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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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The final pressure of the gas is 9.94 atm.

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In a second experiment, you decide to connect a string which has length L from a pivot to the side of block A (which has width d
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Answer:

The answer is in the explanation

Explanation:

A)

i) The blocks will come to rest when all their initial kinetic energy is dissipated by the friction force acting on them. Since block A has higher initial kinetic energy, on account of having larger mass, therefore one can argue that block A will go farther befoe coming to rest.

ii) The force on friction acting on the blocks is proportional to their mass, since mass of block B is less than block A, the force of friction acting on block B is also less. Hence, one might argue that block B will go farther along the table before coming to rest.

B) The equation of motion for block A is

m_{A}\frac{\mathrm{d} v}{\mathrm{d} t} = -m_{A}g\nu_{s}\Rightarrow \frac{\mathrm{d} v}{\mathrm{d} t} = -\nu_{s}g \quad (1)

Here, \nu_{s} is the coefficient of friction between the block and the surface of the table. Equation (1) can be easily integrated to get

v(t) = C-\nu_{s}gt \quad (2)

Here, C is the constant of integration, which can be determined by using the initial condition

v(t=0) = v_{0}\Rightarrow C = v_{0} \quad (3)

Hence

v(t) = v_{0} - \nu_{s}gt \quad (4)

Block A will stop when its velocity will become zero,i.e

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C) We can see that the expression for the distance travelled for block A is independent of its mass, therefore if we do the calculation for block B we will get the same result. Hence the reasoning for Student A and Student B are both correct, the effect of having larger initial energy due to larger mass is cancelled out by the effect of larger frictional force due to larger mass.

D)

i) The block A is moving in a circle of radius L+\frac{d}{2} , centered at the pivot, this is the distance of pivot from the center of mass of the block (assuming the block has uniform mass density). Because of circular motion there must be a centripetal force acting on the block in the radial direction, that must be provided by the tension in the string. Hence

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The speed of the block decreases with time due to friction, hence the speed of the block is maximum at the beginning of the motion, therfore the maximum tension is

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ii) The forces acting on the block are

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b) Friction: Acting tangentially, in the direction opposite to the velocity of the block at any given time, therefore it decreases the speed of the block.

The speed decreases linearly with time in the same manner as derived in part (C), using the expression for tension in part (D)(i) we can see that the tension in the string also decreases with time (in a quadratic manner to be specific).

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