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3 years ago

9

What practical function is provided by the ionosphere? Produces auroras charges atoms heats oxygen and nitrogen permits worldwid

e radio communication

1 answer:

Karolina [17]3 years ago

6 0

Answer:

Worldwide Radio Communication

Explanation:

The ionosphere is important because it is through the ionosphere that world wide radio communication is possible.

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2. A ball is released from a vertical height of 20 cm. It rolls down a "perfectly

Paraphin [41]

Answer:

20 cm

Explanation:

Given that a ball is released from a vertical height of 20 cm. It rolls down a "perfectly frictionless" ramp and up a similar ramp. What vertical height on the second ramp will the ball reach before it starts to roll back down?

Since it is perfectly frictionless, the Kinetic energy in which the ball is rolling will be equal to the potential energy at the edge of the ramp.

Therefore, the ball will reach 20 cm before it starts to roll back down.

7 0

3 years ago

A bus leaves at 9 am with a group of tourists. They travel 350 km before they stop for lunch. Then they travel an additional 250

Anettt [7]

Average speed = Distance traveled / time taken

In this case Time taken = Difference in hours between 3 PM and 9 AM

= 6 hours

Total distance traveled = 350 km + 250 km

= 600 kilometers

So average speed = 600/6 = 100 km/hr

Average speed of bus = 27.78 m/s

So the bus's average speed = 27.78 m/s or 100 km/hr.

8 0

3 years ago

Give an example of intense aerobics activity. Prompt must be accurate.

Lubov Fominskaja [6]

Answer:

Explanation:

An example of an intense aerobic activity would be running/ sprinting sprinting targets six specific muscle groups: hamstrings, quadriceps, glutes, hips, abdominals and calves. Sprinting is a total body workout featuring short, high-intensity repetitions and long, easy recoveries.

7 0

3 years ago

Two small insulating spheres with radius 3.50×10−2 m are separated by a large center-to-center distance of 0.555 m. One sphere i

xz_007 [3.2K]

Answer:

$E = 7.83 \times 10^5 N/C$

Explanation:

Since we know that two sphere is oppositely charged so net electric field at the mid point of two balls will be sum of the electric field due to each ball at the mid point

So we know that

$E = \frac{kq_1}{r^2} + \frac{kq_2}{r^2}$

here we know that

$q_1 = 4.20 \mu C$

$q_2 = 2.50 \mu C$

$r = \frac{0.555}{2}$

so we have

$E = \frac{(9\times 10^9)(4.20 + 2.50) \times 10^{-6}}{0.2775^2}$

$E = 7.83 \times 10^5 N/C$

4 0

3 years ago

Need help with this question. Thirty points.

Anon25 [30]

$1)\\\\\text{Kinetic energy,}\\\\E_k = \dfrac 12 mv^2 \\\\\implies m = \dfrac{2E_k}{v^2} = \dfrac{2\times 4500}{30^2}=10~ kg\\ \\2)\\\\\text{Kinetic energy,}\\\\E_k = \dfrac 12mv^2 \\\\\implies v^2 = \dfrac{2E_k}m\\\\\implies v = \sqrt{\dfrac{2E_k}{m}}=\sqrt{\dfrac{2 \times 320}{\dfrac{20}{9.8}}}= 17.709 ~ ms^{-1}\\\\3)\\\\\text{Kinetic energy,}\\\\E_k=\dfrac 12 mv^2=\dfrac 12 \times 50 \times 10^2=2500 ~J\\\\\\$

$4)\\\\\text{Potential energy,}\\\\E_p =mgh = 5\times 9.8\times 1.5 = 73.5 ~J$

5 0

2 years ago