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3 years ago

List 5 possible effects of not adhering to standards of measurement

Physics

1 answer:

Bond [772]3 years ago

3 0

Explanation:

Without the metric system, we'd have a different International System of Units, the metric system is important because 1mm is 0.1cm, 1 cm is 0.01m, with the imperial system the conversion is tedious. The most important feature of the metric system is its base in scientific fact and repeatable standards of measurement

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3 years ago

A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch

ANTONII [103]

Answer:

A. A = 0.913 m

B. amax = 132.24m/s^2

C. Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

$v_{max}=A\omega$ (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

$\omega=\sqrt{\frac{k}{m}}$ (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

$\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}$

Next, you solve the equation (1) for A and replace the values of vmax and w:

$A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m$

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

$a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}$

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

$F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N$

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

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3 years ago

3. A model rocket is launched straight upward at 58.8 m/s.

kolezko [41]

Answer:

a). 6 seconds

b). 12 seconds

c). 176.4 meters

Explanation:

a). Equation to be applied to calculate the time taken by the rocket to reach at the peak height,

v = u - gt

where v = final velocity

u = initial velocity = 58.8 m per sec

g = gravitational pull = 9.8 m per sec²

t = duration of the flight

At the peak height,

v = 0

Therefore, 0 = 58.8 - (9.8)(t)

t = $\frac{58.8}{9.8}$

= 6 seconds

b). Total time of flight = 2(Time taken to go up)

= 2×6

= 12 sec

c). Formula to get the peak height is,

$h=ut-\frac{1}{2}gt^2$

h = (58.8)6 - $\frac{1}{2}(9.8)(6)^2$

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3 years ago

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