Answer:
The horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m
Explanation:
Given;
height above the ground, h = 16.4 m
speed of the eagle, v = 9.3 m/s
The time it will take the pumpkin to fall at the given height is calculated as;
![t = \sqrt{\frac{2h}{g} }\\\\t = \sqrt{\frac{2*16.4}{9.8} }\\\\t = 1.83 \ s](https://tex.z-dn.net/?f=t%20%3D%20%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%20%7D%5C%5C%5C%5Ct%20%3D%20%20%5Csqrt%7B%5Cfrac%7B2%2A16.4%7D%7B9.8%7D%20%7D%5C%5C%5C%5Ct%20%3D%201.83%20%5C%20s)
The horizontal distance traveled at this time is given by;
x = vt
x = (9.3)(1.83)
x = 17.02 m
Therefore, the horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m
The coefficient is needed in front of NaNO3 to balance the equation is 2.
<h3>What is balanced equation?</h3>
When in a chemical reaction, both the product elements or compounds should have number of moles equal to that of the elements or compounds of reactants.
In the given chemical reaction equation,
Na2S + Zn(NO3)2 → ZnS + _NaNO3
Number of moles of S is 1 on both the product and reactant side. Zn has 1 mole on both side. NO3 has 2 mole and Na has 2 mole on reactant side. So to balance the equation, Na and NO3 both must have 2 moles on product side.
Thus, the coefficient is needed in front of NaNO3 to balance the equation is 2.
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Answer:
A) car - 37500 kg*m/s, minivan - 13332 kg*m/s
B) 50832 kg*m/s
C) 18.83 m/s
Explanation:
Realize that sticky collisions are modeled by: m1v1+m2v2=(m1+m2) vf
conevert to m/s....car going 25 m/s, minivan going 11.11 m/s
A) p=mv
p(car)=(1500)(25)
p(car)=37500 kg*m/s
p(minivan)=(1200)(11.11)
p(minivan)=13332 kg*m/s
B) 37500+13332=50832 kg*m/s
C) 37500+13332=(1500+1200) vf
50832=2700(vf)
18.83 m/s = vf
An good example would be using a Bunsen burner to heat water in a tin container. The flame initially heats the tin can by radiating heat. Conduction then transfers heat from the tin can to the water. The convection process causes the hot water to then climb to the top.
<h3>What is Bunsen Burner?</h3>
A laboratory piece of equipment known as a Bunsen burner, which bears Robert Bunsen's name, produces a single open gas flame and is used for heating, sterilization, and combustion. Natural gas, liquefied petroleum gas, such as propane, butane, or a combination, are all acceptable choices for the gas.
<h3>What three types of flames can you get from a Bunsen burner?</h3><h3>Bunsen burner flames come in three primary categories:</h3>
- Because it is simple to notice in a well-lit space, a yellow flame is also referred to as a safety flame.
- Flaming Blue. A burner's specific flame can reach temperatures of 500 degrees.
- Blue Flame in Flames The roaring blue flame setting on a Bunsen burner produces the hottest flames.
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Potential, if the bow is released kinetic