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chubhunter [2.5K]
3 years ago
9

The moon orbits our Earth. The opposite would never be true; the Earth would never orbit the moon. One reason is due to the forc

e of gravity. The object ____________ has a greater gravitational force and that determines what satellites orbit around it. A) with the greater mass B) containing mostly water C) with a thicker atmosphere D) with the greater diameter
Physics
2 answers:
ANEK [815]3 years ago
7 0
The object with the greater mass has a greater gravitational force and that determines what satellites orbit around it. An object with more mass will never orbit an object with less.
Snowcat [4.5K]3 years ago
5 0

A)With the greater mass

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An instrument used to measure atmospheric pressure is called a
Sonbull [250]

Answer:

Barometer is an instrument used to measure the atmospheric pressure.

Explanation:

  • Torricelli did an experiment by filling a one meter long test tube completely with mercury. He took a beaker with some amount of mercury in it and inverted the test tube filled with mercury by closing its open end into the mercury of beaker ensuring that no bubble is trapped into the mercury of test tube during the process.
  • This experiment was performed at normal temperature and pressure. Now some of the mercury from the test tube fell down into the beaker through the open end until the level becomes stagnant and this denotes the atmospheric pressure in terms of mercury column.
  • There was a vacuum created at the top of the test tube called torricellian vacuum which holds back the remaining mercury height and the rest of the mercury level in the beaker is facing the atmospheric pressure.
6 0
4 years ago
Read 2 more answers
A 24-V battery is powering a light bulb with a resistance of 3.0 ohms. What is the current flowing through the bulb? A) 7.20 A B
Usimov [2.4K]

According to Ohm's law for a portion of the circuit we have:

U=RI=>I=U/R=24/3=8 A

The correct answer is  B


3 0
3 years ago
An elevated water tank holds 1 million gallons of water. Determine the force (in lbf) that the tower structure must be able to w
bixtya [17]

Answer:

8,345,925 lb-f

Explanation:

We know pressure P = F/A = ρgh ⇒ F = ρghA = ρgV

F = ρgV where F = force, ρ = density of water = 1000 kg/m³ , g = acceleration due to gravity = 9.8 m/s² and V = 1 × 10⁶ gallons = 1 × 10⁶ gallons × 1m³/264.2 gallons = 3785 m³

So, F = ρgV = 1000 kg/m³ × 9.8 m/s² × 3785 m³ = 37,093,000 N

We now convert Newtons to pound-force.

1 N = 0.225 lb-f

So, F = 37,093,000 N = 37,093,000 N × 0.225 lb-f/1 N = 8,345,925 lb-f

So, the tower must be able to withstand 8,345,925 lb-f

5 0
3 years ago
A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Su
aniked [119]

Answer:

The planet´s orbital period will be one-half Earth´s orbital period.

Explanation:

The planet in orbit, is subject to the attractive force from the sun, which is given by the Newton´s Universal Law of Gravitation.

At the same time, this force, is the same centripetal force, that keeps the planet in orbit (assuming to be circular), so we can put the following equation:

Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r

As we know to find out the orbital period, as it is the time needed to give a complete revolution around the sun, we can say this:

ω = 2*π / T (rad/sec), so replacing this in the expression above, we get:

Fg = Fc ⇒   G*mp*ms / r² = mp*(2*π/T)²*r

Solving for T²:

T² = (2*π)²*r³ / G*ms (1)

For the planet orbiting the sun in Andromeda, we have:

Ta² = (2*π)*r³ / G*4*ms (2)

As the radius of the orbit (distance to the sun) is the same for both planets, we can simplify it in the expression, so, if we divide both sides in (1) and (2), simplifying common terms, we finally get:

(Te / Ta)² =  4  ⇒ Te / Ta = 2 ⇒ Ta = Te/2

So, The planet's orbital period will be one-half Earth's orbital period.

7 0
3 years ago
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 39 m in front of you. Your reaction tim
miss Akunina [59]

Answer:

a) 10.8 m

b) 24.3 m/s

Explanation:

a)

  • In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.
  • The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.
  • We can find this distance simply applying the definition of average velocity, as follows:

       \Delta x_{1} = v_{1o} * t_{react} = 20 m/s * 0.5 s = 10 m (1)

  • The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.
  • We can find this part using the following kinematic equation (assuming that the deceleration keeps the same all time):

       v_{1f} ^{2}  - v_{1o} ^{2} = 2* a* \Delta x  (2)

  • where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².
  • Solving for Δx, we get:

       \Delta x_{2} = \frac{-(20m/s)^{2}}{2*(-11m/s)} = 18.2 m (3)

  • So, the total distance traveled was the sum of (1) and (3):
  • Δx = Δx₁ + Δx₂ = 10 m + 18.2 m = 28.2 m (4)
  • Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.

b)

  • We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.
  • The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.
  • ⇒Δx = Δx₁ + Δx₂ = 39 m. (5)
  • Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:

       \Delta x_{1} = v_{omax} * t_{react} = 0.5* v_{omax} (6)

  • Δx₂, is the distance traveled while decelerating, and can be obtained  using (2):

        v_{omax} ^{2} = 2* a* \Delta x_{2} (7)

  • Solving for Δx₂, we get:

       \Delta x_{2} = \frac{-v_{omax} ^{2} x}{2*a}  = \frac{-v_{omax} ^{2}}{(-22m/s2)} (8)

  • Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.
  • Taking the positive root in the quadratic formula, we get the following value for vomax:
  • v₀max = 24.3 m/s.
6 0
3 years ago
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