1) the weight of an object at Earth's surface is given by

, where m is the mass of the object and

is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is

2) On Mars, the value of the gravitational acceleration is different:

. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth:

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as

<span>6) On Earth, the gravity acceleration is </span>

<span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
</span>

<span>
</span>
Answer:
Fy=107.2 N
Explanation:
Conceptual analysis
For a right triangle :
sinβ = y/h formula (1)
cosβ = x/h formula (2)
x: side adjacent to the β angle
y: opposite side of the β angle
h: hypotenuse
Known data
h = T = 153.8 N : rope tension
β= 44.2°with the horizontal (x)
Problem development
We apply the formula (1) to calculate Ty : vertical component of the rope force.
sin44.2° = Ty/153.8 N
Ty = (153.8 N ) *(sen44.2°)= 107.2 N directed down
for equilibrium system
Fy= Ty=107.2 N
Fy=107.2 N upward component of the force acting on the stake
Answer:
El punto de referencia más usado es el NORTE. Los puntos de referencia sirven para orientarnos, de esta manera podemos saber nuestra ubicación y luego desplazarnos. En este caso tenemos que el punto cardinal NORTE es el más usado, se tiene como referencia universal para lograr ubicarnos.
Explanation:
gracias