A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force
of static friction f on the block?
A) f > mg
B) f = mgsin(?)
C) f > mgcos(?)
D) f = mgcos(?)
E) f > mgsin(?)
A) f > mg
B) f = mgsin(?)
C) f > mgcos(?)
D) f = mgcos(?)
E) f > mgsin(?)
1 answer:
Answer:
Option B
Explanation:
For a system of block on inclined ramp shown in the attached image. From the attached image, the Normal force N, weight mg and frictional force f act on the block. The sum of vertical forces should be zero just as sum of vertical forces should be zero when the system is in equilibrium condition.
Taking sum of forces along the inclined plane we deduce that
[tex]f=mgsin \theta [tex]
Therefore, option B is the correct option.
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Answer:
The net emissions rate of sulfur is 1861 lb/hr
Explanation:
Given that:
The power or the power plant = 750 MWe
Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:
= 3988126.8 J
= 3.99 MJ
Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.
i.e.
The mass of the coal that is burned per sec
The mass of the coal that is burned per sec = 187.97 lb/s
The mass of sulfur burned
= 2.067 lb/s
To hour; we have:
= 7444 lb/hr
However, If a scrubber with 75% removal efficiency is utilized,
Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr
= 0.25 × 7444 lb/hr
= 1861 lb/hr
Hence, the net emissions rate of sulfur is 1861 lb/hr
To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.
Planet gravitational force
Distance between planet and star
Gravitational force is
Applying the new distance,
Replacing with the previous force,
Replacing our values
Therefore the magnitude of the force on the star due to the planet is