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4 years ago

1. Kepler's 1st law: law of ellipses

Physics

2 answers:

kenny6666 [7]4 years ago

7 0

Here are the choices that fill in the blanks:
7.
1.
4.
6.
3.
2.
5.

poizon [28]4 years ago

5 0

Answer:

1.) Kepler's 1st law: law of ellipses

____The orbit of planets is not circular but elliptical.

2.) Kepler's 2nd law: law of equal areas

____For every action, there is an equal and opposite reaction.

3. Kepler's 3rd law: harmonic law

____Planets closer to the Sun orbit it in a shorter time period than planets further from the Sun.

4. Newton's 1st law of motion

____An imaginary line joining a planet to the Sun sweeps out equal areas in equal amounts of time.

5. Newton's 2nd law of motion

____Acceleration of an object depends on its mass and applied force.

6. Newton's 3rd law of motion

____Stationary objects resist movement, and moving objects resist stopping.

7. Newton's law of universal gravitation

____ The greater the mass, the stronger the gravitational force. The greater the distance, the weaker the weaker the gravitational force.

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3 years ago

The process of making alloys involves

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Answer:

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Explanation:

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3 years ago

Four examples of first class lever

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Answer:

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3 years ago

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.0 ft/s,

natka813 [3]

Answer:

The angle's rate of change is: -0.125 (degree/feet).

Explanation:

In this case of problem we need to find the angle's rate (α) of change when x=6 ft. First we need to relate (α) with x and y and the expression that do it is: $\alpha =Arctan(\frac{y}{x})$ where (α) is the angle between the ladder and the ground, x is the horizontal distance and y the vertical distance, now we need to have the variable y at function of x, so we can do it using the Pythagorean theorem and gets: $y^{2} +x^{2} =10^{2}$ solving for y(x) we get: $y(x)=\sqrt{100-x^{2}}$ . Replacing all we have got in the first equation: $\alpha =Arctan(\frac{\sqrt{100-x^{2} } }{x})$ . Finally we derivate this equation at function of variable x and gets this result: $\frac{d\alpha }{dx} =\frac{-1}{\sqrt{100-x^{2} } }$ evaluating at x=6 ft we get: -0.125(degree/feet). The negative signal means that the angle is decreasing.

4 0

3 years ago

Read 2 more answers

A beaker of mass 1.3 kg containing 2.8 kg of water rests on a scale. A 3.7 kg block of a metallic alloy of density 4600 kg/m3 is

Oksana_A [137]

1) Force read on the upper scale: 33.4 N

2) Force read on the lower scale: 43.0 N

Explanation:

The reading on the upper scale is equal to the net force acting on the block of metallic alloy. The net force is given by:

$F=W-B$ (1)

where

W is the weight of the block (downward)

B is the buoyant force (upward)

The weight of the block is given by

$W=mg$

where m = 3.7 kg is the mass of the block and $g=9.8 m/s^2$ is the acceleration of gravity.

The buoyant force is given by

$B=\rho_w V g$

where

$\rho_w=1000 kg/m^3$ is the water density

V is the volume of the block

The volume of the block can be written as

$V=\frac{m}{\rho_b}$

where $\rho_b=4600 kg/m^3$ is the density of the block.

Substituting everything into eq.(1), we find:

$F=mg-\rho_w \frac{m}{\rho_b}g=(3.7)(9.8)-(1000)\frac{1.3}{4600}(9.8)=33.4 N$

Here we want to find the force on the lower scale.

The force on the lower scale is equal to the difference between the total weight of the system (given by the weight of the beaker + the weight of the water + the weight of the block) and the upper net force exerted on the upper scale, therefore:

$F' = m_B g + m_wg+m_b g - F$