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3 years ago

5

The diagram below shows the relative positions of Earth and the Sun at a certain time of the year. Based on the diagram, which s

eason is occurring in the Southern Hemisphere of Earth?
Question 4 options:

Fall

Winter

Spring

Summer

1 answer:

AlladinOne [14]3 years ago

7 0

Answer:

summer

Explanation:

Notice the higher density of the rays of the sun hitting straight the latitudes below the equator.

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the cycle of the moon through its phases, or the synodic month, is a. 21 days long. b. 27 1/3 days long. c. 29 1/2 days long. d.

olasank [31]

It takes 29.5 days long

7 0

3 years ago

Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini

amid [387]

Answer:

Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

$E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}$

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

$E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}$

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$

= $\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]$

∴NOTE: Graph is attached

8 0

4 years ago

20 Accelerated motion is represented by a<br> line on a nonlinear distance time graph.

Varvara68 [4.7K]

Curved line

Explanation:

Acceleration of motion is represented by a curved line on a non-linear distance-time graph.

The acceleration of a non-linear motion is depicted using a parabola which is a curve. This implies that the velocity is constantly changing and the distance covered by the body is also changing with equal amount of time.

A plot of this will give a parabola. This can be further established using one of the equations of motion below:

x = u + $\frac{1}{2}$ at ²

This is a quadratic function where:

x is the distance

u is the initial velocity

t is the time

a is acceleration

A quadratic function gives a curved line which is a parabola.

Learn more:

Acceleration brainly.com/question/10932946

#learnwithBrainly

3 0

3 years ago

Kim is ice-skating going 4.6 m/s. What is her velocity after 10 seconds ?

MArishka [77]

This is a uniform rectilinear motion (MRU) exercise.

To start solving this exercise, we obtain the following data:

<h3><u>Data:</u></h3>

v = 4.6 m/s
d = ¿?
t = 10 sec

To calculate distance, speed is multiplied by time.

We apply the following formula: d = v * t.

We substitute the data in the formula: the <u>speed is equal to 4.6 m/s,</u> the <u>time is equal to 10 s</u>, which is left as follows:

$\bf{d=4.6\dfrac{m}{\not{s}}*10\not{s} }$

$\bf{d=46 \ m}$

Therefore, the speed at 10 seconds is 46 meters.

$\huge \red{\boxed{\green{\boxed{\boldsymbol{\purple{Pisces04}}}}}}$

6 0

2 years ago

A pipe open at both ends has a fundamental frequency of 220 Hz when the temperature is 0°C. (a) What is the length of the pipe?

Allushta [10]

Explanation:

It is given that,

Fundamental frequency, f = 220 Hz

(a) We know that at 0 degrees, the speed of sound in air is 331 m/s.

For open pipe, $\lambda=2l$

l is the length of pipe

Also,

$v=f\lambda$

$l=\dfrac{v}{2f}=\dfrac{331}{2\times 220}=0.75\ m$

(b) Let f' is the fundamental frequency of the pipe at 30 degrees and v' is its speed.

$v'=331\sqrt{1+\dfrac{T}{273}}$

$v'=331\sqrt{1+\dfrac{30}{273}}$

v' = 348.71 m/s

So, $f'=\dfrac{v'}{\lambda}$

$f'=\dfrac{348.71}{2\times 0.75}$

f' = 232.4 Hz

Hence, this is the required solution.

8 0

3 years ago

Read 2 more answers