Answer:
15.6m/s
Completed Question;
For a short period of time, the frictional driving force acting on the wheels of the 2.5-Mg van is N= 600t^2 , where t is in seconds. If the van has a speed of 20 km/h when t = 0, determine its speed when t = 5
Explanation:
Mass m = 2500kg
Speed v1 = 20km/h = 20/3.6 m/s = 5.556 m/s
To determine speed v2;
Using the principle of momentum and impulse;
mv1 + ∫₀⁵ F dt = mv2
The energy stored in a capacitor is given by:
where
U is the energy
C is the capacitance
V is the potential difference
The capacitor in this problem has capacitance
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
The gravitational force is inversely proportional to the
square of the distance between their centers. So the
force is greatest when the distance is zero.
Answer:
(a) Vf = 128 ft/s
(b) K.E = 122.8 Btu
Explanation:
(a)
In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:
2gh = Vf² - Vi²
where,
g = 32.2 ft/s²
h = height = 253 ft
Vf = Final Velocity = ?
Vi = Initial Velocity = 10 ft/s
Therefore,
(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²
16293.2 ft²/s² + 100 ft²/s² = Vf²
Vf = √(16393.2 ft²/s²)
<u>Vf = 128 ft/s</u>
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(b)
The kinetic energy of the object before it hits the surface of earth is given by:
K.E = (0.5)(m)(Vf)²
where,
m = mass of object = 375 lb
K.E = Kinetic energy of object before it strikes the surface of earth = ?
Therefore,
K.E = (0.5)(375 lb)(128 ft/s)²
K.E = 3073725 lb.ft²/s²
Now, converting this to Btu:
K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)
<u>K.E = 122.8 Btu</u>
