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Aleksandr-060686 [28]
3 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.
Physics
1 answer:
vodka [1.7K]3 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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Estimate the peak wavelength for radiation from ice at 273 k.
Andrews [41]
<h2>Answer: 10615 nm</h2>

Explanation:

This problem can be solved by the Wien's displacement law, which relates the wavelength  \lambda_{p} where the intensity of the radiation is maximum (also called peak wavelength) with the temperature T of the black body.

In other words:

<em>There is an inverse relationship between the wavelength at which the emission peak of a blackbody occurs and its temperature.</em>

Being this expresed as:

\lambda_{p}.T=C    (1)

Where:

T is in Kelvin (K)

\lambda_{p} is the <u>wavelength of the emission peak</u> in meters (m).

C is the <u>Wien constant</u>, whose value is 2.898(10)^{-3}m.K

From this we can deduce that the higher the black body temperature, the shorter the maximum wavelength of emission will be.

Now, let's apply equation (1), finding \lambda_{p}:

\lambda_{p}=\frac{C}{T}   (2)

\lambda_{p}=\frac{2.898(10)^{-3}m.K}{273K}  

Finally:

\lambda_{p}=10615(10)^{-9}m=10615nm  This is the peak wavelength for radiation from ice at 273 K, and corresponds to the<u> infrared.</u>

8 0
3 years ago
Which situation is the best analogy for the doppler effect?
Rudiy27
The best scenario to describe the doppler effect would be listening to the siren of a passing ambulance or fire truck

then it is coming towards you, the pitch is higher, it gets higher as it approaches and peaks as it gets right in front of you. then it drop at once when it passes you and continues to drop till it fades away. this is a classic descrption of the doppler effect
8 0
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Two students designed an experiment to study the effect of solar radiations on four cities of Earth. They used a globe to repres
Furkat [3]
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I hope the students remembered to tilt the axis of the globe.  If they didn't,
and instead kept it straight up and down, then each city had pretty much
the same amount of bulb-light all the way around, and there were no seasons.

If the axis of the globe was tilted, then City-D had the least variation in
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3 0
3 years ago
The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m
Sonbull [250]

Answer:

density is 10^{6} Mg/µL

Explanation:

given data

density of nuclear = 10^{18} kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

so

1 m³ is equal to 10^{6} cm³

and here 1 cm³ is equal to  1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = 10^{18} kg/m³

density = 10^{18} kg/m³ × \frac{10^{-3} }{10^{6} }  Mg/µL

density =  10^{6} Mg/µL

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