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3 years ago

Galileo held that measurements of time were _ and that measurements of distance were __. absolute; relative

Physics

1 answer:

DanielleElmas [232]3 years ago

6 0

Answer:

option A

Explanation:

the correct answer is option A

The given blanks in the question will be filled with Absolute and relative receptively.

time is absolute means the numerical representation of time example 12:00 pm is absolute time.

Measurement of distance is relative. A relative distance is one which is represented with the reference of some other place, it is not an exact value.

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If Scoobie could drive a Jetson's flying car at a constant speed of 160.0 km/hr across oceans and space, approximately how long

lara [203]

Your first step is to find the circumference of the earth, with the numbers given. You can do that by putting the radius of 6200 kilometres into the 2πr equation. That should get you a circumference of 12400π, or about 38,955.75 kilometres.

Next, you can use the rate the Jetson's car is going (180km/h) and divide the 38,955.75 by it to see how many hours it would take at that constant speed.

38,955.75 / 180 = 216.42 hours

Then you can divide that by 24 to get how many days

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3 years ago

Anyone please ??? ASAP 15 points ???

Fed [463]

Answer:

1) you could put more force behind it. (increase) 2) have another object interact with that object. (increase or decrease) 3) the object could hit a wall and stop or slow down (decrease)

Sorry if wrong

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3 years ago

What is the electric flux φ through each of the six faces of the cube? Use ϵ0 for the permittivity of free space?

exis [7]

If total charge Q is enclosed in the surface of cube

then we will have say that total flux linked with all surfaces of the cube will be given by

$\phi = \frac{Q}{\epsilon_0}$

since the position of charge is symmetric with respect to the center of the cube so here this whole flux will be equally linked with each of the face of the cube

So here total 6 faces of the cube is there and the total flux of the charge will equally divide in these 6 faces

so flux linked with each face is given by formula

$\phi = \frac{Q}{6\epsilon_0}$

so each face will have above flux due to central position of charge Q

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3 years ago

Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much l

alex41 [277]

a) 4F0

b) Speed of planet B is the same as speed of planet A

Speed of planet C is twice the speed of planet A

Explanation:

The magnitude of the gravitational force between two objects is given by the formula

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the 2 objects

r is the separation between the objects

For the system planet A - Star A, we have:

$m_1=M_p\\m_2 = M_s\\r=R$

So the force is

$F_A=G\frac{M_p M_s}{R^2}=F_0$

For the system planet B - Star B, we have:

$m_1 = 4 M_p\\m_2 = M_s\\r=R$

So the force is

$F=G\frac{4M_p M_s}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C - Star C, we have:

$m_1 = M_p\\m_2 = 4M_s\\r=R$

So the force is

$F=G\frac{M_p (4M_s)}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

$G\frac{mM}{r^2}=m\frac{v^2}{r}$

where

m is the mass of the planet

M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

$v=\sqrt{\frac{GM}{r}}$

For System A,

$M=M_s\\r=R$

So the tangential speed is

$v_A=\sqrt{\frac{GM_s}{R}}$

For system B,

$M=M_s\\r=R$

So the tangential speed is

$v_B=\sqrt{\frac{GM_s}{R}}=v_A$

So, the speed of planet B is the same as planet A.

For system C,

$M=4M_s\\r=R$

So the tangential speed is

$v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A$

So, the speed of planet C is twice the speed of planet A.

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3 years ago

If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal

iVinArrow [24]

Answer:

xcritical = d− m1 /m2 ( L /2−d)

Explanation: the precursor to this question will had been this

the precursor to the question can be found online.

ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)

. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces

smallest possible value of x such that the bar remains stable (call it xcritical)

∑τA = 0 = m2g(d− xcritical)− m1g( −d)

xcritical = d− m1 /m2 ( L /2−d)

6 0

3 years ago

Galileo held that measurements of time were _____ and that measurements of distance were ______. absolute; relative

Galileo held that measurements of time were _ and that measurements of distance were __. absolute; relative