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Scrat [10]
3 years ago
8

Calculate earth's mass given the acceleration due to gravity at the north pole is 9.830 m/s2 and the mean radius of the earth at

the north pole is 6371 km.
Physics
1 answer:
valentinak56 [21]3 years ago
5 0

Answer: M = 5.98\times 10^{24} kg

Explanation:

We know that force acting on an object due to Earth's gravity on the surface is given by:

mg = G\frac{Mm}{r^2}\\ \Rightarrow g = \frac{GM}{r^2}

where g is the acceleration due to gravity, r would be radius of Earth, M is the mass of Earth and G is the gravitational constant.

It is given that at pole, g = 9.830 m/s² and r = 6371 km = 6371 × 10³ m

\Rightarrow M = \frac{g\times r^2}{G}

M = \frac{9.830 m/s^2 \times (6371 \times 10^3 m )^2}{6.67 \times 10^{-11} m^3 kg^{-1} s^{-2}}

\Rightarrow M = 5.98\times 10^{24} kg

Hence, Earth's mass is 5.98\times 10^{24} kg

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Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

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v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

Where:

v_{a}, v_{b} - Initial and final velocities, measured in meters per second.

t_{a}, t_{b} - Initial and final times, measured in seconds.

a(t) - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

v_{4} = 2\,\frac{m}{s}  +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt

v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)

v_{4} = -6\,\frac{m}{s}

Region II (t = 4 s to t = 6 s)

v_{6} = -6\,\frac{m}{s}  +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt

v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)

v_{6} = -4\,\frac{m}{s}

Region III (t = 6 s to t = 10 s)

v_{10} = -4\,\frac{m}{s}  +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt

v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)

v_{10} = 4\,\frac{m}{s}

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3 years ago
Suppose Joe, who can type 40 words per minute, takes 2.5 hours to type his report. To speed things up, his friend Mark offers to
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Explanation:

When Joe works alone, the total number  of words he typed can be given by:

Total words = (40 words per minute) x (60 minutes per hour) x (2.5 hours)

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Now, when Joe and Mark work together, let 'y' be the number of hours for which they both work simultaneously:

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The answer is false.

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We see from the formula that the force of gravity acting on an object depends on the mass: the larger the mass of the object, the stronger the gravitational force acting on it, and the smaller the mass, the weaker the force of gravity.

The factor that does not change is the acceleration of gravity, which is constant (9.8 m/s^2) if we are near the Earth's surface, and implies that all the objects in free fall accelerate at the same rate towards the ground, regardless of their size and weight.

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