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bija089 [108]
2 years ago
10

What is the action force and the reaction force when you sit down on a chair

Physics
2 answers:
VladimirAG [237]2 years ago
7 0

Your weight pushing down on the chair is the action force. The reaction force is the force exerted by the chair that pushes up on your body.

press thanks if thankful

Harlamova29_29 [7]2 years ago
3 0
Your weight pushing down on the chair is the action force. The reaction force is the force exerted by the chair that pushes up on your body.
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kirill115 [55]

Answer:

8100 g

Explanation:

8.1 kg × 1000

= 8100 g

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Solve for M₂
soldi70 [24.7K]

Explanation:

M₂ = Fr²/GM₁

M₂ = [(132N)(.243m)²]/[(6.67*10^-11N*m²/kg)(1.175*10^4kg)]

M₂ = (7.79N*m²)/(7.84*10^-7N*m²)

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As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes thro
LiRa [457]

Answer:

x = t

y = \frac{1}{3}t

z =t

Explanation:

Given

r(t) = f(t)i + g(t)j + h(t)k at t = 0

Point: (f(t0), g(t0), h(t0))

r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk, t0 = 1 -- Missing Information

Required

Determine the parametric equations

r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk

Differentiate with respect to t

r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k

Let t = 1 (i.e t0 = 1)

r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k

r'(1) = i +\frac{3}{3^2}j + (0 + 1)k

r'(1) = i +\frac{3}{9}j + (1)k

r'(1) = i +\frac{1}{3}j + (1)k

r'(1) = i +\frac{1}{3}j + k

To solve for x, y and z, we make use of:

r(t) = f(t)i + g(t)j + h(t)k

This implies that:

r'(1)t = xi + yj + zk

So, we have:

xi + yj + zk  = (i +\frac{1}{3}j + k)t

xi + yj + zk  = it +\frac{1}{3}jt + kt

By comparison:

xi = it

Divide by i

x = t

yj = \frac{1}{3}jt

Divide by j

y = \frac{1}{3}t

zk = kt

Divide by k

z = t

Hence, the parametric equations are:

x = t

y = \frac{1}{3}t

z =t

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