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3 years ago

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What is the distance between a 900 kg compact car and a 1600 kg pickup truck if the gravitational force between them is about 0.

0001 N?

2 answers:

Sidana [21]3 years ago

5 0

Answer:

<em>The distance between the compact car and pickup truck is 0.96048 m</em>

Explanation:

The gravitational force is directly proportional to the product of the masses of the interacting object, it is also inversely proportional to the square of the distance between them. This is shown in equation 1;

$F =G \frac{m_{1} X m_{2} }{d^{2} }$ ............ 1

Where F is the gravitational force = 0.0001 N

G is the gravitational constant = 6.673 x $10^{-11} Nm^{2} kg^{-2}$

$m_{1}$ is the mass of the compact car = 900kg

$m_{2}$ is the mass of the pickup truck = 1600kg

d is the distance and its unknown ?

Let us make d the subject formula in equation 1

$d = \sqrt{G\frac{m_{1} m_{2} }{F } }$ .... 2

Substituting into equation 2 we have

$d = \sqrt{\frac{6.673x10^{-11} x 900 x 1600}{0.0001N} }$

d = 0.96048m

Therefore the distance between the compact car and pickup truck is 0.96048 m

elena-14-01-66 [18.8K]3 years ago

3 0

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

$r= \sqrt{\frac{Gm1m2}{F} }$

$r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m$

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required:

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

$r= \sqrt{\frac{Gm1m2}{F} }$

$r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m$

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Answer:

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Explanation:

Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;

v² = u² - 2·g·h

Where;

v = The final velocity

u = The initial velocity

g = The acceleration due to gravity ≈ 9.8 m/s²

h = The height she jumps

At the maximum height, $h_{max}$ = 0.500 m, she jumps, v = 0, therefore, we have;

0² = u² - 2·g· $h_{max}$

u² = 2 × 9.8 × 0.5 = 9.8

u = √9.8 ≈ 3.13

u = 3.13 m/s

Her initial jumping velocity ≈ 3.13 m/s

Escape velocity, $v_e = \sqrt{\dfrac{2 \cdot G \cdot M}{r} }$

Where;

M = The mass of the asteroid

G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

r = The radius of the asteroid

The average density of the Earth = 5515 kg/m³

The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³

The escape velocity, she has, $v_e$ ≈ 3.13 m/s is therefore;

$3.13 = \sqrt{\dfrac{2 \times 6.67408 \times 10^{-11} \times 5515 \times \frac{4}{3} \times \pi \times r^3}{r} } = r \times \sqrt{3.084 \times 10^{-6}}$

$r = \dfrac{3.13}{ \sqrt{3.084 \times 10^{-6}}} \approx 1782.45$

Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.

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An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

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Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

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Answer: A

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Explanation:

Using right hand rule, the magnetic force is perpendicular to the plane form by the magnetic field of a charged particle and its speed. Which will be into the screen.

But the negative charged particle moves in the opposite direction of the positive charged particle. Therefore, the magnetic force direction will be out of the screen

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Answer:

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Now the problem also states that the kinetic friction of the desk in this problem is 370 N

Since the pulley is smooth, the weight of Dwayne Johnson being transferred fully, and pulls the desk with a force of 1157.58 N. The frictional force of the desk is resisting this motion by a force of 370 N. Subtracting both forces we get the resultant force on the desk to be: 1157.58 - 370 = 787.58 N

Now lets use F = ma to calculate for the acceleration of the desk:

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$v^2 - u^2 = 2*a*s$

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a = 12.501 m/s

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Solving this we get:

$v^2 - 0 = 2 * 12.501 * 10$

$v = 15.812 m/s$

Since the desk and Mr. Dwayne Johnson are connected by a taught rope, they are travelling at the same speed. Thus, Dwayne also travels at 15.812 m/s when the desk reaches the window.

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