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3 years ago

A canon is tilled back 30.0 degrees and shoots a cannon ball at 155 m/s. What is the

Physics

1 answer:

Umnica [9.8K]3 years ago

6 0

Answer:

$= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m$

Therefore, highest point that the cannon ball reaches is 168.7m

Explanation:

the cannon is fired at an angle 30 o to the horizonatal with a speed of 155 m/s

highest point that the cannon ball reaches?

$H_{max}=\frac{V^2\sin ^2 \theta}{2g}$

g = 9.8m/s2

$= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m$

Therefore, highest point that the cannon ball reaches is 168.7m

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You observe a distant galaxy. You find that a spectral line, resulting from an electron transition in hydrogen, is shifted from

alexandr1967 [171]

Answer:

The galaxy is moving away from the observer

Explanation: when a galaxy is moving away from us, the light we percieve from it is "streched". Since the wavelength has an inverse raltionship whith frequency, the longer the wavelength is, the lower the frequency. And lower frequencies correspond to red and infrarred light.

So when we see the light has shifted to the infrarred part of the spectrum, it means the source is traveling away from us, making the light waves we percieve streched and move from visible light to infrarred.

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3 years ago

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

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3 years ago

If the voltage output of a digital manifold absolute pressure (MAP)sensor is 2 volts, the approximate engine vacuum is _______ i

LenaWriter [7]

The correct answer to this question is 15

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3 years ago

What would be the final temperature if you mixed a liter of 40C water with 2 liters of 20C water?

riadik2000 [5.3K]

Answer:

I found

33 ∘ C

Explanation:

i hope this helps :)

5 0

3 years ago

Imagine that you have a 6.50 l gas tank and a 4.50 l gas tank. you need to fill one tank with oxygen and the other with acetylen

Margarita [4]

V₁(O2) = 6.50<span> L
</span>p₁(O2) = 155 atm
V₂(acetylene) = <span>4.50 L
</span>p₂(acetylene) =?

According to Boyle–Mariotte law (At constant temperature and unchanged amount of gas, the product of pressure and volume is constant) we can compare two gases that have ideal behavior and the law can be usefully expressed as:

V₁/p₁ = V₂/p₂

6.5/155 = 4.5/p₂

0.042 x p₂ = 4.5

p₂ = 107.3 atm

7 0

3 years ago