Acceleration/gravity problem?

Question

You place a 7.50 kg tv on a spring scale. If the scale reads 78.4N, what is the acceleration due to gravity at that location?

Like what equation do you use?

F=ma


so...


78.4N=7.5kg(a)

=10.5?

Answer 1

This is best done by ratios. Let W = mG equal the weight of mass k = 7.5 kg on Earth's surface and G = 9.81 m/sec^2. Let w = mg = 78.4 N be the weight of your mass m on some other location.

Thus, w/W = 78.4/(7.5*9.81) = mg/mG = g/G; so that g = 9.81 (78.4/73.6) = 10.4 m/sec^2 at the new location.

Using ratios is always a good starting point when comparing like characteristics (e.g., the g values) under different conditions. This works because many time a lot of factors (like m) cancel out in the ratio.

Answer 2

this could be a query that relates time, speed, acceleration and relativity. interior the 1st project the two articles start to enhance up on the comparable time and for this reason their velocities at a given ingredient in time will additionally be comparable. for this reason relative to a minimum of one yet another, there will be no speed distinction. for this reason the gap between the two bodies will proceed to be comparable. interior the 2d occasion the bodies initiate from the comparable place yet at diverse situations. At any given occasion, their relative velocities will continuously be diverse via fact the 1st ball having been sped up over a larger quantity of time, will continuously be at a larger speed. subsequently the distances between the two bodies will proceed to extend.

Answer 3

Your formula and answer 10.5 m/s² are correct.
Double check input data, though.
They probably tried to trick you.

Answer 4

F = ma

Answer 5

f=m*a
78.4=7.5*a