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3 years ago

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a rock dropped on the moon will increases its speed from 0 m/s (its starting speed when first dropped) to 4.9 m/s in about 3 sec

onds. What value do your calculate for the magnitude of the acceleration of gravity on the moon?

1 answer:

Tanzania [10]3 years ago

5 0

Answer:

a = 1.63 m/s²

Explanation:

Given in the question that,

initial speed of the rock = u = 0 m/s

final speed of the rock = v = 4.9 m/ s

time taken = t = 3 sec

We will use Newton's law of motion

Formula to use to calculate the acceleration of gravity on moon

<h3>v - u = a(t)</h3>

4.9 - 0 = a(3)

4.9 = a(3)

a = 4.9/3

a = 1.63 m/s²

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Vector ~A has a negative x-component 3.07 units in length and a positive y-component 3.17 units in length. When a vector ~B = b1

luda_lava [24]

Answer:

a. 3.07 b. 1.26

Explanation:

Given that A = -3.07i + 3.17j and B = b1i + b1j and C = A + B = 0i + 4.43j

Since A + B = -3.07i + 3.17j + b1i + b2j

= (-3.07 + b1)i + (3.17 + b2)j

So,(-3.07 + b1)i + (3.17 + b2)j = 0i + 4.43j

Comparing components,

-3.07 + b1 = 0 (1) and 3.17 + b2 = 4.43 (2)

a. From (1), b1 = 3.07

b. From(2) b2 = 4.43 - 3.17 = 1.26

4 0

3 years ago

Sarah and Maisie are analysing data from their school sports day. Looking at the 1500 m results for Stephen, Maisie believes tha

Dahasolnce [82]

Answer:

Sarah is right

Explanation:

This is an exercise that differentiates between scalars and vectors.

A scalar is a number, instead a vector is a number that represents the module in addition to direction and sense.

In this case, the distance (scalar) traveled is a number, which is why it is worth 1500m, but the displacement is a vector and since the point where it leaves is the same point where the vector's modulus arrives is zero, so the DISPLACEMENT VECTOR is zero

consequently Sarah is right

4 0

3 years ago

Just wanna make sure im right

NikAS [45]

It is right have a good day

explanation step by step

6 0

3 years ago

Which of the following is not an example of osmosis? Multiple Choice

NNADVOKAT [17]

Answer:

Sugar in coffee dissolves.

Explanation:

In this case we have a solution, which is defined as separating what was bound in some way, by homogeneously mixing the molecules of a substance into a liquid. Therefore, it is the resulting homogeneous mixture after dissolving any substance in a liquid.

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3 years ago

Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

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3 years ago