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vovikov84 [41]
3 years ago
12

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

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Describe the factors that cause static friction between two surfaces to increase.
maw [93]

Answer : The static friction depends on two factors

1. Roughness of the surface

2. Force

Explanation : Friction occur when surface is not smooth.

The formula of friction is

F= \mu mg

Static friction depends on the roughness of the surface and force which is trying to push to object along the surface.

Static friction is caused by the attraction between two surfaces that are in contact.  when the surface will rough and the object will heavier then the force will be larger.

4 0
3 years ago
Calculate the east component of a resultant 32.5 m/s, 35.0° east of north.
ValentinkaMS [17]

Answer:

East component is: 18.64 m/s

Explanation:

If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:

East component = 32.5 m/s * sin(35) = 18.64 m/s

4 0
3 years ago
A 50-kg block is at rest on a 15o slope. A force of 250 N is acting on the block up the slope parallel to it. If the block does
Iteru [2.4K]

The minimum value of the coefficient of static friction between the block and the slope is 0.53.

<h3>Minimum coefficient of static friction</h3>

Apply Newton's second law of motion;

F - μFs = 0

μFs = F

where;

  • μ is coefficient of static friction
  • Fs is frictional force
  • F is applied force

μ = F/Fs

μ = F/(mgcosθ)

μ = (250)/(50 x 9.8 x cos15)

μ = 0.53

Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.

Learn more about coefficient of friction here: brainly.com/question/20241845

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6 0
2 years ago
(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r
Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface U_1=\frac{GM_em}{R_e}

Gravitational Potential Energy at height h is U_2=\frac{GM_em}{R_e+h}

Energy required to lift the satellite E_1=U_1-U_2

E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}

Now Energy required to orbit around the earth

E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}

\Delta E=E_1-E_2

\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}

E_1=E_2  (given)

\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0

\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0

h=\frac{R_e}{2}

h=3.19\times 10^6\ m

(b)For greater height E_1  is greater than E_2

thus energy to lift the satellite is more than orbiting around earth

4 0
3 years ago
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