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3 years ago

14

for circular orbits the potential energy of the companion star is constant throughout the orbit. if the radius of the orbit doub

les, what is the new value of velocity of the campanion star

1 answer:

Vlada [557]3 years ago

7 0

Explanation:

the new value is 9.8*10⁷ m/s²

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A person standing on a scale feels a normal force of 655 N pushing up on him. What is his mass? (Unit=kg)

dimulka [17.4K]

Answer:

Approximately 66,8kg

Hope this help :D

Explanation:

F=mg. So 655N= 9.8 m/s² × mass.

Mass = 655 / 9.8

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3 years ago

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Why do we generally hear a thunder when there is a Bolt of lightning

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Explanation:

When a lightning bolt travels from the cloud to the ground it actually opens up a little hole in the air, called a channel. Once then light is gone the air collapses back in and creates a sound wave that we hear as thunder. The reason we see lightning before we hear thunder is because light travels faster than sound!

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4 years ago

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Why do areas with low altitude have warmer air than areas with high altitude

barxatty [35]

Answer:

Air at higher altitude is under less pressure than air at lower altitude because there is less weight of air above it, so it expands (and cools), while air at lower altitude is under more pressure so it contracts (and heats up).

Explanation:

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2 years ago

Figure 8-56 shows a solid, uniform cylinder of mass 7.00 kg and radius 0.450 m with a light string wrapped around it. A 3.00-N t

AVprozaik [17]

Answer:

a) The cylinder has an angular acceleration of 3.810 radians per square second, b) The frictional force has a magnitude of 9 newtons and has the same direction of tension force.

Explanation:

The external force exerted on string creates a tension force that tries to move the cylinder in translation, but it is opposed by the friction force between cylinder and ground that generates rolling on cylinder. The Free Body Motion on cylinder-string system is presented below as attachment. Given that cylinder is a rigid body in planar motion, two equations of equilibrium for translation and an equation of equilibrium for rotation are needed to represent the system, which are now described:

$\Sigma F_{x} = T + f = M\cdot R\cdot \alpha$

$\Sigma F_{y} = N - M\cdot g = 0$

$\Sigma M_{G} = (T-f)\cdot R = I_{G}\cdot \alpha$

Where:

$T$ - Tension, measured in newtons.

$f$ - Friction force, measured in newtons.

$M$ - Mass of the cylinder, measured in kilograms.

$R$ - Radius of the cylinder, measured in meters.

$\alpha$ - Angular acceleration, measured in radians per square second.

$N$ - Normal force from ground exerted on cylinder, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$I_{G}$ - Moment of inertia of the cylinder with respect to its center of mass, measured in kilogram-square meters.

The moment of inertia of the cylinder is:

$I_{G} = \frac{1}{2}\cdot M\cdot R^{2}$

a) The angular acceleration is determined by solving on first and third equation after eliminating friction force:

$f = M\cdot R \cdot \alpha - T$

$(T-M\cdot R\cdot \alpha+T) \cdot R = I_{G}\cdot \alpha$

$2\cdot T\cdot R = (I_{G} + M\cdot R^{2})\cdot \alpha$

$\alpha = \frac{2\cdot T\cdot R}{I_{G}+M\cdot R^{2}}$

$\alpha = \frac{2\cdot T \cdot R}{\frac{1}{2}\cdot M\cdot R^{2}+M\cdot R^{2} }$

$\alpha = \frac{4\cdot T}{3\cdot M\cdot R}$

If $T = 3\,N$ , $M = 7\,kg$ and $R = 0.45\,m$ , then:

$\alpha = \frac{4\cdot (3\,N)}{(7\,kg)\cdot (0.45\,m)}$

$\alpha = 3.810\,\frac{rad}{s^{2}}$

The cylinder has an angular acceleration of 3.810 radians per square second.

b) The magnitude of the frictional force can be determined with the help of the following expression:

$f = M\cdot R \cdot \alpha - T$

Given that $T = 3\,N$ , $M = 7\,kg$ , $R = 0.45\,m$ and $\alpha = 3.810\,\frac{rad}{s^{2}}$ , the magnitude of the friction force is:

$f = (7\,kg)\cdot (0.45\,m)\cdot \left(3.810\,\frac{rad}{s^{2}} \right)-3\,N$

$f = 9\,N$

The frictional force has a magnitude of 9 newtons and has the same direction of tension force.

5 0

4 years ago

Two electric charges are moved so that they are twice as far apart as they had originally been. Is the force they experience fro

Natasha_Volkova [10]

Answer:

No.

Explanation:

The force that two particle experience is inversely proportional to the sqare of the distance, this is:

$F \ \alpha \ \frac{1}{D^{2}}$ for a distance D

If we move them so that D is doubled:

$\frac{1}{2^{2}.D^{2} }$ = $\frac{1}{4} \eq \frac{1}{.D^{2} } \eq$

Then the force they experience is one fourth of the original.

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3 years ago