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oksian1 [2.3K]
3 years ago
5

Science Will give brainless and 22 points

Physics
2 answers:
fenix001 [56]3 years ago
4 0

Answer:

Mercury 0.39 AU, 36 million miles

57.9 million km

Venus 0.723 AU

67.2 million miles

108.2 million km

Earth 1 AU

93 million miles

149.6 million km

Mars 1.524 AU

141.6 million miles

227.9 million km

Jupiter 5.203 AU

483.6 million miles

778.3 million km

Saturn 9.539 AU

886.7 million miles

1,427.0 million km

Uranus 19.18 AU

1,784.0 million miles

2,871.0 million km

Neptune 30.06 AU

2,794.4 million miles

4,497.1 million km

Pluto (a dwarf planet) 39.53 AU

3,674.5 million miles

5,913 million km

otez555 [7]3 years ago
3 0

Answer:

Mercury 0.39 AU, 36 million miles 57.9 million km

Venus 0.723 AU 67.2 million miles 108.2 million km

Earth 1 AU 93 million miles 149.6 million km

Mars 1.524 AU 141.6 million miles 227.9 million km

Jupiter 5.203 AU 483.6 million miles 778.3 million km

Saturn 9.539 AU 886.7 million miles 1,427.0 million km

Uranus 19.18 AU 1,784.0 million miles 2,871.0 million km

Neptune 30.06 AU 2,794.4 million miles 4,497.1 million km

Pluto (a dwarf planet) 39.53 AU 3,674.5 million miles 5,913 million km

Explanation:

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A wave traveling in the positive x-direction with a frequency of 50.0 Hz is shown in the figure below. Find the following values
Klio2033 [76]

Answer:

Explanation:

a. The amplitude is the measure of the height of the wave from the midline to the top of the wave or the midline to the bottom of the wave (called crests). The midline then divides the whole height in half. Thus, the amplitude of this wave is 9.0 cm.

b. Wavelength is measured from the highest point of one wave to the highest point of the next wave (or from the lowest point of one wave to the lowest point of the next wave, since they are the same). The wavelength of this wave then is 20.0 cm. or \lambda=20.0cm

c. The period, or T, of a wave is found in the equation

f=\frac{1}{T} were f is the frequency of the wave. We were given the frequency, so we plug that in and solve for T:

50.0=\frac{1}{T} so

T=\frac{1}{50.0} and

T = .0200 seconds to the correct number of sig fig's (50.0 has 3 sig fig's in it)

d. The speed of the wave is found in the equation

f=\frac{v}{\lambda} and since we already have the frequency and we solved for the wavelength already, filling in:

50.0=\frac{v}{20.0} and

v = 50.0(20.0) so

v = 1.00 × 10³ m/s

And there you go!

5 0
2 years ago
An astronaut exploring a distant solar system lands on an unnamed planet with a radius of 2530 km. When the astronaut jumps upwa
Natali [406]

Answer:

1.38*10^18 kg

Explanation:

According to the Newton's law of universal gravitation:

F=G*\frac{m_a*m_p}{r^2}

where:

G= Gravitational constant (6.674×10−11 N · (m/kg)2)

ma= mass of the astronaut

mp= mass of the planet

F=m_a.a\\(v_f )^2=(v_o)^2+2.a.\Delta y\\\\a=\frac{(v_f)^2-(v_o)^2}{2.\Delta y}\\\\a=\frac{(0)^2-(4.29m/s)^2}{2.0.64m}=14.38m/s^2\\\\F=m_a*14.38m/s^2

so:

m_a*14.38m/s^2=(6.674*10^{-11}N.(m/kg)^2)*\frac{m_a.m_p}{(2.530*10^3m)^2}\\m_p=\frac{14.38m/s^2(2.530*10^3m)^2}{(6.674*10^{-11}N.(m/kg)^2)}\\\\m_p=1.38*10^{18}kg

7 0
3 years ago
A space shuttle takes off from FL and circles Earth several times, finally landing in CA. While the shuttle is in flight, a phot
mixer [17]

Answer:

Both the astronauts and photographer have the same displacement

Explanation:

Displacement is the minimum distance between two point. The initial point of both the astronauts and the photographer was Florida and the final point was California. So, the minimum distance for both of the astronauts and the photographer would be the distance between Florida and California would be the same.

Hence, both the astronauts and photographer will have the same displacement.

3 0
3 years ago
Physical properties of matter can be observed without changing the identity of the substance . The phases of matter are physical
ZanzabumX [31]

Solids have molecules that do not vibrate.

4 0
3 years ago
Need help solving this question.
MatroZZZ [7]

Answer:

See the answers below.

Explanation:

to solve this problem we must make a free body diagram, with the forces acting on the metal rod.

i)

The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.

We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.

For the summation of forces we will take the forces upwards as positive and the negative forces downwards.

ΣF = 0

-15+T-W=0\\T-W=15

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.

ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.

ΣM = 0

(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]

7 0
2 years ago
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