2 years ago

Question 8

Physics

1 answer:

FinnZ [79.3K]2 years ago

8 0

Answer:

A hope this helps

Explanation:

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Part A Determine the magnitude of the x component of F using scalar notation. Fx F x = nothing lb Request Answer Part B Determin

maksim [4K]

As we know that force F makes an angle of 60 degree with X axis

so the X component is given as

$cos60 = \frac{F_x}{F}$

now we have

$F_x = F cos60$

$F_x = 0.50 F$

Similarly we know that force F makes an angle of 45 degree with Y axis

so the X component is given as

$cos45 = \frac{F_y}{F}$

now we have

$F_y = F cos45$

$F_y = 0.707 F$

Now for the component along z axis we know that

$F_x^2 + F_y^2 + F_z^2 = F^2$

now plug in all components

$(0.707 F)^2 + (0.50 F)^2 + F_z^2 = F^2$

$0.5 F^2 + 0.25 F^2 + F_z^2 = F^2$

$F_z^2 = F^2(1 - 0.75)$

$F_z^2 = 0.25 F^2$

$F_z = 0.5 F$

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2 years ago

Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it

erica [24]

<u>Answer:</u>

2N/cm

<u>Step-by-step explanation:</u>

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

$F=kx$

where, $F$ is the force which is stretching or compressing the spring,

$k$ is the spring constant; and

$x$ is the distance the spring is stretched.

Substituting the given values to find the elastic constant $k$ to get:

$F=kx$

$4=k(2)$

$k=\frac{4}{2}$

$k=2$

Therefore, the elastic constant is 2 Newton/cm.

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A.) A control variable

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