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3 years ago

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What is the gravitational force of attraction between a planet and a 17-kilogram mass that is falling freely toward the surface

of the planet at 8.8 meters per second squared?a. 150 Nb. 8.8 Nc. 1.9 Nd. 0.52 N

1 answer:

PolarNik [594]3 years ago

5 0

Answer:

a. 150 N

Explanation:

Gravitational Force: This is the force that act on a body under gravity.

The gravitational force always attract every object on or near the earth's surface. The earth therefore, exerts an attractive force on every object on or near it.

The S.I unit of gravitational force is Newton(N).

Mathematically, gravitational force of attraction is expressed as

(i) F = GmM/r² ........................ Equation 1 ( when it involves two object of different masses on the earth)

(ii) F = mg ............................... Equation 2 ( when it involves one mass and the gravitational field).

Given: m = 17 kg, g = 8.8 m/s²

Substituting into equation 2,

F = 17(8.8)

F = 149.6 N

F ≈ 150 N.

Thus the gravitational force = 150 N

The correct option is a. 150 N

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Find the position of the center of mass of the system of the sun and Jupiter? (Since Jupiter is more massive than the rest of th

8090 [49]

Answer:

$r_{cm} = 0.074 m$ from the position of the center of the Sun

Explanation:

As we know that mass of Sun and Jupiter is given as

$M_s = 1.98 \times 10^{30} kg$

$M_j = 1.89 \times 10^{27} kg$

distance between Sun and Jupiter is given as

$r = 7.78 \times 10^{11} m$

now let the position of Sun is origin and position of Jupiter is given at the position same as the distance between them

so we will have

$r_{cm} = \frac{M_s r_1 + M_j r_2}{M_s + M_j}$

$r_{cm} = \frac{1.98 \times 10^{30} (0) + (1.89 \times 10^{27})(7.78 \times 10^{11})}{1.98 \times 10^{30} + 1.89 \times 10^{27}}$

$r_{cm} = 0.074 m$ from the position of the center of the Sun

3 0

3 years ago

A 0.10 g honeybee acquires a charge of +23 pC while flying.

kari74 [83]

Answer:

a) $\frac{F}{w} =2.347\times 10^{-6}\ N$

b) $E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, $m=10^{-4}\ kg$

charge acquired by the bee, $q_2=23\times 10^{-12}\ C$

a.

Electrical field near the earth surface, $E=100\ N.C^{-1}$

Now the electric force on the bee:

we know:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}$

$F=E.q_2$

$F=100\times 23\times 10^{-12}$

$F=23\times 10^{-10}\ N$

The weight of the bee:

$w=m.g$

$w=10^{-4}\times 9.8$

$w=9.8\times10^{-4}\ N$

Therefore the ratio :

$\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}$

$\frac{F}{w} =2.347\times 10^{-6}\ N$

b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

$F=9.8\times10^{-4}\ N$

$E.q_2=9.8\times10^{-4}\ N$

$E\times 23\times 10^{-12}=9.8\times10^{-4}\ N$

$E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

3 0

3 years ago

Express the distance between the moon and earth in meters with a meter prefix (km).

tankabanditka [31]

Answer;

= 3.86 × 10^8 Meters

Explanation;

-The distance between the Earth and the moon is 386000 km

But; 1 km = 1000 m

Therefore; 386000 km will be equivalent to;

= 386000 × 1000

= 386000000 m

= 3.86 × 10^8 meters

3 0

3 years ago

Read 2 more answers

A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the pl

Firdavs [7]

Answer:

d=0.137 m ⇒13.7 cm

Explanation:

Given data

m (Mass)=3.0 kg

α(incline) =34°

Spring Constant (force constant)=120 N/m

d (distance)=?

Solution

F=mg

F=(3.0)(9.8)

F=29.4 N

As we also know that

Force parallel to the incline=FSinα

F=29.4×Sin(34)

F=16.44 N

d(distance)=F/Spring Constant

d(distance)=16.44/120

d(distance)=0.137 m ⇒13.7 cm

4 0

3 years ago

Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

Nuetrik [128]

Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(c) $A = 5.51 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

(c) For face centered cubic lattice:

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

6 0

3 years ago