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2 years ago

Which of the following is NOT an example of the 3rd law of motion (for every Action, there is an equal but opposite reaction?

Physics

2 answers:

Vlad1618 [11]2 years ago

6 0

Answer:

C a basketball player pushes into another one and they both fall to the left

Explanation:

I believe his is the answer because I don't see any force and not enough reaction

igor_vitrenko [27]2 years ago

4 0

The answer is C, or A basketball player pushes into another one and they both fall to the left.

Explanation:

Isaac Newton’s third law states that for every action, there is an equal and opposite reaction. In this instance, there is no reaction in opposition to the basketball players being pushed into each other. Both players fall to the left, and this means that there is no equal and opposite reaction. Therefore, this example does not display Newton’s third law of motion.

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A lamp draws a current of 0.50 A when it is connected to a 120 V source? What is the resistance of the lamp?

emmainna [20.7K]

Given,
Current (I) = 0.50A
Voltage (V) = 120 volts
Resistance (R) =?
We know that:-
Voltage (V) = Current (I) x Resistance (R)
→Resistance (R) = Voltage (V) / Current (I)
= 120/0.50
= 24Ω
∴ Resistance (R) = 24Ω

8 0

3 years ago

Read 2 more answers

Why are there only two elements in the first period of the periodic table?(1 point)

svetlana [45]

Answer:

because only two electrons can fit in the first orbit around the nucleus, and each period on the table is organized by number of orbits

5 0

3 years ago

Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

3 years ago

On both the spring and fall equinoxes, the sun's rays are vertically overhead at the _______

Anni [7]

It is overhead at the equator, it is because the sun ray’s will be moving vertically as this will be directed at the equator. It is because if it moves vertically, it will hit or overhead the equator and this usually happens in spring and fall.

5 0

3 years ago

A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra

blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

$y=\sqrt{1+x^3}$