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3 years ago

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (37pts)

Physics

2 answers:

sweet [91]3 years ago

8 0

Answer:

the answer is b.

=》distance = 10 miles

=》displacement = 0 miles

Explanation:

=》total distance = 2 + 3 + 5 = 10 miles

( sum of total miles covered )

=》total displacement = shortest distance from starting point to the destination. but here both starting point and destination are same so, total displacement = 0 miles

Artemon [7]3 years ago

6 0

Answer:

Explanation:

Distance is 10 because 2+3+5=10

Displacement =0.

Bf?

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a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$

$\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}$

Using the formulas for $V_{ox}, V_{oy}:$

$\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}$

Simplifying

$4cos\theta sin\theta=3sin^2\theta$

Dividing by $sin\theta$

$4cos\theta=3sin\theta$

Rearranging

$tan\theta=\frac{4}{3}$

$\theta=arctan\frac{4}{3}$

$\theta=53.13^o$

4 0

4 years ago

A man travles at an average velocity of 12 m/s for 4 seconds. what is the total distance travled by the man?

gizmo_the_mogwai [7]

48 meters.
12 m/s and 4 seconds, so 4*12=48.

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4 years ago

According to Newton's 3rd law of motion, an action creates _____.

miv72 [106K]

C.an equal and opposite reaction

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3 years ago

Read 2 more answers

3. Calculate the wavelength of wave that has a frequency of 4.75 x 1012Hz.

Klio2033 [76]

Frequency=v=4.75×10^12Hz
Wavelength=?

We know

$\boxed{\sf \lambda=\dfrac{C}{V}}$

$\\ \sf\longmapsto \lambda=\dfrac{3\times 10^8ms^{-1}}{4.75\times 10^{12}s^{-1}}$

$\\ \sf\longmapsto \lambda=0.631\times 10^{-4}m$

$\\ \sf\longmapsto \lambda=6.31\times 10^{-5}m$

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3 years ago

Two satellites are orbiting earth at different altitudes. Which satellite orbits at a higher speed v around earth? Assume that t

alexandr1967 [171]

Answer:

a) the one with a lower orbit b) the one with a higher orbit

Explanation:

Let's consider orbital mechanics. To get an object in orbit, we need it to fall to earth parallel to the earth's surface. To understand it easily imagine a projectile thrown horizontally further and further away, at one point, the projectile hits the cannon from behind. Considering there is no wind resistance, that would be a projecile in orbit.

In other words, the circular orbits of some objects around a massive body are due to the equality between centrifugal acceleration and gravity acceleration.

$\frac{v^2}{r} = \frac{GM}{r^2}$ .

so the velocity is

$v = \sqrt{\frac{GM}{r} }$

where "G" is the gravitational constant, "M" the mass of the massive body and "r" the distance between the object and the center of gravity of mass M. As you can note, if "r" increase, "v" decrease.

The orbital period of any object in orbit is

$T = 2\pi \sqrt{\frac{a^3}{GM} }$

where "a" is length of semi-major axis (a = r in circular orbits). So if "r" increase, "T" increase.

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4 years ago