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3 years ago

A trampoline has a spring constant of 3430 N/m how far will the trampoline sink when a 70 kg person steps on it

Physics

2 answers:

aalyn [17]3 years ago

4 0

Answer:

20 cm

Explanation:

Force = mass * acceleration F = ma (1)

Spring equation is given as Force = Spring constant*Distance (2)

combine 1 & 2

F = ma = Spring constant*Distance

70 * 9.81 = 3430 * d

d = 0.2 m or 20 cm

Dominik [7]3 years ago

3 0

F = ma = -kx

a = 9.81 m/s²
k = 3430 N/m
m = 70 kg

x = - ma/ k = 0.2m

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Calculate the force of gravity between planet X and planet y if both planets are 3.75 X 10^11 m apart, planet X has a mass of 1.

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So, the force of gravity that the asteroid and the planet have on each other approximately $\boxed{\sf{2.9 \times 10^{17} \: N}}$

<h3>Introduction</h3>

Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :

$\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}$

With the following condition :

F = gravitational force (N)
G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
$\sf{m_1}$ = mass of the first object (kg)
$\sf{m_2}$ = mass of the second object (kg)
r = distance between two objects (m)

<h3>Problem Solving</h3>

We know that :

G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
$\sf{m_X}$ = mass of the planet X = $\sf{1.55 \times 10^{22}}$ kg.
$\sf{m_Y}$ = mass of the planet Y = $\sf{3.95 \times 10^{28}}$ kg.
r = distance between two objects = $\sf{3.75 \times 10^{11}}$ m.

What was asked :

F = gravitational force = ... N

Step by step :

$\sf{F = G \times \frac{m_X \times m_Y}{r^2}}$

$\sf{F = 6.67 \cdot 10^{-11} \times \frac{1.55 \cdot 10^{22} \cdot 3.95 \times 10^{28}}{(3.75 \times 10^{11})^2}}$

$\sf{F \approx \frac{40.84 \times 10^{-11 + 22 + 28}}{14.0625 \times 10^{22}}}$

$\sf{F \approx 2.9 \times 10^{39 - 22}}$

$\sf{F \approx 2.9 \times 10^{17} \: N}$

<h3>Conclusion</h3>

So, the force of gravity that the asteroid and the planet have on each other approximately

$\boxed{\sf{2.9 \times 10^{17} \: N}}$

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