The string is assumed to be massless so the tension is the sting above the 12.0 N block has the same magnitude to the horizontal tension pulling to the right of the 20.0 N block. Thus,
1.22 a = 12.0 - T (eqn 1)
and for the 20.0 N block:
2.04 a = T - 20.0 x 0.325 (using µ(k) for the coefficient of friction)
2.04 a = T - 6.5 (eqn 2)
[eqn 1] + [eqn 2] → 3.26 a = 5.5
a = 1.69 m/s²
Subs a = 1.69 into [eqn 2] → 2.04 x 1.69 = T - 6.5
T = 9.95 N
Now want the resultant force acting on the 20.0 N block:
Resultant force acting on the 20.0 N block = 9.95 - 20.0 x 0.325 = 3.45 N
<span>Units have to be consistent ... so have to convert 75.0 cm to m: </span>
75.0 cm = 75.0 cm x [1 m / 100 cm] = 0.750 m
<span>work done on the 20.0 N block = 3.45 x 0.750 = 2.59 J</span>
Given:
The expression is
![\sqrt[3]{48}=\sqrt[3]{8\cdot \_\_}=](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D%5Csqrt%5B3%5D%7B8%5Ccdot%20%5C_%5C_%7D%3D)
To find:
The simplified form of the expression.
Solution:
We have,
![\sqrt[3]{48}=\sqrt[3]{8\cdot \_\_}=](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D%5Csqrt%5B3%5D%7B8%5Ccdot%20%5C_%5C_%7D%3D)
The expression
can be written as
![\sqrt[3]{48}=\sqrt[3]{8\cdot 6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D%5Csqrt%5B3%5D%7B8%5Ccdot%206%7D)
![[\because \sqrt[3]{ab}=\sqrt[3]{a}\sqrt[3]{b}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Csqrt%5B3%5D%7Bab%7D%3D%5Csqrt%5B3%5D%7Ba%7D%5Csqrt%5B3%5D%7Bb%7D%5D)
![\sqrt[3]{48}=2\cdot \sqrt[3]{6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D2%5Ccdot%20%5Csqrt%5B3%5D%7B6%7D)
![\sqrt[3]{48}=2\sqrt[3]{6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D2%5Csqrt%5B3%5D%7B6%7D)
Therefore,
.
Answer:
C (y=5x-7)
Step-by-step explanation:
First, you find the slope between the 2 points which is 5, then you use one of the points to form an equation.
You then find it's y=5x-7.
Answer:
Step-by-step explanation:
Hope this helps
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The sum of 5 times a number and -8:
5x - 8 = y