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wolverine [178]
1 year ago
11

Which of the following is NOT a requirement for a planet? A. must orbit a star, but is not a star or satellite of another planet

B. must be roundC. must clear its orbit of debrisD. must have a moon
Physics
1 answer:
Ket [755]1 year ago
6 0

Answer:

D. must have a moon

Explanation:

Scientists say that a planet must orbit a star, must be round, and must be big enough so its gravity clears its orbit of other objects. Therefore, the statement that is not a requirement for a planet is

D. must have a moon

For example, Mercury and Venus are planets with no moons.

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Temperature of a substance is proportional to the average kinetic energy of the motion of the molecules of that substance.
MrRissso [65]
True, according to the Kinetic Molecular Theory.
4 0
3 years ago
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Suppose that you are swimming in a river while a friend watches from the shore. In calm water, you swim at a speed of 1.25 m/s .
aliya0001 [1]

Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.

Explanation:

  • Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
  • Let S_{0} be the speed of the river's current given as 1.00 m/s.

  • Note that this speed is the magnitude of the velocity which is a vector quantity.
  • The direction of the swimmer is upstream.

Hence the resultant velocity is given as, S_{R} = S — S 0S_{0}

S_{R} = 1.25 — 1

S_{R} = 0.25 m/s.

Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.

6 0
3 years ago
A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
gulaghasi [49]

A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

R_1 = 240 \Omega is the resistance of the bulb

Substituting numbers, we get

P_1 = (0.77 A)^2 (240 \Omega)=142.3 W

C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0
2 years ago
An object is moving in a straight line along the y axis. As a function of time, its position is given by the equation y=3.0+4.8x
romanna [79]

Answer:

Explanation:

<u>Instant Velocity and Acceleration </u>

Give the position of an object as a function of time y(x), the instant velocity can be obtained by

v(x)=y'(x)

Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by

a(x)=v'(x)=y''(x)

We are given the function for y

y(x)=3.0+4.8x +6.4x^2

Note we have changed the last term to be quadratic, so the question has more sense.

The velocity is

v(x)=y'(x)=4.8+12.8x

And the acceleration is a(x)=v'(x)=12.8

5 0
3 years ago
An object has a weight of 500 on earth what is the mass of this object
erastova [34]

550! OBVY! lol! ope this helps1

7 0
3 years ago
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