Answer:
There is absolutely No relationship between the weight of an object (which is constant) and the frictional force. If a block is sliding on a surface, that surface will be exerting a force on the block. That force can be resolved into a component parallel to the surface (which we call the frictional component), and a component perpendicular to the surface (called the normal component). For many situations, we find experimentally that the frictional component is approximately proportional to the normal component. The frictional component divided by the normal component is defined to be a quantity called the coefficient of kinetic or sliding friction. The coefficient of kinetic friction obviously depends on the nature of the surfaces involved. The normal component on an object can be decreased if you pull in the direction of the normal component (the weight does not change). However pulling this way on the object not only decreases the normal component, but it also decreases the frictional component since they are proportional. This is why it is easier to slide something if you pull up on it while you push it. If you push down, the normal and frictional components increase so it is harder to slide the object. The weight of an object is the downward force exerted by Earth’s gravity on that object, and it does not change no matter how you push or pull on the object.
consider the motion along the horizontal direction :
v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s
X = horizontal displacement of the ball = 2.0 m
a = acceleration along the horizontal direction = 0 m/s²
t = time taken to land = ?
using the kinematics equation
X = v₀ t + (0.5) a t²
2.0 = 3.0 t + (0.5) (0) t²
t = 2/3
consider the motion of the ball along the vertical direction
v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s
Y = vertical displacement of the ball = height of the table = h
a = acceleration along the vertical direction = 9.8 m/s²
t = time taken to land = 2/3
using the kinematics equation
Y = v₀ t + (0.5) a t²
h = 0 t + (0.5) (9.8) (2/3)²
h = 2.2 m
C 2.2 m
Coulomb's law:
Force = (<span>8.99×10⁹ N m² / C²<span>) · (charge₁) · (charge₂) / distance²
= (</span></span><span>8.99×10⁹ N m² / C²<span>) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C) / (1.0 m)²
= (8.99×10⁹ x 1×10⁻¹² / 1.0) N
= 8.99×10⁻³ N
= 0.00899 N repelling.
Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.
-- '4.0 kg masses'; don't need it.
Mass has no effect on the electric force between them.
-- 'frictionless table'; don't need it.
Friction has no effect on the force between them,
only on how they move in response to the force.
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Answer:
Explanation:
a ) Thermal efficiency = work output / heat input
= .38 MW / 1 MW = .38
OR 38%
Heat rejected at cold reservoir = heat input - work output
1 MW - .38 MW
= 0.62 MW.
b ) For reversible power output
efficiency = T₂ - T₁ / T₂ ; T₂ is temperature of hot reservoir and T₁ is temperature of cold reservoir.
= 1200 - 300 / 1200 = 900 / 1200
= .75
or 75%
rate at which heat is rejected
= 1 - .75 x 1
= .25 MW .
Hey there,
So first we need to know how much molecules are there in 4 grams of hydrogen right?
In one gram of hydrogen there are 6.0221415 multiplied by 10^25 molecules right?(this number is known as avagadors constant )
And to make one molecule of water u need two molecules of hydrogen and one of oxygen.
And oxygen is unlimited right?
So when we have 6.0221415 grams of hydrogen tow if these are required to make one water.
So divide that by two and you get 3.0110704 multiplied by 10^25.
So the answer is you can make the above number of water molecules.
Hope this helps!!!
