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Angelina_Jolie [31]
3 years ago
14

Is kicking a ball a conservation of momentum ?

Physics
1 answer:
Katarina [22]3 years ago
7 0

Answer:

Yes, When the momentum and kinetic energy are conserved, as they are approximately conserved when kicking a ball, there is no loss in momentum and energy before or after the collision.

Hope this helps!

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A river flowing steadily at a rate of 175 m3/s is considered for hydroelectric power generation. It is determined that a dam can
Alex_Xolod [135]

Answer:

137200000 watts or 137200 kilowatts

Explanation:

The formula for power is P= dhrg

Where P = Power in watts

d = density of water (1000 kg/m^3)

h = height in meters

r = flow rate in cubic meters per second,

g = acceleration due to gravity of 9.8 m/s^2,

Plugging in the known values,

we get

P = 1000 kg/m^3 * 80 m * 175 m^3/s * 9.8 m/s^2

P = 80000 kg/m^2 * 175 m^3/s * 9.8 m/s^2

P = 14000000 kg m/s * 9.8 m/s^2

P = 137200000 kg m^2/s^3

P = 137200000 watts or 137200 kilowatts

The above figure assumes 100% efficiency which is impossible. A good efficiency would be 90% so the actual power available would be close to 0.90 * 137200 = 123480 kilowatts

6 0
3 years ago
Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th
Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of aluminium = 0.90J/g^oC

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of aluminum = 0.500 kg = 500 g

m_2 = mass of water = 0.250 kg  = 250 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of aluminum = 150^oC

T_2 = initial temperature of water = 20^oC

Now put all the given values in the above formula, we get:

500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC

T_f=59.10^oC

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

8 0
3 years ago
A diver stands on a diving platform 10.0 m above the surface of a pool and leaps upward with an initial speed of 2.5 m/s. how fa
Alexus [3.1K]
<span>The diver is heading downwards at 12 m/s Ignoring air resistance, the formula for the distance under constant acceleration is d = VT - 0.5AT^2 where V = initial velocity T = time A = acceleration (9.8 m/s^2 on Earth) In this problem, the initial velocity is 2.5 m/s and the target distance will be -7.0 m (3.0 m - 10.0 m = -7.0 m) So let's substitute the known values and solve for T d = VT - 0.5AT^2 -7 = 2.5T - 0.5*9.8T^2 -7 = 2.5T - 4.9T^2 0 = 2.5T - 4.9T^2 + 7 We now have a quadratic equation with A=-4.9, B=2.5, C=7. Using the quadratic formula, find the roots, which are -0.96705 and 1.477251164. Now the diver's velocity will be the initial velocity minus the acceleration due to gravity over the time. So V = 2.5 m/s - 9.8 m/s^2 * 1.477251164 s V = 2.5 m/s - 14.47706141 m/s V = -11.97706141 m/s So the diver is going down at a velocity of 11.98 m/s Now the negative root of -0.967047083 is how much earlier the diver would have had to jump at the location of the diving board. And for grins, let's compute how fast he would have had to jump to end up at the same point. V = 2.5 m/s - 9.8 m/s^2 * (-0.967047083 s) V = 2.5 m/s - (-9.477061409 m/s) V = 2.5 m/s + 9.477061409 m/s V = 11.97706141 m/s And you get the exact same velocity, except it's the opposite sign. In any case, the result needs to be rounded to 2 significant figures which is -12 m/s</span>
7 0
3 years ago
About what fraction of earth's freshwater is in the form of ice
Bingel [31]
68% So 68/100 of freshwater is found in ice
3 0
3 years ago
How many cubic centimeters of area does 762 g of distilled water occupy?
BartSMP [9]
1 kg of water = 1 L = 1 dm³
762 g of water = 762 cm³
8 0
3 years ago
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