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timofeeve [1]
3 years ago
5

30 POINTS!!! I need help ASAP, if someone comments just for points, or is incorrect just for points I will report! Please help!!

!
1.) A cue ball rolls with a speed of 1.8 m/s at an angle of 24 degrees relative to the edge of the billiards table. How large is the component of the velocity parallel to the edge of the table?
A) 0.73 m/s
B) 1.6 m/s
C) 1.8 m/s
D) 13 m/s
2.) A snowboarder travels 150m down a mountain slope that is 65 degrees above horizontal. What is his vertical displacement?
A) 32.5m
B) 58.9m
C) 63.4m
D) 135.9m
3.) A tennis player hits a ball at an angle of 50 degrees above horizontal so that it has an acceleration of 12 m/s². What is the horizontal component of the acceleration?
A) 7.7 m/s²
B) 9.2 m/s²
C) 10.4 m/s²
D) 48.9 m/s²
​

Physics
1 answer:
Anika [276]3 years ago
7 0

Answer:

1.) answer B

2.) answer D

3.) answer A

Explanation:

In all of these problems, it is essential to draw pictures in order to understand which trigonometric function to use according to the angle that the vector in question forms with the component requested. For all of them try to picture a right angle triangle with the vector as the hypotenuse, and the components as the triangle's shorter sides. Please refer to the three pictures attached as image for this answer a,d notice that the vector quantity known for all cases is represented in red, and the component to find is represented in green.

Problem 1) : the vector velocity makes an angle of 24 degrees with the edge of the table. So picture that vector as the hypotenuse of a right angle triangle for which you know the value: 1.8 m/s

So in this case, where you know the angle, the hypotenuse, and need to find the adjacent side to the angle, you use the cosine function as follows:

requested component =1.8 * cos (24) =1.644 \frac{m}{s}

which we round to 1.6 to match answer C).

For problem 2.) wee need to find the component opposite to the given angle in the triangle for which we also know the hypotenuse. So we use the sine function as follows:

requested component =150 * sin(65) = 135.94 m

which we round to 135.9 m to match answer D).

For problem 3.) we need to find the horizontal component to the acceleration which corresponds to the adjacent side to the known angle, so we use the cosine function as follows:

requested component =12 *cos(50) = 7.71 \frac{m}{s^2}

which we round tp 7.7 to match answer A).

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<h3>Calculation:</h3>

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Read 2 more answers
At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
bogdanovich [222]

Answer:

magnitude of force on charge 2Q  = \frac{KQ^{2} }{I^{2} }

Direction of force on charge = 61 ⁰

Explanation:

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║F║ = \sqrt{f_{x}^{2} + f_{y}^{2}    }  =  after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW

magnitude of force acting on 2Q = \frac{KQ^{2} }{I^{2} }

The direction of the force on charge 2Q is calculated as

tan ∅ = \frac{f_{y} }{f_{x} } = 1.8284

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