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2 years ago

Does the orbital period of a planet depend on the mass of the planet or on the mass of the star that it orbits?

Physics

1 answer:

jasenka [17]2 years ago

8 0

Answer:

The orbital period of a planet depends on the mass of the planet.

Explanation:

A less massive planet will take longer to complete one period than a more massive planet.

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Which of the four fundamental forces is responsible for holding together the molecules of the pizza dough as it is spinning in t

defon

Answer:

None.

Explanation:

Molecules are formed by an element's need or excess of electrons. For example, in nature oxygen generally exists as 02. Other molecules are formed via chemical reaction. The example here is the burning of gasoline. Gasoline's two main byproducts are water and carbon dioxide.

Hydrogen as an atom has one electron making it unstable. Put a second hydrogen atom next to the first and the two atoms will share electrons to fill the first energy level the atom needs to be stable.

5 0

2 years ago

The velocity of an object is equal to the distance divided by time. The equation is velocity = distance/time. If you wanted to c

Korolek [52]

Answer: C) divide: distance ÷ velocity

Explanation:

The velocity $V$ equation is distance $d$ divided by time $t$ :

$V=\frac{d}{t}$

If we isolate $t$ we will have:

$t=\frac{d}{V}$

Hence, the correct option is C: distance divided by velocity.

7 0

2 years ago

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0

2 years ago

A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s

sweet [91]

This question is in complete.The question is

A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.

Answer:

distance=0.124 m

Explanation:

$mgh=mglSin\alpha =(1/2)Iw_{i}^{2}+(1/2)mv^{2}\\ v=wR\\Solve for L\\L=((1/2)(1/2)0.015^{2}*60^{2}+(1/2)(60*0.015^{2} ))/9.8Sin30\\ L=0.124m$

6 0

3 years ago

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam

iris [78.8K]

Answer:

Explanation:

Given:

U1 = 1.6 m/s

U2 = -1.1 m/s

M1 = 1850 kg

M2 = 1400 kg

V1 = 0.27 m/s

Using momentum- collision equation,

M1U1 + M2U2 = M1V1 + M2V2

1850 × 1.6 - 1400 × 1.1 = 1850 × 0.27 + 1400 × V2

1420 = 499.5 + 1400V2

V2 = 0.6575 m/s

KE = 1/2 × MV^2

KEa1 + KEa2 = KEb1 + KEb2

Delta KE = KE2 - KE1

KEa1 = 2368 J

KEb1 = 847 J

KEa2 = 67.433 J

KEb2 = 302.6 J

KE1 = KEa1 + KEb1

= 3215 J

KE2 = 370.033 J

Delta KE = -2845 J.

5 0

3 years ago