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3 years ago

Earth’s orbital speed is approximately _____.

2 answers:

5 0

The answer that best fits the blank provided above is 30 KILOMETERS PER SECOND. The orbital speed of the Earth as recorded is 30 kilometers per second or <span>107000 km/h or</span> 18½ miles per second. The orbital period is 365 days which makes one year. Hope this helps.

liraira [26]3 years ago

4 0

THE CORRECT ANSWER IS 107,000 KILOMETER PER HOUR

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A strong magnetic field surrounds Earth. How does Earth’s magnetic field help humans?

elena-14-01-66 [18.8K]

A strong magnetic field surrounds Earth and (C) It keeps the solar wind from reaching Earth. If all the particles of the solar wind contact the earth, they would make the ozone layer disappear. The ozone layer is the one in charge of protecting the earth from the ultraviolet radiation.

Explanation:

5 0

2 years ago

Please helpp this is due in 10 minutes?!!!!

Sati [7]

Answer:

1. at the top of the coaster.

2. at the bottom of the coaster.

3. when the car is moving

4. when the car is moving

Explanation:

there is the most amount if potential energy at the top, and the keast at the bottom.

5 0

3 years ago

If a wave traveling through air decreases its wavelength by half, describe what happens to the wave speed and frequency.

natka813 [3]

The frequency, the speed and the wavelength of a wave are related by the following equation:
$f= \frac{v}{\lambda}$ (1)
where
f is the frequency
$\lambda$ is the wavelength
v is the wave speed

The speed of the wave does depend only on the properties of the medium, so since the wave is still traveling in air, the medium has not changed and therefore the speed remains the same. We see instead from eq.(1) that the frequency is inversely proportional to the wavelength, so if the wavelength is decreased by half, we see that the frequency will double.

5 0

3 years ago

Scientists want to place a 3400.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2480.0

BaLLatris [955]

Answer:

Part a)

$r = 6.96 \times 10^6 m$

Part b)

$F_g = 3.004 \times 10^3 N$

Part c)

$a = 0.88 m/s^2$

Part d)

$v = \frac{2480}{2} = 1240 m/s$

Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

$v = 2480 m/s$

now we will have

$v = \sqrt{\frac{GM_{mars}}{r}}$

now we have

$M_{mars} = 6.4191 \times 10^{23} kg$

$R_{mars} = 3.397 \times 10^6 m$

now we have

$2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}$

$r = 6.96 \times 10^6 m$

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

$F_g = \frac{GM_{mars} m}{r^2}$

$F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}$

$F_g = 3.004 \times 10^3 N$

Part c)

Acceleration of satellite is the ratio of gravitational force and mass of the satellite

$a = \frac{F_g}{m}$

$a = \frac{3004}{3400}$

$a = 0.88 m/s^2$

Part d)

As we know by III law of kepler

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

here we know that T2 = 8 T1

$(\frac{1}{8})^2= \frac{r_1^3}{r_2^3}$

$\frac{r_1}{r_2} = (\frac{1}{2})^2$

so we have

$r_2 = 4r_1$

as we know that speed is given as

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

so we can say since radius is orbit becomes 4 times so the orbital speed must be half

$v = \frac{2480}{2} = 1240 m/s$

7 0

4 years ago

Put the balloon near (BUT NOT TOUCHING) the wall. Leave about as much space as the width of your pinky finger between the balloo

natka813 [3]

Answer:

Move towards the wall.

Explanation:

When the balloon is kept near to the wall not touching the wall, there is a force of electrostatic attraction so that the balloon moves towards the wall and stick to it.

As there is some charge on the balloon and the wall is uncharged so the force is there due to which the balloon moves towards the wall.

3 0

3 years ago