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3 years ago

6

2. What gravitational attraction does the moon have on the earth? (mass of earth is 6 x

1 answer:

lions [1.4K]3 years ago

7 0

Answer:

Given that,

Mass of the Earth m

1

=6×10

24

Kg

Mass of the Moon m

2

=7.4×10

22

kg

Distance between the Earth and the Moon d=3.84×10

5

km=3.84×10

8

m

Gravitational Constant G=6.7×10

−11

Nm

2

/kg

2

Now, by using Newton’s law of gravitation

F=

r

2

Gm

1

m

2

F=

(3.84×10

8

)

2

6.7×10

−11

×6×10

24

×7.4×10

22

F=

14.8225×10

16

297.48×10

35

F=20.069×10

19

F=20.1×10

19

N

Hence, the gravitational force of attraction is 20.1×10

19

N

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If the statement is true, select True. If it is false, select False.

RUDIKE [14]

Answer:

false

Explanation:

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3 years ago

Why do temperatures on the moons surface vary ncyb more than temperatures on earth

ollegr [7]

Answer:

The moon has no atmosphere

Explanation:

The temperatures on the surface of the Moon vary much more than those on Earth because the moon has no atmosphere (third answer in the list), and therefore there are no molecules that could retain residual heat and make the change from day to night a softer transition.

5 0

3 years ago

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

3 0

3 years ago

A student of mass 40kg takes 10s to run up a flight of 50steps. If each step is 15cm high, calculate the Potential Energy of the

astra-53 [7]

Answer:

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Explanation:

Potential energy = mass x acceleration due to gravity x height in meters

(P.E. = mgh)

Substitute the given numbers: (I take acceleration due to gravity as 9.81 m s^-2)

PE = 40 x 9.81 x (0.15x50)

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Luda [366]

The gravitational force will double as well because
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7 0

3 years ago