Jack pushed and pushed, and finally moved the new refrigerator into the kitchen. What is the BEST explanation of what happened?

The sound is perceived as louder if the amplitude increases, and softer if the amplitude decreases. ( for amplitude)

Since sound at all frequencies has the same speed in air, a change in frequency means a change in wavelength. ( for frequency)

7 0

2 years ago

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Two vehicles approach a right angle intersection and then suddenly collide. After the collision, they become entangled. If their

geniusboy [140]

Answer:

a. 11 m/s at 76° with respect to the original direction of the lighter car.

Explanation:

In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:

$m_{1} v_{x1}+m_{2}v_{x2}=(m_{1}+m_{2})v_{fx}\\m_{1} v_{y1}+m_{2}v_{y2} =(m_{1}+m_{2})v_{fy}$

Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.

$1*13 + 4*0 = (1+4) v_{fx}\\v_{fx}=\frac{13}{5} =2.6\\1*0 + 4*13 = (1+4) v_{fy}\\v_{fy}=\frac{13*4}{5} =10.4\\$

The magnitude of the final velocity of the wreck can be found as:

$v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72$

The final velocity has an intensity of roughly 11 m/s

As for the angle, it can be determined in respect to the lighter car (x axis) as follows:

$\theta = cos^{-1}(\frac{v_{fx} }{v_{f}} )\\\theta = cos^{-1}(\frac{2.6}{10.7} )\\\theta = 76^{o}$

Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.

5 0

3 years ago

A bottle dropped from a balloon reaches the ground in 20 s. determine the height of the balloon if (a) it was at rest in the air

romanna [79]

<span>a) 1960 m b) 960 m Assumptions. 1. Ignore air resistance. 2. Gravity is 9.80 m/s^2 For the situation where the balloon was stationary, the equation for the distance the bottle fell is d = 1/2 AT^2 d = 1/2 9.80 m/s^2 (20s)^2 d = 4.9 m/s^2 * 400 s^2 d = 4.9 * 400 m d = 1960 m For situation b, the equation is quite similar except we need to account for the initial velocity of the bottle. We can either assume that the acceleration for gravity is negative, or that the initial velocity is negative. We just need to make certain that the two effects (falling due to acceleration from gravity) and (climbing due to initial acceleration) counteract each other. So the formula becomes d = 1/2 9.80 m/s^2 (20s)^2 - 50 m/s * T d = 1/2 9.80 m/s^2 (20s)^2 - 50m/s *20s d = 4.9 m/s^2 * 400 s^2 - 1000 m d = 4.9 * 400 m - 1000 m d = 1960 m - 1000 m d = 960 m</span>

6 0

3 years ago