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jek_recluse [69]
3 years ago
5

Use at least two to three completed content related sentences to explain why physics is considered the basic science. Provide at

least one example to explain how physics relates to other sciences
Physics
2 answers:
Virty [35]3 years ago
6 0
Idk your answer sorry
Lyrx [107]3 years ago
4 0
It is considered the basic science because other sciences depend on it to develop .For example , studying the space and the motion of planets need physics and we can't go sometimes to the space and do our experiment , also it uses in X-ray which consists of electromagnetic waves .
physics is needed in more sciences like chemisrty , geology ect..
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The answer can't be D because I got it wrong when I took my test so it has to be A,B,or C
Cerrena [4.2K]
To what question please put the question ill be glad to help
5 0
3 years ago
Over a span of 6.0 seconds, a car changes it's speed from 89 km/h to 37 km/h. What is its average acceleration in meters per sec
scoundrel [369]

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) - (speed at the beginning)

change in speed = (37 km/hr) - (89 km/hr) = -52 km/hr

Acceleration = (-52 km/hr) / (6 sec)

Acceleration = (-26/3) km/(hr·sec)

Units: (1/hr·sec) · (hr/3600 sec) = 1 / 3600 sec²

(-26/3) km/(hr·sec) = (-26/3) km/(3600 sec²)

= -26,000/(3 · 3600) m/s²

<em>Acceleration = -2.41 m/s²</em>

3 0
3 years ago
Describe the initial horizontal and vertical velocity of a horizontally launched projectile on Earth, as well as what happens to
Jobisdone [24]

Answer:

Explained below

Explanation:

To explain this, let's consider a tennis ball being launched from the top of a very high building.

Now, if the tennis ball is launched horizontally without any upward angle but with an initial velocity of 10 m/s. In this motion, If there is no gravity, the tennis ball would continue in motion at that same speed of 10 m/s in the horizontal direction. However, in reality, gravity causes the tennis ball to accelerate downwards at a rate of 9.8 m/s for every second. This implies that the vertical velocity component is changing at the rate of 9.8 m/s every second.

Thus, after 1 second, horizontal velocity component will remain 10 m/s and vertical component will be 9.8 m/s × 1 = 9.8 m/s downwards.

Also, after 2 seconds, the vertical velocity component will remain 10 m/s, however the vertical component will now be 9.8 × 2 = 19.6 m/s downwards.

Same procedure is repeated as t increases by 1 second.

5 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
mrs_skeptik [129]

Answer:

1. Largest force: C;  smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

F=K\dfrac{ q_{1}q_{2}}{r^{2}}

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write  

F=\dfrac{k}{r^{2}}

1. Net force on each particle

Let's

  • Call the distance between adjacent charges d.
  • Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}}  - \dfrac{k}{(2d)^{2}}  +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(b) Force on B

\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}}  - \dfrac{k}{d^{2}}  + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(C) Force on C

\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}}  + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(d) Force on D

\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}}  - \dfrac{k}{(2d)^{2}}  - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

F_{A} : F_{B} : F_{C} : F_{D}  =  \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\

2. Ratio of largest force to smallest

\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}

7 0
3 years ago
Longer wavelengths of light, such as _______, have ________ energy than shorter wavelengths, such as _________
Ilya [14]

Answer:

Micro and radio waves.

Lower energy.

Gamma rays.

Explanation:

The electromagnetic spectrum is the range of frequencies of electromagnetic radiation and their respective wavelengths.

Ionising radiation os defined as the energy required of photons of a wave to ionize atoms, causing chemical reactions.

The energy of the wave depends on both the amplitude and the frequency. If the energy of each wavelength is a discrete packet of energy, a high-frequency wave will deliver more of these packets per unit time than a low-frequency wave. In summary, the longer the wavelength, the lower the energy to ionise.

The velocity of a wave is directly proportional to the frequency of that wave.

c = f * lambda

Where,

c = velocity of the wave

f = frequency of the wave = 1/time

Lambda = wavelength.

From the above expression, the longer the wavelength, lambda the shorter the frequency.

Examples of waves with longer wavelengths are, micro and radio waves, while radiations with shorter wavelengths like gamma rays.

8 0
3 years ago
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