Answer:
h = 2 R (1 +μ)
Explanation:
This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the
let's use the mechanical energy conservation agreement
starting point. Lower, just at the curl
Em₀ = K = ½ m v₁²
final point. Highest point of the curl
= U = m g y
Find the height y = 2R
Em₀ = Em_{f}
½ m v₁² = m g 2R
v₁ = √ 4 gR
Any speed greater than this the body remains in the loop.
In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law
X axis
-fr = m a (1)
Y Axis
N - W = 0
N = mg
the friction force has the formula
fr = μ N
fr = μ m g
we substitute 1
- μ mg = m a
a = - μ g
having the acceleration, we can use the kinematic relations
v² = v₀² - 2 a x
v₀² = v² + 2 a x
the length of this zone is x = 2R
let's calculate
v₀ = √ (4 gR + 2 μ g 2R)
v₀ = √4gR( 1 + μ)
this is the speed so you must reach the area with fricticon
finally have the third part we use energy conservation
starting point. Highest on the ramp without rubbing
Em₀ = U = m g h
final point. Just before reaching the area with rubbing
= K = ½ m v₀²
Em₀ = Em_{f}
mgh = ½ m 4gR(1 + μ)
h = ½ 4R (1+ μ)
h = 2 R (1 +μ)
Inverse Square Law of Light states that light intensity falls off rapidly with distance from its source. T<span>he intensity varies with the square of the flash-to-subject distance.
intensity at distance 1/intensity at distance 2=distance2^2/distance1^2
In our case, distance1=10m and distance2=20m, 20^2/10^2= 400/100=4, which means that t</span>he intensity of the wave<span> has increased by a factor of four.
Answer:D</span>
In elastic
collision, both the kinetic energy and momentum are conserved. Conservation
means that both the kinetic energy and momentum will have the same values
before and after elastic collision.
<span>As the
object A has low mass than object B. Hence upon collision, object B moves
forward, while object A will move backward. So option "C" is correct. </span>
If we assume also that the temperature of the air does not change, we can use Boyle's Law:
p₁V₁ = p₂V₂
Now, we know:
p₁ = 100kPa
V₂ = 100cm³ (the volume of the tyre)
V₁ = 120cm³ (becuse the air is contained inside the tyre AND the pump)
We can solve for p₂:
p₂ = (p₁V₁)/V₂
= (100×120)/100
= 120kPa
Therefore your answer is: 120kPa
Answer:
a) The schematic illustrating is attached
b) The heat transfer to the heat engine is 2142.86 kJ, the heat transfer from the heat engine is 1392.86 kJ
c) The heat transfer to the heat engine is 1648.35 kJ, the heat transfer from the heat engine is 898.35 kJ
Explanation:
b) The heat transfer to the engine and the heat transfer from the engine to the air is:
Where
W = 750 kJ
n = 35% = 0.25
Replacing:
c) The efficiency of Carnot engine is:
The heat transfer to the heat engine is:
The heat transfer from the heat engine is:
