Which of the following is a metal?<br> Calcium (Ca)<br> Iron (Fe)<br> Sodium (Na)<br> all of these

Which best explains how an object at rest deep in space and far from any massive body behaves compared to an object in free fall

7nadin3 [17]

Answer: They behave the same because, according to the principle of equivalence, the laws of physics work the same in all frames of reference.

Explanation:

According to the equivalence principle postulated by Einstein's Theory of General Relativity, acceleration in space and gravity on Earth have the same effects on objects.

To understand it better, regarding to the equivalence principle, Einstein formulated the following:

A gravitational force and an acceleration in the opposite direction are equivalent, both have indistinguishable effects. Because the laws of physics must be accomplished in all frames of reference.

Hence, according to general relativity, gravitational force and acceleration in the opposite direction (an object in free fall, for example) have the same effect. This makes sense if we deal with gravity not as a mysterious atractive force but as a geometric effect of matter on spacetime that causes its deformation.

3 0

3 years ago

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis

Bumek [7]

Answer:

a)3.5s

b)28.57m/S

c)34.33m/S

d)44.66m/S

Explanation:

Hello!

we will solve this exercise numeral by numeral

a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

$Y= VoT+0.5gt^{2}$

where

Vo = Initial speed =0

T = time

g=gravity=9.81m/s^2

y = height=60m

solving for time

$Y=0.5gt^2\\t=\sqrt{\frac{Y}{0.5g} } \\t=\frac{60}{0.5(9.81)} \\$

T=3.5s

b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

$V=\frac{x}{t} =\frac{100}{3.5}=28.57m/s$

c)

to find the final vertical velocity we use the equations for motion with constant velocity as follows

Vf=Vo+g.t

Vf=0+(9.81 )(3.5)=34.335m/S

d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

$V=\sqrt{Vx^2+Vy^2} =\sqrt{34.33^2+28.57^2} =44.67m/S$