With an acceleration of 4.0 m/s², the bottle attains a speed of
(4.0 m/s²) * (9.0 s) = 36 m/s
so that the slide has length ∆<em>x</em> such that
(36 m/s)² - 0² = 2 * (4.0 m/s²) * ∆<em>x</em>
==> ∆<em>x</em> = 162 m ≈ 160 m
Alternatively, we know the bottle covers a distance ∆<em>x</em> with acceleration <em>a</em> at time <em>t</em> according to
∆<em>x</em> = 1/2 <em>a</em> <em>t</em>²
so that
∆<em>x</em> = 1/2 * (4.0 m/s²) * (9.0 s)² = 162 m ≈ 160 m
Force is a vector quantity and denotes any interaction that change the motion of an object in a certain direction. When influence by force an object can change its velocity or
The two things that are a included when describing force are: magnitude and direction. To fully describe the force acting upon an object, you must describe both the magnitude (size or numerical value) and the direction.
The tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
<h3>Tension in the cable</h3>
Apply the principle of moment and calculate the tension in the cable;
Clockwise torque = TL sinθ
Anticlockwise torque = ¹/₂WL
TL sinθ = ¹/₂WL
T sinθ = ¹/₂W
T = (W)/(2 sinθ)
T = (29 x 9.8)/(2 x sin57)
T = 169.43 N
<h3>Vertical component of the force</h3>
T + F = W
F = W - T
F = (9.8 x 29) - 169.43
F = 114.77 N
Thus, the tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
Learn more about tension here: brainly.com/question/24994188
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Mass divided by volume is density, while mass times density is volume. you cannot calculate density without volume and you cannot calculate volume without density.
i believe that's the answer.
oh gosh i didn't realize my middle school education involved high school stuff.
Answer:
Yes, the value will be the same.
Explanation:
Yes, or at least to some degree, that value of K will remain the same. You're looking for a difference in absorbance, and the difference should be visible at all wavelengths, not only at the limit. That being said, resolution varies, and if we don't read the value to the maximum, we can get a less accurate reading.
