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3 years ago

When measuring the distance between two objects to find universal gravitation where must you measure from?

1 answer:

6 0

The distance is measured between the 'centers of mass',
or 'centers of gravity' of both objects. That's the point inside
the object where there's the same amount of mass in every
direction away from it.

If the object is something like a sphere, or a cube, or a uniform
wooden board, or anything "symmetrical", then the center of mass /
center of gravity is the actual center of the object. If the object has
some irregular shape, then it's harder to calculate or measure where
its center is.

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Identify two factors that determine the intensity of sound

anastassius [24]

Amplitude and distance.

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3 years ago

An object has an acceleration of 12.0 m/s/s. The mass of the object is doubled while the net force on the object is held constan

Aloiza [94]

Answer:

The net force acting on the object is doubled while the mass of the object is held constant. What will be the new acceleration? An object has an acceleration of 12.0 m/s^2. The net force acting on the object is halved (decreased to one half its original value) while the mass of the object is held constant.

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3 years ago

What will a spring scale read for the weight of a 75.0-kg woman in an elevator that moves upward with constant speed of 5.8 m/s

Sholpan [36]

<span>On the scale the only external forces are the man's weight acting downwards and the normal force which the scale exerts back to support his weight. So F = Ma = mg + Fs The normal force Fs (which is actually the reading on the scale) = Ma + Mg But a = 0 So Fs = Mg which is just his weight. Fs = 75 * 9.8 = 735N</span>

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4 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

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3 years ago

You have 2 minutes to get to PE from science class before you get a tardy. If PE is 100m away and you walk at a speed of 1.1m/s

mart [117]

Answer:

Explanation:

4 0

3 years ago

Read 2 more answers