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VLD [36.1K]
3 years ago
6

A load of 54 N attached to a spring that ishanging vertically stretches the spring 0.15 m.What is the spring constant?Answer in

units of N/m.
Physics
2 answers:
beks73 [17]3 years ago
8 0

Answer:

300 N/m

Explanation:

given,

Load attached to the spring, W = 54 N

length of stretch of the spring, x = 0.15 m

spring constant= ?

Force applied on the spring is calculated by the equation

F = k x

where k is the spring constant

x is the displacement of the spring due to applied load

now,

54 = k × 0.15

k = \dfrac{54}{0.15}

k =300\ N/m

hence, the spring constant is equal to 300 N/m

myrzilka [38]3 years ago
6 0

Answer: The spring constant is 360 N/m

Explanation:

To calculate the spring constant, we use the equation:

F=kx       ......(1)

where,

F = force exerted on the spring = 54 N

k = spring constant = ?

x = length of the spring = 0.15 m

Putting values in equation 1, we get:

54N=k\times 0.15m\\\\k=\frac{54N}{0.15m}=360N/m

The spring constant is 360 N/m

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Answer:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>

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Explanation:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:  

a) The vectorial product, a×b  is:

a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k

b) The escalar product a·b is:

a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43

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(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17

d) The component of a along the direction of b is:

a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66

I hope it helps you!                        

5 0
3 years ago
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