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2 years ago

Earth orbits the sun once every 365.25 days. Find the average angular speed of earth about the sun. Answer in units of rad/s.

Physics

1 answer:

Talja [164]2 years ago

7 0

Answer:

2.0x10-7 rad/s.

Explanation:

Calculating the angular velocity of the Earth is a deceptively easy task. The reason for this is simple - the angular velocity is defined as the angle subtended in a certain time.

We know the Earth goes round the Sun, all the way around is 2 radians (360 degrees). We also know that it takes a year (approx 365 days) which is therefore about 3.2x107 secs.

Therefore = 2 / 3.2x107 = 2.0x10-7 rad/s. We have calculated the angular velocity.

However, if we can measure the distance to the Sun we can also calculate the velocity of the Earth relative to the Sun. Although unless we define a direction this is more technically known as the speed. This can be done by looking at the definition of the radian. The radian is a unit which conects the radius of an arc, the length of the arc and the angle subtended by the arc. The formula for this is s = r x (where s is the length of the arc, r is the radius and the angle). So if we know the radius of the Earth's orbit (1.5x10¹¹m) we can substitute the angular velocity from our previous equation to give v = x r (where v is the velocity, the angular velocity and r the radius).

So, the Earth travels through space (relative to the Sun) at: v = 2.0x10-7 x 6.4x106 = 3.0x104m/s

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While dangling a hairdryer by its cord, you observe that the cord is vertical when the hairdryer isoff and, once it is turned on

jekas [21]

Answer:

The speed of the air leaving the hairdryer is 10m/s.

Explanation:

The thrust that the dryer produces is what keeps it elevated at an angle of 5° from the vertical; therefore, from the force diagram we get

$(1).\: tan (5^o) = \dfrac{F_t}{Mg}$

putting in $M =0.420kg$ , $g = 9.8m/s^2$ and solving for $F_t$ we get:

$F_t = Mg\:tan(5^o)$

$F_t = (0.420kg)(9.8m/s^2)\:tan(5^o)$

$\boxed{F_t = 0.3601N.}$

Now, this thrust produced is related to to the air ejection speed $v$ by the relation

$(2).\: F_t = v\dfrac{dM}{dt}$

where $dM/dt$ is the rate of air ejection which we know is

$0.06m^3/2s = 0.03m^3/s$

and since $1m^3 = 1.2kg$ ,

$0.03m^3/s \rightarrow 0.036kg/s$

$\dfrac{dM}{dt} = 0.036kg/s,$

putting this into equation (2) and the value of $F_t$ we get:

$0.3601N = 0.036v$

which gives

$v= \dfrac{0.3601}{0.036}$

$\boxed{v =10m/s.}$

which is the speed of the air ejected.

6 0

2 years ago

A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou

attashe74 [19]

Answer:

Explanation:

Impulse = Force x time = change in momentum

F x t = m ( v - u )

In both cases, u are same and v=0

So change in momentum is same

hence , impulse is same.

F = Change in momentum / time

In case of air mattress , time increases

Hence average force decreases .

Option e is correct .

8 0

2 years ago

A proposed space elevator would consist of a cable stretching from the earth's surface to a satellite, orbiting far in space, th

NISA [10]

To solve this problem we will apply the concepts related to energy conservation. Here we will use the conservation between the potential gravitational energy and the kinetic energy to determine the velocity of this escape. The gravitational potential energy can be expressed as,

$PE= \frac{GMm}{d}$

The kinetic energy can be written as,

$KE= \frac{1}{2} mv^2$

Where,

$G = 6.67*10^{-11}m^3/kg\cdot s^2$ Gravitational Universal Constant

$m = 5.972*10^{24}kg$ Mass of Earth

$h = 56*10^6m$ Height

$r = 6.378*10^6m$ Radius of Earth

From the conservation of energy:

$\frac{1}{2} mv^2 = \frac{GMm}{d}$

Rearranging to find the velocity,

$v = \sqrt{\frac{2Gm}{d}} \rightarrow$ Escape velocity at a certain height from the earth

If the height of the satellite from the earth is h, then the total distance would be the radius of the earth and the eight,

$d = r+h$

$v = \sqrt{\frac{2Gm}{r+h}}$

Replacing the values we have that

$v = \frac{2(6.67*10^{-11})(5.972*10^{24})}{6.378*10^6+56*10^6}$

$v = 3.6km/s$

Therefore the escape velocity is 3.6km/s

3 0

2 years ago

Think of something from everyday life that follows a two-dimensional path. It could be a kicked football, a bus that's turning a

OLEGan [10]

Answer:

Let us consider the case of a bus turning around a corner with a constant velocity, as the bus approaches the corner, the velocity at say point A is Va, and is tangential to the curve with direction pointing away from the curve. Also, the velocity at another point say point B is Vb and is also tangential to the curve with direction pointing away from the curve. Although the velocity at point A and the velocity at point B have the same magnitude, their directions are different (velocity is a vector quantity), and hence we have a change in velocity. By definition, an acceleration occurs when we have a change in velocity, so the bus experiences an acceleration at the corner whose direction is away from the center of the corner.

The acceleration is not aligned with the direction of travel because the change in velocity is at a tangent (directed away) to the direction of travel of the bus.

4 0

2 years ago

If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it

Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of 27.0 m/s2 is

$\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$