Why in example multiplying the length by 2

Calculate the force of gravity on the 0.60-kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth r

KIM [24]

The force of gravity between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the mass of the object is $m_1=0.60 kg$ , while the Earth's mass is $m_2=5.97 \cdot 10^{24} kg$ . Their separation is $r=1.3 \cdot 10^7 m$ , therefore the gravitational force exerted on the object is
$F=(6.67 \cdot 10^{-11}m^3 kg^{-1} s^{-2}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}=1.4 N$

5 0

3 years ago

The lowest surface temperature in the solar system (-200°c occurs on

zimovet [89]

The lowest surface temperature in the solar system was recorded on Uranus (-224 degrees Celsius). The temperature of a planet does not only depend on the amount of solar radiation that it receives but also on the amount of heat that it gives off. Because of Uranus' orientation it absorbs little radiation which makes it colder than Neptune although Neptune is further away from the Sun. <span />

3 0

3 years ago

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$