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STALIN [3.7K]
3 years ago
12

A cannon sitting on level ground is aimed at 45.0 degrees relative to the horizontal. It fires a test shot at a target located 1

00.0 meters away from the cannon on the same level ground. The test overshoots the target by 20.0 meters. Which of the following angles can the cannon be adjusted to to hit the target. You may neglect air resistance and assume the cannon always delivers the same initial velocity to the cannonball .
A. 35.9 deg
B. 49.1 deg
C. 28.2 deg
D. 52.8 deg
E. 22.7 deg
Physics
1 answer:
Mrac [35]3 years ago
7 0

Answer:

C. 28.2 deg

Explanation:

The horizontal range of a projectile is given as:

R = \frac{v^2Sin2\theta}{g}

where,

R = Range

v = speed

θ = angle of launch

g = acceleration due to gravity = 9.81 m/s²

First, we will find the launch speed (v) by using the initial conditions:

R = 120 m

θ = 45°

Therefore,

120\ m = \frac{v^2Sin 90^o}{9.81\ m/s^2}\\\\v = \sqrt{(120\ m)(9.81\ m/s^2)}\\\\v = 34.31\ m/s

Now, consider the second scenario to hit the target:

R = 100 m

Therefore,

100\ m = \frac{(34.31\ m/s)^2Sin2\theta}{9.81\ m/s^2}\\\\Sin2\theta = \frac{(100\ m)(9.81\ m/s^2)}{(34.31\ m/s)^2}\\\\2\theta = Sin^{-1}(0.833)\\\\\theta = \frac{56.44^o}{2}\\\theta = 28.22^o

Hence, the correct option is:

<u>C. 28.2 deg</u>

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A flask contains equal masses of f2 and cl2 with a total pressure of 2.00 atm at 298k. What is the partial pressure of cl2 in th
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Answer:

Partial Pressure of F₂ = 1.30 atm

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Explanation:

Partial pressure for gases are given by Daltons law.

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Pt = P(f₂) + P(cl₂)

Partial pressure = mole fraction × total pressure

Let the mass of each gas present be m

Number of moles of F₂ = m/38 (molar mass of fluorine = 38 g/Lol

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Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.How
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Answer:

The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.

Explanation:

Given;

speed of the faster car, v₁ = 60 mi/h

speed of the slower car, v₂ = 55 mi/h

Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles

\frac{x}{55} - \frac{x}{60}   = \frac{15}{60}

Note: divide 15 mins by 60 to convert to hours for consistency in the units.

\frac{x}{55} - \frac{x}{60}   = \frac{15}{60}\\\\multiple \ through \ by \ 660\\\\12x - 11x = 165\\\\x = 165 \ miles

Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.

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