How fast must a 100 kg object be going in order for it to stop a 200 kg object traveling at 10 km/hr when the two objects collid
1 answer:
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Answer:
The value is
Explanation:
From the question we are told that
The velocity which the rover is suppose to land with is
The mass of the rover and the parachute is
The drag coefficient is
The atmospheric density of Earth is
The acceleration due to gravity in Mars is
Generally the Mars atmosphere density is mathematically represented as
=>
=>
Generally the drag force on the rover and the parachute is mathematically represented as
=>
=>
Gnerally this drag force is mathematically represented as
Here A is the frontal area
So
=>
=>
Answer:
V_{a} - V_{b} = 89.3
Explanation:
The electric potential is defined by
= - ∫ E .ds
In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.
V_{b} - V_{a} = - ∫ E ds
We substitute
V_{b} - V_{a} = - ∫ (α + β/ y²) dy
We integrate
V_{b} - V_{a} = - α y + β / y
We evaluate between the lower limit A 2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m
V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)
V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33
V_{b} - V_{a} = - 89.3 V
As they ask us the reverse case
V_{b} - V_{a} = - V_{b} - V_{a}
V_{a} - V_{b} = 89.3