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ladessa [460]
3 years ago
9

A person wants to fire a water balloon cannon such that it hits a target 100m100m away. if the cannon can only be launched at 45

â45â above the horizontal, what should be the initial speed at which it is launched?
Physics
1 answer:
vladimir2022 [97]3 years ago
3 0
<span>31.3 m/s Since the water balloon is being launched at a 45 degree angle, the horizontal and vertical speeds will be identical. Also the time the balloon takes to reach its peak altitude will match the time it takes to fall. So let's create a few expressions about what we know. Distance the water balloon travels at velocity v for time t d = vt Total time required for the entire trip is double since the balloon goes up, then goes down t = 2v/a Now let's plug in the numbers we have, assuming the acceleration due to gravity is 9.8 m/s^2 t = 2v/9.8 100 = vt Substitute 2v/9.8 for t in the 2nd formula 100 = v(2v/9.8) Solve for v. 100 = v(2v/9.8) 100 = 2v^2/9.8 980. = 2v^2 490 = v^2 22.13594 = v So we now know that both the horizontal velocity and vertical velocity needed is 22.13594 m/s. Let's verify that 2*22.13594 / 9.8 = 4.51754 So it will take 4.51754 second for the balloon to hit the ground after being launched. 4.51754 * 22.13594 = 100 And during that time it will travel 100 meters horizontally. But we need to know the total velocity. And the Pythagorean theorem comes to the rescue. Just square the 2 velocities, add them together, and take the square root. We already know the square is 490 from the work above, so sqrt(490+490) = sqrt(980) = 31.30495 m/s</span>
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Answer : The energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

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Putting the values, in above equation, we get:

\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )

\lambda=6.56\times 10^{-7}m

Now we have to calculate the energy.

E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 6.56\times 10^{-7}m

Putting the values, in this formula, we get:

E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}

E=3.03\times 10^{-19}J

Therefore, the energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

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Answer:

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see further explanation

Explanation:for clarification

Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination. Also, each spring must exert the same force. Do you see why?

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