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3 years ago

15

When writing a lab report where do your data tables belong?

2 answers:

SashulF [63]3 years ago

7 0

Answer:

When writing a lab report where do your data tables belong?

Results.

Mamont248 [21]3 years ago

5 0

Results is the answer

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If an object has a constant acceleration, it must be changing velocity at the same rate over and over again..TrueFalse

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If its accel is constant and not changing from its number then the velocity is changing so True

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3 years ago

1. Which will speed up the rate of a chemical

marusya05 [52]

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Consider a light, single-engine airplane such as the Piper Super Cub. If the maximum gross weight of the airplane is 7780 N, the

viva [34]

Answer:

V=19.08 m/s

Explanation:

Airplane gross weight w=7780 N

Airplane wing area S=16.6 m²

Air density p=1.2250 kg/m³

Maximum lift coefficient CL=2.1

To find

Stalling Speed

Solution

The equation to find stalling speed is given below

$W=(1/2)S_{area}(V_{Stalling-speed} )^{2} (P_{Air-density} )(C_{Lmax} )\\ so\\V_{Stalling-speed}=\sqrt{\frac{2W}{S_{area}*(P_{Air-density} )(C_{Lmax} )} }\\V_{Stalling-speed}=\sqrt{\frac{2*7780}{16.6*2.1*1.225} }\\ V_{Stalling-speed}=19.08 m/s$

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3 years ago

10. John does 176 J of work lifting himself a distance of 0.40 m. How

ch4aika [34]

The answer is 440N

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3 years ago

A 11.3-kg object oscillates at the end of a vertical spring that has a spring constant of 2.20 ✕ 104 N/m. The effect of air resi

3241004551 [841]

Answer:

(a) the frequency of the dampened oscillation is 7.02 Hz

(b) percentage decrease in amplitude of the oscillation in each cycle is 2%

Explanation:

Given;

mass of the object = 11.3 kg

the spring constant = 2.2 X 10⁴ N/m.

damping coefficient b = 3.00 N · s/m

Part (a) the frequency of the dampened oscillation

The oscillation frequency is calculated as follows;

$\omega _D = \sqrt{\omega_o^2 -(\frac{b}{2m})^2}\\\\\omega_o^2 = \frac{k}{m} =\frac{2.2X10^{4}}{11.3} = 1946.903rad/s\\\\thus, \omega _D = \sqrt{1946.903-(\frac{3}{2*11.3})^2} =44.12 rad/s$

The damped frequency $= \frac{\omega _D}{2\pi } = \frac{44.12}{2\pi } = 7.02 Hz$

Part (b) percentage decrease in amplitude of the oscillation in each cycle

The amplitude of the oscillation depends on the damping coefficient (b) and period (T), and it is given as;

$A(t) = e^{-\frac{b}{2m}(t)}$

After one cycle, the amplitude changes from A(t) to A(t+T), where T is the period of the oscillation.

$A(t +T) = e^{-\frac{b}{2m}(t+T)}$

Percentage decrease in amplitude is gotten by dividing A(t) by A(t+T)

$= \frac{e^{-\frac{b}{2m}(t)}}{e^{-\frac{b}{2m}(t+T)}} =e^{-\frac{b}{2m}(T)}$

But T = 1/f

Substituting the values of the parameters in the above equation, we will have;

$=e^{-\frac{b}{2m}(T)} = e^{-\frac{3}{2X11.3}(\frac{1}{7.02})} = 0.98$

Percentage decrease = 1 - 0.98 = 0.02 = 2%

4 0

3 years ago