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GenaCL600
18 days ago
7

In general, how do you find the average velocity of any object falling in a vacuum?

Physics
1 answer:
Softa [2K]18 days ago
5 0
To determine the average velocity of an object descending in a vacuum, assuming you know its final speed, multiply the final result by the overall time. 3. The equation d = v • t expresses how distance, average velocity, and time relate to each other.
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An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte
Ostrovityanka [2208]

Answer:

35.79 meters

Explanation:

We have an archer, and there is a target. Denote the distance between them as d.

The bowman releases the arrow, which travels the distance d at a velocity of 40 m/s until it hits the target. We establish the equation as:

v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d

Right after this, the arrow produces a muffled noise, traveling the same distance d at a speed of 340 m/s in time t_{sound}. Thus, we can derive:

v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d.

Consequently, the sound reaches the archer, precisely 1 second post-firing the bow, resulting in:

t_{arrow} + t _{sound} = 1 s.

Using this relationship in the distance formula for sound allows us to write:

340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d.

Substituting the value of d from the first equation yields:

40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow}).

Now, after some calculations, we can proceed further:

40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s.

Finally, the value is inserted into the initial equation:

40 \frac{m}{s} * t_{arrow} = d

40 \frac{m}{s} * 340/380 s = 35,79 s = d

6 0
1 day ago
A 2.0-kg projectile moves from its initial position to a point that is displaced 20 m horizontally and 15 m above its initial po
Yuliya22 [2438]

Answer:

W = 294 J

Explanation:

provided,

mass of the projectile = 2 Kg

horizontal displacement = 20 m

vertical displacement = 15 m

work performed by the gravitational force =?

the work done by gravitational force only accounts for vertical motion.

force due to gravity =  m g

= 2 x 9.8 = 19.6 N

work is equal to force x displacement

W = F x s

W = 19.6 x 15

W = 294 J

7 0
12 days ago
Read 2 more answers
Below you are given data about a wave in three different substances.
inna [2210]

1) The wave's period remains constant across different media

2) The wave's velocity varies depending on the medium it travels through

3) As a wave transitions between media, its speed, direction, and wavelength can change, while its frequency stays unchanged

Clarification:

1)

The period of a wave signifies the duration it takes for one full oscillation.

The wave's period is the inverse of its frequency:

T=\frac{1}{f}

where

T denotes the period

f is the frequency

The provided table illustrates that the frequency remains consistent across the three media; hence, the period is unchanged as it solely relies on frequency. We can compute it as we know that

f = 350 Hz

thus the period equals

T=\frac{1}{350}=2.86\cdot 10^{-3} s = 2.86 ms

2)

The velocity of a wave can be derived from the wave equation:

v=f \lambda

where

f indicates the frequency

\lambda is the wavelength

<pin the="" first="" medium="">

f=350 Hz, \lambda = 0.75 m, resulting in a speed of

v_1 = (350)(0.75)=262.5 m/s

In the second medium,

f=350 Hz, \lambda = 0.70 m, leading to a speed of

v_2 = (350)(0.70)=245 m/s

In the third medium,

f=350 Hz, \lambda = 0.65 m, showing a speed of

v_3 = (350)(0.65)=227.5 m/s

As a result, we conclude that the wave's speed varies with the medium.

3)

<pwhen a="" wave="" shifts="" from="" one="" medium="" to="" another="" the="" following="" occurs:="">

- The wave's direction alters. Specifically, if the subsequent medium is of greater optical density, the wave bends towards the normal; conversely, it bends away if the second medium is of lesser optical density.

- The wave's speed is affected. The wave decelerates in media with higher optical density and accelerates in those with lower optical density.

- The wave's frequency remains unchanged.

- Ultimately, the wave's wavelength is modified. If moving into a medium of greater optical density, the wavelength decreases, while it increases in one of lower optical density.

Discover more about waves here:

</pwhen></pin>
7 0
16 days ago
4. We have 4 identical strain gauges of the same initial resistance (R) and the same gauge factor (GF). They will be used as R1,
Softa [2035]
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3 0
14 days ago
The structural diversity of carbon-based molecules is determined by which properties?
Yuliya22 [2438]

Explanation:

The diverse structures of carbon-based compounds are influenced by several factors:

1. the capacity of bonds to rotate freely,

2. the ability of carbon to create four covalent bonds,

3. the spatial arrangement of bonds resembling a tetrahedron.

5 0
22 days ago
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