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Aleksandr-060686 [28]

4 years ago

6

Three cannons are located at the top of a cliff above a level plain. Cannon A is aimed at an angle of 25° above the horizontal a

nd shoots cannonballs with a speed of 85 m/s. Cannon B is aimed directly horizontally and shoots with a speed of 100 m/s. Cannon C is aimed 25° below the horizontal and shoots with a speed of 75 m/s. Which of the cannonballs will hit the ground with the greatest speed? The three cannonballs are not necessarily the same mass.

1 answer:

vodka [1.7K]4 years ago

3 0

Answer:

The canon B hits the ground fast.

Explanation:

Given that,

Speed of cannon A = 85 m/s

Speed of cannon B= 100 m/s

Speed of cannon C = 75 m/s

We need to calculate the cannonballs will hit the ground with the greatest speed

Using conservation of energy

The final kinetic energy of canon depends on initial kinetic energy and potential energy.

The final velocity depends upon initial velocity and initial height.

So, the initial velocity of canon B is high.

Hence, The canon B hits the ground fast.

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1. How do energy and matter recycle in a closed-loop system?

Makovka662 [10]

Answer:

In a closed-loop system, matter is used to generate energy, and the energy generated is used to produce matter, and the cycle goes on without end.

Explanation:

A closed-loop system is one in which materials or energy is recycled without end through a production cycle. This means that a raw material is used to produce a finished product, and the finished product at the end of its use cycle is converted back and used as a raw material to produce more of it again. Energy and matter can also be cycled in the same way in an energy and matter closed-loop system, converting matter to energy, and the energy is put back into the production of more of the matter.

7 0

3 years ago

A whale swims due east for a distance of 6.9 km, turns around and goes due west for 1.8 km, and finally turns around again and h

Amiraneli [1.4K]

Answer:

Explanation:

a) Total distance traveled by the whale = 6.9 km + 1.8 km + 3.7 km = 12.4 km

b) the magnitude and the direction of the displacement of the whale = 6.9 km due east - 1.8 km due west = 5.1 due east

it final displacement = 5.1 km due east + 3.7 k due east = 8.8 km due east

c) average speed in km / hr = total distance traveled / total time = 12.4 km / 0.5 hr ( 30 min/ 60 min × 1 hr) = 24.8 km / hr

d) average velocity in km / h = total displacement / total time = 8.8 km / 0.5 h = 17.6 km/h

6 0

3 years ago

Two capacitors, C1 = 25.0 μF and C2 = 31.0 μF, are connected in series, and a 6.0-V battery is connected across them.a. Find the

Brrunno [24]

Answer:

$13.83928\times 10^{-6}\ F$

$249.10704\times 10^{-6}\ J$

$137.89848\times 10^{-6}\ J$

$111.20842\times 10^{-6}\ J$

2.98273 V

Explanation:

$C_1=25\ mu F$

$C_2=31\ mu F$

V = Voltage = 6 V

Equivalent capacitance is given by

$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}\\\Rightarrow C=\dfrac{C_1C_2}{C_1+C_2}\\\Rightarrow C=\dfrac{25\times 10^{-6}\times 31\times 10^{-6}}{(25+31)\times 10^{-6}}\\\Rightarrow C=13.83928\times 10^{-6}\ F$

Equivalent capacitance is $13.83928\times 10^{-6}\ F$

Energy stored is given by

$E=\dfrac{1}{2}CV^2\\\Rightarrow E=\dfrac{1}{2}\times 13.83928\times 10^{-6}\times 6^2\\\Rightarrow E=249.10704\times 10^{-6}\ J$

Total energy stored is $249.10704\times 10^{-6}\ J$

Charge is given by

$Q=CV\\\Rightarrow Q=13.83928\times 10^{-6}\times 6\\\Rightarrow Q=83.03568\times 10^{-6}\ C$

Voltage is given by

$V_1=\dfrac{Q}{C_1}\\\Rightarrow V_1=\dfrac{83.03568\times 10^{-6}}{25\times 10^{-6}}\\\Rightarrow V_1=3.3214272\ V$

$E_1=\dfrac{1}{2}C_1V_1^2\\\Rightarrow E_1=\dfrac{1}{2}\times 25\times 10^{-6}\times 3.3214272^2\\\Rightarrow E_1=137.89848\times 10^{-6}\ J$

Energy strored in C1 is $137.89848\times 10^{-6}\ J$

$V_2=\dfrac{Q}{C_2}\\\Rightarrow V_2=\dfrac{83.03568\times 10^{-6}}{31\times 10^{-6}}\\\Rightarrow V_2=2.67857\ V$

$E_2=\dfrac{1}{2}C_2V_2^2\\\Rightarrow E_2=\dfrac{1}{2}\times 31\times 10^{-6}\times2.67857^2\\\Rightarrow E_2=111.20842\times 10^{-6}\ J$

Energy stored in C2 is $111.20842\times 10^{-6}\ J$

$E=E_1+E_2\\\Rightarrow E=137.89848\times 10^{-6}+111.20842\times 10^{-6}\\\Rightarrow E=249.107\times 10^{-6}\ J$

So, the energy is equivalent

Equivalent capacitance

$C=C_1+C_2\\\Rightarrow C=25+31\\\Rightarrow C=56\times 10^{-6}\ F$

$E=\dfrac{1}{2}CV^2\\\Rightarrow V=\sqrt{\dfrac{2E}{C}}\\\Rightarrow V=\sqrt{\dfrac{2\times 249.107\times 10^{-6}}{56\times 10^{-6}}}\\\Rightarrow V=2.98273\ V$

The voltage would be 2.98273 V

8 0

4 years ago

creativ13 [48]

The answer is B- the process of cells making new cells

8 0

4 years ago

A horizontal pipe narrows from a radius of 0.250 m to 0.1000 m. If the speed of the water in the pipe is 1.00 m/s in the larger-

umka21 [38]

Answer:

6.25 m/s

196 kg/s

Explanation:

The areas of the pipe from big to small:

$A = \pi R^2 = \pi*0.25^2 = 0.196 m^2$

$a = \pi r^2 = \pi*0.1^2=0.0314m^2$

As the product of speed and cross-section area is constant, the speed in the smaller pipe would be

$AV = av$

$v = \frac{AV}{a} = \frac{0.196 * 1}{0.0314} = 6.25 m/s$

The mass flow rate would be:

$\dot{m} = \ro AV = 1000 * 0.196 * 1 = 196 kg/s$

4 0

3 years ago