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3 years ago

All objects are in motion relative to the sun. a. True b. False

1 answer:

6 0

That depends on what "objects" refer to.
If "objects" refer to the ones on Earth, then it is TRUE. These objects are in motion relative to the sun same as with Earth itself.

If "objects" refer to the ones outside earth, then it may be a TRUE or FALSE depending on how far or near they are from the sun.

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Two sacks contain the same number of identical apples and are separated by a distance r. The two sacks exert a gravitational for

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Explanation:

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2 years ago

The distance between two electric pole is 800m then express this in terms of mm

Lana71 [14]

Answer:

Mm stands for milimetres

Explanation:

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2 years ago

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The velocity of a ball changes from ‹ 9, −6, 0 › m/s to ‹ 8.96, −6.12, 0 › m/s in 0.02 s, due to the gravitational attraction of

Alenkasestr [34]

Answer:

a) $a=(-2,-6,0)m/s^2$ , with a magnitude of $6.3m/s^2$

b) $\frac{\Delta p}{\Delta t}=(-0.24,-0.72,0)Kgm/s^2$ , with a magnitude of $0.76Kgm/s^2$

c) $F=(-0.24,-0.72,0)N$ , with a magnitude of $0.76N$

Explanation:

We have:

$v_{ix}=9m/s, v_{iy}=-6m/s, v_{iz}=0m/s\\v_{fx}=8.96m/s, v_{fy}=-6.12m/s, v_{fz}=0m/s\\t=0.02s, m=0.12Kg$

We can calculate each component of the acceleration using its definition $a=\frac{\Delta v}{\Delta t}$

$a_x=\frac{v_{fx}-v_{ix}}{t} = \frac{(8.96m/s)-(9m/s)}{0.02s} =-2m/s^2\\a_y=\frac{v_{fy}-v_{iy}}{t} = \frac{(-6.12m/s)-(-6m/s)}{0.02s} =-6m/s^2\\a_y=\frac{v_{fz}-v_{iz}}{t} = \frac{(0m/s)-(0m/s)}{0.02s} =0m/s^2\\$

The rate of change of momentum of the ball is $\frac{\Delta p}{\Delta t} = \frac{\Delta mv}{\Delta t} = \frac{m\Delta v}{\Delta t} = ma$

So for each coordinate:

$\frac{\Delta p_x}{\Delta t}=-0.24Kgm/s^2\\\frac{\Delta p_y}{\Delta t}=-0.72Kgm/s^2\\\frac{\Delta p_z}{\Delta t}=0Kgm/s^2$

And these are equal to the components of the net force since F=ma.

If magnitudes is what is asked:

$a=\sqrt{a_x+a_y+a_z} =6.3m/s^2\\F=ma=\frac{\Delta p}{\Delta t}=0.76N$

(N and $Kgm/s^2$ are the same unit).

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2 years ago

What is the differential equation governing the growth of current in the circuit as a function of time after t=0? express the ri

Reika [66]

Answer:

$v_{b}=ir+L\frac{di}{dt}$

Explanation:

A differential equation that contain a term with di(t)/dt is in a RL circuit. Here we have

$v_{b}=v_{r}+v_{i}$

where vr is the voltage in the resistance, vi is the voltage in the inductance and vb is the source voltage. But also we have that

$v_{r}=ir\\v_{i}=L\frac{di}{dt}$

where L is the inductance of the circuit, r is the resistance an i is the current. By replacing we have the differential equation

$v_{b}=ir+L\frac{di}{dt}$

I hope this is useful for you

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2 years ago

Please help, What does a Transverse and longitude wave combine to form?

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3 years ago

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