"Energy and Momentum" is always conserved in an inelastic condition
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A, B, and C are all incorrect because they are talking about springs. I'm assuming you're talking about a 2 dimensional, side to side pendulum. In that case, the answer is D because as the pendulum goes up from one side, gravity accelerates it downward until it swings to the other side. If you're not talking about this type of pendulum, leave a comment and ill come back.
The advantage is having new cells to clone or help different organisms with.
Hope this helps:)
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Answer:
(a) F = 1500 N.
(b) Ratio force to the antelepe's weight = 3.40
Explanation:
Force : This can be defined as the product of mass and the distance moved by a body. Its S.I unit is Newton. It can be represented mathematically as
F = Ma
Where F= force, M = mass (Kg) and a = Acceleration (m/s²)
Weight: This can be defined as the force on a body due to gravitation field. It is also measured in Newton (N). It can be represented mathematically as
W = Mg
Where W = weight of the body, M = mass of the body (Kg), g = Acceleration due to gravity.
(a)
F = Ma
Where M = 45kg,
a = unknown.
But we can look for acceleration Using one of the equation of motion,
v² = u² + 2gs
Where v= final velocity(m/s), u = initial velocity (m/s) g = 0 m/s, g = 9.8m/s² and s = height = 2.5m.
∴ v² = 2gs
v = √2gs = √(2×9.8×2.5)
v= √49 = 7m/s
With the force applied, the impala’s velocity must increase from 0 m/s to 7 m/s in 0.21 second
∴ a = (v-u)/t
a = (7-0)/0.21 = 7/0.21
a = 33.33 m/s².
F = 45 × 33.33 ≈ 1500
F = 1500 N.
(b)
Where F = Force = 1500 N
and W = Weight = Mg = 45 × 9.8 = 441 N
∴Ratio force to the antelepe's weight = F/W = 1500/441 = 3.40
Ratio force to the antelepe's weight = 3.40
Answer:
The specific heat addition is 773.1 kJ/kg
Explanation:
from table A.5 we get the properties of air:
k=specific heat ratio=1.4
cp=specific heat at constant pressure=1.004 kJ/kg*K
We calculate the pressure range of the Brayton cycle, as follows
n=1-(1/(P2/P1)^(k-1)/k))
where n=thermal efficiency=0.5. Clearing P2/P1 and replacing values:
P2/P1=(1/0.5)^(1.4/0.4)=11.31
the temperature of the air at state 2 is equal to:
P2/P1=(T2/T1)^(k/k-1)
where T1 is the temperature of the air enters the compressor. Clearing T2
11.31=(T2/290)^(1.4/(1.4-1))
T2=580K
The temperature of the air at state 3 is equal to:
P2/P1=(T3/T4)^(k/(k-1))
11.31=(T3/675)^(1.4/(1.4-1))
T3=1350K
The specific heat addition is equal to:
q=Cp*(T3-T2)=1.004*(1350-580)=773.1 kJ/kg
