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2 years ago

A ____ is a region where jets of gas from young stars impact and heat the gas surrounding the young star.

Physics

2 answers:

Tomtit [17]2 years ago

7 0

Bipolar outflow is the blank

podryga [215]2 years ago

3 0

Answer:

Hi, There! my name is Jay And I'm here to help!

<h2>Question</h2>

A ____ is a region where jets of gas from young stars impact and heat the gas surrounding the young star.

<h2>Answer</h2>

bipolar outflow

Explanation:

A bipolar outflow is a stream of matter in two opposing directions from a object, Most likely a star.

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Hope this Helps!

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Have a great day!

$-Jay-$

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An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane

lubasha [3.4K]

Answer:

$\rho = 12580.7 kg/m^3$

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

$F = \frac{GMm}{(r + h)^2}$

here we have

$F =\frac {mv^2}{(r+ h)}$

$\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}$

here we have

$v = \sqrt{\frac{GM}{(r + h)}}$

now we can find time period as

$T = \frac{2\pi (r + h)}{v}$

$T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}$

$1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}$

$M = 4.54 \times 10^{23} kg$

Now the density is given as

$\rho = \frac{M}{\frac{4}{3}\pi r^3}$

$\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}$

$\rho = 12580.7 kg/m^3$

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3 years ago

What is an essential characteristic of an object in equilibrium?

bulgar [2K]

Answer:

2) zero acceleration

Explanation:

Motion can be defined as a change in the location (position) of a physical object or body with respect to a reference point.

This ultimately implies that, motion would occur as a result of a change in location (position) of an object with respect to a reference point or frame of reference i.e where it was standing before the effect of an external force.

Mathematically, the motion of an object is described in terms of time, distance, speed, velocity, position, displacement, acceleration, etc.

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

Generally, an object is said to be in equilibrium when neither the energy possessed by the object not state of motion changes with respect to time. Thus, the vector sum of all the forces acting upon an object that's in equilibrium is zero.

In conclusion, an essential characteristic of an object in equilibrium is zero (0) acceleration because there's no change in its velocity with respect to time.

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3 years ago

The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.

S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

$\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }$

Where:

$\omega$ - Angular frequency, measured in radians per second.

$m$ - Mass of the physical pendulum, measured in kilograms.

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

$d$ - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

$I_{O}$ - Moment of inertia with respect to pivot point, measured in $kg\cdot m^{2}$ .

In addition, frequency and angular frequency are both related by the following formula:

$\omega =2\pi\cdot f$

Where:

$f$ - Frequency, measured in hertz.

If $f = 0.658\,hz$ , then angular frequency of the physical pendulum is:

$\omega = 2\pi \cdot (0.658\,hz)$

$\omega = 4.134\,\frac{rad}{s}$

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

$\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}$

$I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}$

Given that $m = 1.15\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $d = 0.425\,m$ and $\omega = 4.134\,\frac{rad}{s}$ , the moment of inertia associated with the physical pendulum is:

$I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}$

$I_{o} = 0.280\,kg\cdot m^{2}$

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

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3 years ago

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What happens when a rock is physically or chemically changed?

Ierofanga [76]

Weathering the process that breaks down exposed stone and rock causing it to split apart or wear away. Different forces can cause rocks to become weathered physical weathering is caused by purely mechanical changes to the rock while chemical weathering is caused by chemical reactions

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4 years ago