28%5Cphi%29%20-%20tan%28%5Ctheta%29%5D%20%3D%20d%20%5C%2C%20tan%28%5Cphi%29%20%5C%5C%20x%3D%20%5Cfrac%7Bd%20tan%28%5Cphi%29%7D%7Btan%28%5Cphi%29-tan%28%5Ctheta%29%7D%20)
From (1), obtain the required expression for y.
Answer:
We want to find y in terms of d, φ and θ.
By definition,
Therefore
y = x tan(θ) (1)
y = (x - d) tan(φ) (2)
Equate (1) and (2).
From (1), obtain the required expression for y.
Answer:
Answer : The specific heat of unknown sample 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 unknown sample = ?
= specific heat of water =
= specific heat of copper =
= mass of unknown sample = 72.0 g = 0.072 kg
= mass of water = 203 g = 0.203 kg
= mass of copper = 187 g = 0.187 kg
= final temperature of calorimeter =
= initial temperature of unknown sample =
= initial temperature of water and copper =
Now put all the given values in the above formula, we get
Therefore, the specific heat of unknown sample is,