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Brums [2.3K]
3 years ago
11

What other experiments can you do with bubble gum?

Physics
1 answer:
Elanso [62]3 years ago
7 0
What happens when you leave it in differnt soda pops.
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During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0
madreJ [45]

Answer:

Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4orbits}

Kilometers\ in\ 1\ Orbit=3.46*10^4 Km/Orbit

Explanation:

Given Data:

Numbers of times Telescope cycled around the earth in 6 years=37,000 times

Total Distance traveled in 6 years by the Hubble Space Telescope=1,280,000,000 Km

Find:

Kilometers in one Orbit=?

Solution:

Kilometers in 37,000 Orbits=1,280,000,000 Km

Kilometers in 1 Orbit=1,280,000,000/37,000

In Scientific Notation:

Kilometers\ in\ 3.7*10^4\ Orbits=1.28*10^9 Km

Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4 orbits}

Kilometers in 1 Orbit=34594.594 Km

Kilometers in 1 Orbit in Scientific notation:

Kilometers\ in\ 1\ Orbit=3.46*10^4 Km

8 0
3 years ago
what is the acceleration of an object that moves at a constant velocity of 2.0 meters per second for 5 seconds?
nexus9112 [7]
Constant velocity means moving in a straight line at a speed that doesn't change. If the object is moving with constant velocity then its acceleration is zero. Acceleration is the rate at which velocity is changing.
3 0
3 years ago
) The square plates of a 5000-pF parallel-plate capacitor measure 50 mm by 50 mm and are separated by a dielectric that is 0.23
notsponge [240]

Answer:

4 x 10⁻⁴ J

Explanation:

C = 5000 pF, V = 400 V

Energy = CV²/2 = 5000 x 10⁻¹² x 400²/2 = 4 x 10⁻⁴ J

6 0
1 year ago
Pls help will give brainlest​
Karo-lina-s [1.5K]

Respon

lqiudos ciopatmibes

ly  apsamtios ccoriendor sabe r

llpop

io.

7 0
2 years ago
Read 2 more answers
A proton moving at 8.00 106 m/s through a magnetic field of magnitude 1.72 T experiences a magnetic force of magnitude 7.20 10-1
gladu [14]

Answer:

19.1 deg

Explanation:

v = speed of the proton = 8 x 10⁶ m/s

B = magnitude of the magnetic field = 1.72 T

q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C

F = magnitude of magnetic force on the proton = 7.20 x 10⁻¹³ N

θ = Angle between proton's velocity and magnetic field

magnitude of magnetic force on the proton is given as

F = q v B Sinθ

7.20 x 10⁻¹³ = (1.6 x 10⁻¹⁹) (8 x 10⁶) (1.72) Sinθ

Sinθ = 0.327

θ = 19.1 deg

4 0
3 years ago
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