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
hence, the spring constant is equal to 300 N/m
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B.) Carbon Dioxide because the carbon is surrounded by oxygen
When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
<h3>
Frictional force between the block and the horizontal surface</h3>
The frictional force between the block and the horizontal surface is determined by applying Newton's law;
∑F = ma
F - Ff = ma
Ff = F - ma
Ff = 4 - 2(1.2)
Ff = 4 - 2.4
Ff = 1.6 N
When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;
F - Ff = ma
5 - 1.6 = 2a
3.4 = 2a
a = 3.4/2
a = 1.7 m/s²
Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
Learn more about frictional force here: brainly.com/question/4618599
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