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3 years ago

Which statements accurately describe the shape of Earth’s orbit around the Sun? Check all that apply.

2 answers:

4 0

The Earth’s orbit is a nearly circular ellipse.
The Sun is located at one of the two focal points.

The Earth moves around the Sun in an orbit that is almost but not quite circular. As Kepler proved in the seventeenth century, the orbit is actually an ellipse. A parameter called the eccentricity (e) defines the degree of departure from a circle. A value of e=0 would indicate a circle whereas a value of e=0.9 would indicate a very elongated ellipse. The eccentricity of the Earth's orbit is currently e=0.0167.

anyanavicka [17]3 years ago

4 0

2. The Earth’s orbit is a nearly circular ellipse.

5. The Sun is located at one of the two focal points.

These are the two correct answers.

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A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between

SashulF [63]

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

$\Delta W = \Delta KE$

$\Delta W = \frac{1}{2} mv^2$

Here,

m = mass

v = Velocity

Our values are given as,

$m = 79.7kg$

$v = 4.77m/s$

Replacing,

$\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2$

$\Delta W = 907J$

Therefore the mechanical energy lost due to friction acting on the runner is 907J

6 0

3 years ago

What causes waves to slow down: change in wave’s wavelength or change in its frequency and y?

Mashutka [201]

Answer:

Although the wave slows down, its frequency remains the same, due to the fact that its wavelength is shorter. When waves travel from one medium to another the frequency never changes.

3 0

3 years ago

A block is at rest on the incline shown in the gure. The coefficients of static and kinetic friction are 0.6 and 0.51, respec- t

Alekssandra [29.7K]

Normal reaction force on the block while it is at rest on the inclined plane is given as

$F_n = mgcos\theta$

here we know that

m = 46 kg

$\theta = 29^o$

now we will have

$F_n = 46*9.8*cos29 = 394.3 N$

now the limiting friction or maximum value of static friction on the block will be given as

$F_s = \mu_s * F_n$

$F_s = 0.6 * 394.3 = 236.56 N$

Above value is the maximum value of force at which block will not slide

Now the weight of the block which is parallel to inclined plane is given as

$F_{||} = mg sin\theta$

here we know that

$F_{||} = 46*9.8 sin29 = 218.55 N$

Now since the weight of the block here is less than the value of limiting friction force and also the block is at rest then the frictional force on the block is static friction and it will just counter balance the weight of the block along the inclined plane.

So here friction force on the given block will be same as its component on weight which is 218.55 N

5 0

3 years ago

You are trying to remove small particles from water. What would be the BEST thing you could do to remove the small particles? *

kozerog [31]

Answer:

filter

Explanation:

5 0

3 years ago

A ball is dropped from the top of a 77 m building. With what speed does the ball hit the ground? _________ m/s

vitfil [10]

Answer:

38.87 m/s

Explanation:

Given that the ball is dropped from a height = 77 m

u = 0 m/s

s = 77 m

a = g = 9.81 m/s²

Applying the expression as:

$v^2-u^2=2as$

Applying values as:

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 77+0^2}\\\Rightarrow v=38.87\ m/s$

The speed with which the ball hit the ground = 38.87 m/s

3 0

3 years ago