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Leokris [45]
3 years ago
15

During a snowball fight two balls with masses of 0.4 kg and 0.6 kg, respectively, are thrown in such a manner that they meet hea

d on and combine to form a single mass of 1 kg. the magnitude of intial velocity foreach is 15 m/s. the final kinetic energy of the system just after the collision is what percentage of kinectic energy just before the collision
Physics
1 answer:
valkas [14]3 years ago
3 0
Considering conservation of momentum;
m1v1 + m2v2 = m3v3

In which,
m1 = mass of snowball 1 = 0.4 kg
v1 = velocity of snowball 1 = 15 m/s
m2 = mass of snowball 2 = 0.6 kg
v2 = velocity of snow ball 2 = 15 m/s
m3 = combined mass = 1 kg
v3 = velocity after comination
Therefore;
0.4*15 + 0.6*15 = 1*v3
v3 = 6+9 = 15 m/s
KE = 1/2mv^2

Then,
KE1 = 1/2*0.4*15^2 = 45 J
KE2 = 1/2*0.6*15^2 = 67.5 J
KE3 = 1/2*1*15^2 = 112.5 J

Therefore, KE3 (kinetic energy after collision) = K1+K2 {kinetic energy before collision). And thus it is 100%.
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Answer:

Φ = 5.589×10⁻⁵  Wb

Explanation:

The inductance of a coil is given as

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Also,

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equating equation 1 and equation 2

e/(di/dt) = NΦ/i

making Φ the subject of the equation,

Φ = (e×i)/N.(di/dt) .................. Equation 3

Given: e = 28.0 mV = 0.028 V, N = 501 turns, di/dt = 12.0 A/s, i = 4.00 a

Substitute into equation 3,

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Φ = 0.112/2004

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6 0
3 years ago
Why do we balance chemical equations
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An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane travel
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Answer:

v=4.44\frac{m}{s}

Explanation:

Given that the airplane starts from the rest (this is initial velocity equals to zero)  and accelerates at a constant rate, position can be described like this: x=v_{0}t +\frac{1}{2} at^{2} where x is the position, t is the time a is the acceleration and v_{0} is initial velocity. In this way acceleration can be found. a=\frac{2(x-v_{0}t) }{t^{2} } =\frac{2(1.11m-0)}{1s^{2} } =2.22\frac{m}{s^{2} }.

Now we are able to found velocity at any time with the formula: v=v_{0} +at = 0\frac{m}{s} +(2.22\frac{m}{s^{2}}.2s)=4.44\frac{m}{s}

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elasticity stretches and can also return to it's normal size ..

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