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Darina [25.2K]
3 years ago
10

Assume that the car at point A and the one at point E are traveling along circular paths that have the same radius. If the car a

t point A now moves twice as fast as the car at point E, how is the magnitude of its acceleration related to that of car E.
Physics
1 answer:
vazorg [7]3 years ago
6 0

Answer:

The magnitude of car A acceleration is 4 times that of car E

Explanation:

Assuming they are traveling at a constant speed. Their (centripetal) acceleration is as the following

a = \frac{v^2}{r}

where v is the (linear) velocity and r is the radius. We can use this to calculate their ratio

a_A / a_E = \frac{v_A^2/r_A}{v_E^2/r_E} = \left(\frac{v_A}{v_E}\right)^2\frac{r_E}{e_A}

Since v_A = 2V_E \rightarrow \frac{v_A}{v_E} = 2 and r_A = r_E

a_A / a_E = 2^2 = 4

So the magnitude of car A acceleration is 4 times that of car E

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Newton has 3 Laws specifically The Three Laws of Motion
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3 years ago
A visible light has a wavelength of 727.3 nm. Determine its frequency, energy per photon, and color.
Snezhnost [94]

Answer:

f=4.12\times 10^{14}\ Hz and E=2.73\times 10^{-19}\ J

Explanation:

The wavelength of a visible light is 727.3 nm.

727.3\ nm=727.3 \times 10^{-9}\ m

The formula is as follows :

c=f\lambda

f is the frequency of the visible light

f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{727.3 \times 10^{-9}}\\\\f=4.12\times 10^{14}\ Hz

Energy of a photon is given by :

E = hf, h is Planck's constant

E=6.63\times 10^{-34}\times 4.12\times 10^{14}\\\\E=2.73\times 10^{-19}\ J

Red color has a frequency of 4.12\times 10^{14}\ Hz and energy per photon is 2.73\times 10^{-19}\ J.

3 0
3 years ago
Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for sl
ikadub [295]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

  • We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).
  • Both movements are independent each other, due to they are perpendicular.
  • In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
  • Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:

       v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)

  • Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

       t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s  (2)

  • In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
  • Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

       \Delta y = v_{oy}  * t - \frac{1}{2} *g*t^{2} (3)

  • In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

       v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)

  • Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

       \Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)

  • Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.
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andreyandreev [35.5K]
Yes For example Neptune
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Read 2 more answers
If 30 N of force is exerted over an area of 20 m2, how much pressure is being applied?
Ludmilka [50]
Pressure= F/A
Given, F=30 N
Area=20 m^2
Putting values in formula,
Pressure = 30/20
=3/2
=1.5 Pascals
4 0
3 years ago
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