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3 years ago

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Objects 1 and 2 attract each other with a gravitational force of 72.0 units. If the mass of Object 1 is one-fourth the original

value AND the mass of object 2 is tripled AND the distance separating Objects 1 and 2 is halved, then the new gravitational force will be _____ units.

1 answer:

Black_prince [1.1K]3 years ago

6 0

Answer:

New gravitational force, F' = 216 units

Explanation:

Given that,

Gravitational force between object 1 and 2, F = 72 N

The gravitational force between two objects is given by :

$F=\dfrac{Gm_1m_2}{r^2}$

$\dfrac{Gm_1m_2}{r^2}=72$ ............(1)

If the mass of object 1, $m_1'=\dfrac{m_1}{4}$

$m_2'=3m_2$

If the distance between objects is halved, $r'=\dfrac{r}{2}$

Let new gravitational force is F'. It is changed to :

$F=\dfrac{Gm'_1m'_2}{r'^2}$

$F'=\dfrac{G\times \dfrac{m_1}{4}\times 3m_2}{(\dfrac{r}{2})^2}$

$F'=3\times \dfrac{Gm_1m_2}{r^2}$

$F'=3\times F$

$F'=3\times 72$

F' = 216 units

So, the new gravitational force will be 216 units.

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umka2103 [35]

Answer:

70 N

21°

1.1 m/s²

Explanation:

Draw a free body diagram of the block. There are three forces:

Weight pulling straight down

Normal force pushing perpendicular to the incline

Friction force pushing parallel to the incline

Part 1

Sum the forces in the perpendicular direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

The block is at rest, so F = N μs:

F = N μs

F = mg μs cos θ

F = (20 kg) (9.8 m/s²) (0.38) (cos 19°)

F = 70 N

Part 2

Sum the forces in the parallel direction (down the incline is positive):

∑F = ma

mg sin θ − F = 0

mg sin θ = N μs

mg sin θ = mg μs cos θ

tan θ = μs

θ = atan μs

θ = atan 0.38

θ = 21°

Part 3

Sum the forces in the parallel direction (this time, acceleration is not 0).

∑F = ma

mg sin θ − F = ma

mg sin θ − N μk = ma

mg sin θ − mg μk cos θ = ma

a = g (sin θ − μk cos θ)

a = (9.8 m/s²) (sin 24° − 0.32 cos 24°)

a = 1.1 m/s²

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3 years ago

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allsm [11]

Answer:

greater than

Explanation:

the answer is greater than

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3 years ago

Which statement is accurate about seasonal temperatures in North carolina

Alex

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3 years ago

Two objects of same material are travelling near you. Object A is a 1.9 kg mass traveling 8 m/s; object B is a 2 kg mass traveli

Alex73 [517]

To determine the object which could give the greatest impact we will apply the concept of momentum. The object that has the highest momentum will be the object that will impact the strongest. Our values are

Mass of Object A

$m_A=1.9 kg$

Velocity of object A

$v_A=8ms$

Mass of object B

$m_B=2 kg$

Velocity of object B

$v_B=5ms$

The general formula for momentum is the product between mass and velocity, then

$p = mv$

For each object we have then,

$p_A=m_Av_A=1.9 kg(8ms)=15.2kg \cdot m/s$

$p_B=m_Bv_B=2 kg(5ms)=10kg \cdot m/s$

Since the momentum of object A is greater than that of object B, then object A will make you feel force upon impact.

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3 years ago

On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s

Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

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3 years ago