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3 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]3 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

You might be interested in

Calculate the Force required to give a bullet of mass 50 g an acceleration of 300 m/s2

NNADVOKAT [17]

Answer:

<h3>The answer is 15 N</h3>

Explanation:

The force acting on an object can be found by using the formula

<h3>Force = mass × acceleration</h3>

From the question

mass = 50 g = 0.05 kg

acceleration = 300 m/s²

We have

force = 0.05 × 300

We have the final answer as

Hope this helps you

7 0

2 years ago

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0

3 years ago

From a lake, water is pumped at a rate of 67 L/s to a storage tank positioned 14 m above while consuming 16.4 kW of electrical p

LenaWriter [7]

Answer:

57 %

Explanation:

input power = 16.4 kW = 16.4 x 10^3 W = 16400 W

Water pumped per second = 67 L/s

Mass of water pumped per second, m = Volume of water pumped epr second x density of water

m = 67 x 10^-3 x 1000 = 67 kg/s

height raised, h = 14 m

Output Power = m x g x h / t = 67 x 10 x 14 = 9380 W

efficiency = output power / input power = 9380 / 16400 = 0.57

% efficiency = 57 %

thus, the efficiency of the pump is 57 %.

3 0

3 years ago

The cheetah can run a distance of 275 m for 9 seconds. Calculate its speed.

AnnZ [28]

Answer: 110000

Explanation:

26/9=30.5555555556

30.5555555556 x 60=1833.33333333

110000 x 60=110000

4 0

3 years ago

Which perspective is most directly involved with the study of how brain

daser333 [38]

Answer:

biopsychlogy

Explanation:https://www.answers.com/Q/Which_perspective_is_most_directly_involved_with_the_study_of_how_brain_development_might_affect_behavior

6 0

3 years ago