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ANSWER:
d. remains a non-zero constant.
STEP-BY-STEP EXPLANATION:
If we consider that there is no air resistance and that the horizontal component would be at x, the velocity remains a non-zero constant
Answer:
13.18m/s²
Explanation:
According to Newton's second law of motion
Force = Mass * acceleration
Given
Mass = 11kg
Force = 145N
Required
acceleration
From the formula
Acceleration = Force/Mass
Acceleration = 145/11
Acceleration = 13.18m/s²
Hence the initial acceleration of the object is 13.18m/s²
Up until the moment the box starts to slip, the static friction is maximized with magnitude <em>f</em>, so that by Newton's second law,
• the net force acting on the box parallel to the ramp is
∑ <em>F</em> = <em>mg</em> sin(<em>α</em>) - <em>f</em> = 0
where <em>mg</em> sin(<em>α</em>) is the magnitude of the parallel component of the box's weight; and
• the net force acting perpendicular to the ramp is
∑ <em>F</em> = <em>n</em> - <em>mg</em> cos(<em>α</em>) = 0
where <em>n</em> is the magnitude of the normal force and <em>mg</em> cos(<em>α</em>) is the magnitude of the perpendicular component of weight.
From the second equation we have
<em>n</em> = <em>mg</em> cos(<em>α</em>)
and <em>f</em> = <em>µn</em> = <em>µmg</em> cos(<em>α</em>), where <em>µ</em> is the coefficient of static friction. Substituting these into the first equation gives us
<em>mg</em> sin(<em>α</em>) = <em>µmg</em> cos(<em>α</em>) ==> <em>µ</em> = tan(<em>α</em>) ==> <em>α</em> = arctan(0.35) ≈ 19.3°
Answer:
1.16cm were cut off the end of the second pipe
Explanation:
The fundamental frequency in the first pipe is,
<em><u>Since the speed of sound is not given in the question, we would assume it to be 340m/s</u></em>
f1 = v/4L, where v is the speed of sound and L is the length of the pipe
266 = 340/4L
L = 0.31954 m = 0.32 m
It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.
<u>So, the length of the second pipe is L – L’</u>
Then, the fundamental frequency in the second pipe is
f2 = v/4(L - L’)
<u>The beat frequency due to the fundamental frequencies of the first and second pipe is</u>
f2 – f1 = 10hz
[v/4(L - L’)] – 266 = 10
[v/4(L – L’)] = 10 + 266
[v/4(L – L’)] = 276
(L - L’) = v/(4 x 276)
(L – L’) = 340/(4 x 276)
(L – L’) = 0.30797
L’ = 0.31954 – 0.30797
L’ = 0.01157 m = 1.157 cm ≅ 1.16cm
Hence, 1.16 cm were cut from the end of the second pipe
