At what condition does a body becomes weightless at the equator?

Physics

1 answer:

ArbitrLikvidat [17]3 years ago

4 0

Answer:

The decrease is due to the bulge at the equator (putting more distance between the rest of the planet and the surface

Explanation:

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Mr. Ben drove from Town A to Town B. For the first 3 h, he traveled at an average speed of

Vanyuwa [196]

Answer:

It’s 7 hours

Explanation:

You have to use the formula your teacher has given to you plug in the numbers then solve be sure to use a calculator made for physics it helps a lot :)

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3 years ago

What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?

sweet [91]

Answer:

The new force becomes 4 times the initial force.

Explanation:

The force of attraction or repulsion is given by the relation as follows :

$F=k\dfrac{q_1q_2}{d^2}$

Where

d is the distance between the interacting charges

F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

$F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F$

So, the new force becomes 4 times the initial force.

4 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

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