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4 years ago

9

What's the energy transfer for a catapult

1 answer:

just olya [345]4 years ago

3 0

Kinetic energy is the energy for a catapult.

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A water trough is 12 feet long, and its cross section is an equilateral triangle with sides 2 feet long. Water is pumped into th

MAXImum [283]

Answer:

The water level rises at 11.76 $ft^{2} /s$

$h=\sqrt{3/2} s$ let be h: height and s: side of an equilateral triangle

Explanation:

The picture shows a diagram of the situation, first we have to determine the height of the trough as follows:

With the Pytagorean Theorem we can find out that:

$h^{2} =s^{2} -s^{2} /4 =\sqrt{3} /2 s$

Then, the area of an equilateral triangle, as any triangle, is a half of its base times its height:

$A=\frac{1}{2} hs=\frac{1}{2} \frac{\sqrt{3} }{2} s^{2} =\frac{\sqrt{3} }{4} s^{2}$

Replacing values, we have:

$A=1.73ft^{2}$

That is the total area of the trough, but the problem specifies that it has been filled until 0.5 ft. Therefore, we have to find the cross section area of the water flow by substracting the total area minus the unfilled area of the trough:

$h_{u}: height of the unfilled triangle\\s_{u} : side of the unfilled triangle\\A_{u}: area of the unfilled triangle\\h_{u}=\sqrt{3} -0.5 =1.23\\s_{u}= \frac{2}{\sqrt{3} } 1.23=1.42\\A_{u}=\frac{\sqrt{3} }{4} 1.42^{2} =0.87ft^{2} A_{u}$

Then, the cross section area of water flow is

$A_{s} =1.73ft^{2} -0.87ft^{2} =0.85ft^{2}$

Finally, to determine the speed of water flow at this point we solve for v, the flow formula:

$Q: water flow\\v: speed\\Q=vA\\v=Q/A =\frac{10ft^{3}/s }{0.85ft^{2} } =11.76 \frac{ft^{2}}{s}$

6 0

3 years ago

What is the force of gravity between a 500kg bal and a 100,000kg basket 5m apart?

klemol [59]

Answer:

10 Mil. J

Explanation:

multiplyed balls and divided by 5

8 0

4 years ago

Read 2 more answers

When using a crowbar to remove a nail, the person should hold it at which of the following spots to increase the IMA of the leve

Ilia_Sergeevich [38]

Answer: the furthest from the end prying out the nail

5 0

4 years ago

A typical maximum speed for a unicycle is about 5 mph. While riding his giraffe unicycle, Mike accelerates from rest to the typi

DIA [1.3K]

Answer:

$s = 9.167\,ft$

Explanation:

Converting units into fundamental units is the first step. The maximum speed is:

$v_{f} = (5\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )$

$v_{f} = 7.333\,\frac{ft}{s}$

Let assume that Mike accelerates at a constant rate. Hence, accelaration is obtained from this formula:

$a = \frac{v_{f}}{t}$

$a = \frac{7.333\,\frac{ft}{s} }{2.5\,s}$

$a = 2.933\,\frac{m}{s^{2}}$

Then, distance travelled by the unicycle can be found easily:

$s = \frac{(7.333\,\frac{m}{s} )^{2}}{2\cdot (2.933\,\frac{m}{s^{2}} )}$

$s = 9.167\,ft$

4 0

3 years ago

During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the trebuc

Snezhnost [94]

Answer:The above given quantities can be calculated in following steps;

Explanation:

7 0

4 years ago