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3 years ago

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a plane travels 204 km, northeast in 15.0 minutes. It also increases elevation by 1.6 km, upward in the same amount of time. Wha

t are the average velocities in the two seperate directions?

2 answers:

oksian1 [2.3K]3 years ago

8 0

<u>Answer:</u> The average speed of the plane in northeast direction is 816 km/hr and that in upward direction is 6.4 km/hr

<u>Explanation:</u>

Average speed is defined as the ratio of total distance traveled to the total time taken.

To calculate the average speed of the runner, we use the equation:

$\text{Average speed}=\frac{\text{Total distance traveled}}{\text{Total time taken}}$ .......(1)

<u>In northeast direction:</u>

Total distance traveled = 204 km

Total time taken = 15.0 mins = 0.25 hr (Conversion factor: 1 hr = 60 mins)

Putting values in equation 1, we get:

$\text{Average speed of plane}=\frac{204km}{0.25hr}=816km/hr$

<u>In upward direction:</u>

Total distance traveled = 1.6 km

Total time taken = 15.0 mins = 0.25 hr

Putting values in equation 1, we get:

$\text{Average speed of plane}=\frac{1.6km}{0.25hr}=6.4km/hr$

Hence, the average speed of the plane in northeast direction is 816 km/hr and that in upward direction is 6.4 km/hr

LUCKY_DIMON [66]3 years ago

5 0

Answer:

230 m/s northeast, 1.8 m/s up

Explanation:

204 kilometres = 204000 metres

15.0 minutes = 900 seconds

Velocity = Distance / Time

= 204000 / 900

= 230 m/s northeast (to 2 sf.)

1.6km = 1600 metres

Velocity = 1600 / 900

= 1.8 m/s up (to 2 sf.)

Read more on Brainly.com - brainly.com/question/13863590#readmore

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Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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With that we can plug the values in to get:

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