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3 years ago

A runner of 60kg accelerates at 2.0 m/s at the start of race calculate the force provided from her legs

Physics

1 answer:

vfiekz [6]3 years ago

4 0

Answer:

120 kgm/s

Explanation:

Force= mass× acceleration

= 60kg × 2.0m/s

= 120 kgm/s

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Consider a horizontal, uniform board of weight 125 N and length 4 m that is supported by vertical chains at each end. A person w

Norma-Jean [14]

Answer:

b. 375 N

Explanation:

System of forces in balance

ΣFy = 0 Equation (1)

Forces acting on the board

T₁: Tension in the left chain , vertical and upward

T₂ = 250 N : Tension in the right chain , vertical and upward

W₁ = 125 N : Weight of the board , vertical and downward

W₂ = 500 N : Weight of the person , vertical and downward

Calculation of the T₁

We apply the Equation (1)

ΣFy = 0

T₁+T₂-W₁-W₂ = 0

T₁ = -T₂+W₁+W₂

T₁ = -250 N+ 125 N+ 500 N

T₁ = 375 N

5 0

3 years ago

Most of the funding for research comes from the federal government or ? And is provided to Principal Investigators (PIs) through

densk [106]

Answer: is a term generally covering any funding for scientific research, in the areas of both "hard" science and technology and social science. The term often connotes funding obtained through a competitive process, in which potential research projects are evaluated and only the most promising receive funding. Such processes, which are run by government, corporations or foundations, allocate scarce funds.

Explanation:

7 0

2 years ago

The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

$m$ - Mass of the person, measured in kilograms.

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

$F = m \cdot g$

Where:

$m$ - Mass of the person, measured in kilograms.

$g$ - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is: