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3 years ago

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What is force derived from

1 answer:

Nata [24]3 years ago

6 0

Answer:

Force can also be described intuitively as a push or a pull. ... It is measured in the SI unit of newtons and represented by the symbol F. The original form of Newton's second law states that the net force acting upon an object is equal to the rate at which its momentum changes with time.

Explanation:

hope this helps : )

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a family backpacking at Yosemite national park took 5 hours to climb a mountain trail 7.5 km long. what was the family's average

nirvana33 [79]

Answer:

The average speed will be 1.5km/hr .

Explanation:

Distance need to be traveled = 7.5Km

Time taken = 5 hours

Average speed refers to total distance traveled with respect to total time taken .

It can be calculated as given below :

Average speed =total distance /total time taken

Substituting the values we get ,as shown below

Average speed =7.5/5=75/50

Average speed = 1.5 Km/hr

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A material's ability to allow heat or electricity to flow through it...

NikAS [45]

Answer:

a

Explanation:

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3 years ago

The equation that is used to solve second law problems is # F= ma.

maw [93]

F= Force
M=Mass
A= acceleration
F=N
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A= m/s

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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