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schepotkina [342]
3 years ago
9

Drag the words in the left-hand column to the appropriate blanks in the right-hand column. Reset Help Newton's first law of moti

on the orbit of each planet about the Sun is an ellipse with the Sun at one focus. Newton's second law of for any force, there is an equal and opposite reaction force. motion Newton's third law of motion a planet moves faster in the part of its orbit nearer the Sun and slower when farther from the Sun, sweeping out equal areas in equal times. Kepler's first law of planetary motion force = mass x acceleration Kepler's second law of planetary motion more distant planets orbit the Sun at slower average speeds, obeying the precise Kepler's third law of mathematical relationship p2 a3 planetary motion an object moves at constant velocity if there is no net force acting upon it
Physics
1 answer:
Tom [10]3 years ago
7 0

Explanation:

In this case, we need to write correct statement about the topics :

1. Kepler's first law of planetary motion = the orbit of each planet about the Sun is an ellipse with the Sun at one focus.

2. Newton's third law of motion = for any force, there is an equal and opposite reaction force.

3. Kepler's second law of planetary motion = a planet moves faster in the part of its orbit nearer the Sun and slower when farther from the Sun, sweeping out equal areas in equal times.

4. Newton's second law of motion = force = mass x acceleration

5. Kepler's third law = more distant planets orbit the Sun at slower average speeds obeying the mathematical relation, p^2=a^3

6. Newton's second law of motion = an object moves at constant velocity if there is no net force acting upon it

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A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is
sp2606 [1]

Answer:

see that the correct one is B

Explanation:

To solve this exercise let us use the kinematic relations

           v² = v₀² - 2 a x

as they indicate that the car stops, therefore the final speed is yield v = 0

          x = v₀² / 2a

let's calculate

          x = 2²/(2 0.8)

         x = 2.5 m / s²

When reviewing the answers we see that the correct one is B

7 0
3 years ago
Residential building codes typically require the use of 12-gauge copper wire (diameter 0.205 cm) for wiring receptacles. Such ci
Alecsey [184]

Given Information:  

Current = I = 20 A

Diameter = d = 0.205 cm = 0.00205 m

Length of wire = L = 1 m

Required Information:  

Energy produced = P = ?

Answer:  

P = 2.03 J/s

Explanation:  

We know that power required in a wire is

P = I²R

and R = ρL/A

Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m

L is the length of the wire and A is the area of the cross-section and is given by

A = πr²

A = π(d/2)²

A = π(0.00205/2)²

A = 3.3x10⁻⁶ m²

R = ρL/A

R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶

R = 5.09x10⁻³ Ω

P = I²R

P = (20)²*5.09x10⁻³

P = 2.03 Watts or P = 2.03 J/s

Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A

8 0
4 years ago
Q5- Knowing now that your mood might affect your memory, what strategies could you employ to assist your memories and make sure
Tema [17]
You can try recalling and rehearsing information about the memory.
4 0
2 years ago
The force that pulls falling objects toward Earth is called *
larisa [96]

Answer:

Gravity

Explanation:

That's easy because gravity is the only thing that can pull us down that hard at that fast without anything helping it.

5 0
3 years ago
The water line from the street to my house is 1 inch diameter and made of PVC (i.e. smooth). The line is roughly 450 ft long. Th
jekas [21]

Answer:

The right solution is "126 Psi".

Explanation:

The given values are:

P₁ = 130 psig

i.e.,

   = 130\times 6.894

   = 896.22 \ Kpa

or,

   = 896.22\times 10^3 \ Pa

Z₂ = 10ft

    = 3.05 m

\delta = 1000 kg/m³

According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒  \frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2

On substituting the values, we get

⇒  \frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2

⇒  \frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05

⇒  P_2=866330 \ P_a

i.e.,

⇒       =866330\times 0.000145

⇒       =126 \ Psi

8 0
3 years ago
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