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Ksju [112]
3 years ago
11

When a force of 20.0N is applied to a spring, it elongates 0.20m. Determine the period of oscillation of a 4.0kg object suspende

d from this spring
Physics
1 answer:
choli [55]3 years ago
7 0
So what we can do is apply the<span> Hooke's law wich states that
F = -kx ( P.S the -ve sign means opposite in direction ) 
Also we will need to determine the spring's constant with the formula:
k = F / x 
Where F = the force ( = 20 N ) 
x = the displacement of the end of the spring from it's position ( = 0.20 m ) 
k = the spring's constant ( = unknown ) 
So this would be: k = 20 / 0.20 = 100 N/m 
The period of oscillation of 4 kg : T = 2 * pi * square root m / k 
T = 2 * pi * square root 4 / 100 
T = 1.256 seconds
Hope it helps</span>
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A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr
Paha777 [63]

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The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

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Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

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T=r*F

Where

F=F_{x}i+(7N)j-(5N)k  and the position vector

r=(2m)i-(3m)j+(2m)k

using the determinant method to expand the cross product in order to determine the torque we have

\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\

by expanding we arrive at

T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

since we have determine the vector value of the toque, we now compare with the torque value given in the question

(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

if we directly compare the j coordinate we have

10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m

8 0
3 years ago
Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist
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Answer:

time=4s

Explanation:

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