The average speed is the ratio between the total space and the total time of the motion:
The total space is
while the total time is
So, the average velocity is
We can also rewrite it in m/s. The total space is
, while the time is
, and so
Answer:
14 × a^3
Explanation:
Product means multiplication
Cubed means to the power of 3
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The force of friction is equal to the product of the vertical force applied by the surface to the object in the coefficient of friction.
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In this question ,
surface vertical force = Weight of the object
Thus ;
svf = ( mass ) × ( gravity acceleration )
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If gravity acceleration is 10 :
svf = 10 × 10 = 100 N
So ;
frictional force = 100 × 0.20
frictional force = 20 N
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If gravity acceleration is 9.8 :
svf = 10 × 9.8 = 98 N
So ;
frictional force = 98 × 0.20
frictional force = 19.6 N
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Answer:
A) 60%
B) p2 = 1237.2 kPa
v2 = 0.348 m^3
C) w1-2 = w3-4 = 1615.5 kJ
Q2-3 = 60 kJ
Explanation:
A) calculate thermal efficiency
Л = 1 - 
where Tl = 300 k
Th = 750 k
hence thermal efficiency ( Л ) = [1 - ( 300 / 750 )] * 100 = 60%
B) calculate the pressure and volume at the beginning of the isothermal expansion
calculate pressure ( P2 ) :
= P3v3 = mRT3 ----- (1)
v3 = 0.4m , mR = 2* 0.287, T3 = 750
hence P3 = 1076.25
next equation to determine P2
Qex = p3v3 ln( p2/p3 )
60 = 1076.25 * 0.4 ln(p2/p3)
hence ; P2 = 1237.2 kpa
calculate volume ( V2 )
p2v2 = p3v3
v2 = p3v3 / p2
= (1076.25 * 0.4 ) / 1237.2
= 0.348 m^3
C) calculate the work and heat transfer for each four processes
work :
W1-2 = mCv( T2 - T1 )
= 2*0.718 ( 750 - 300 ) = 1615.5 kJ
W3-4 = 1615.5 kJ
heat transfer
Q2-3 = W2-3 = 60KJ
Q3-4 = 0
D ) sketch of the cycle on p-V coordinates
attached below
Answer:
the shooting angle ia 18.4º
Explanation:
For resolution of this exercise we use projectile launch expressions, let's see the scope
R = Vo² sin (2θ) / g
sin 2θ = g R / Vo²
sin 2θ = 9.8 75/35²
2θ = sin⁻¹ (0.6)
θ = 18.4º
To know how for the arrow the tree branch we calculate the height of the arrow at this point
X2 = 75/2 = 37.5 m
We calculate the time to reach this point since the speed is constant on the X axis
X = Vox t
t2 = X2 / Vox = X2 / (Vo cosθ)
t2 = 37.5 / (35 cos 18.4)
t2 = 1.13 s
With this time we calculate the height at this point
Y = Voy t - ½ g t²
Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²
Y = 6.23 m
With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch
