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3 years ago

Suppose that 2 scientists use different measurement systems in their work. What problems might arise if they shared their data?

1 answer:

8 0

In assumption that there were two scientists who used different measurement systems in their research. The problems that might arise if they shared their data is obviously and primarily error. Errors are recognized and one element in every measure, system and quantity. Error was already even present in the measurement system alone a scientist used and it will furthermore aggregate, when these two different scales are combined the more error escalates in the process. There are two types: random and systematic error.

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A cube has sides of 11.4 cm. What is its volume? 1,482 cm3 130 cm3 1,443 cm3 34.2 cm3

viktelen [127]

The Volume of a Cube with a side(s) of 11.4 cm is

V = s^3 
 V = (11.4)^3 cm^3 
 V = 1481.544 cm^3 
 V = 1482 cm^3

however the Surface Area of a cube is the combined area of all the sides 
 A = 6s^2 
 A = 6(11.4)^2 cm^2 
 A = 779.76 cm^2

4 0

2 years ago

The tape in a videotape cassette has a total

Softa [21]

Answer:

1.37 rad/s

Explanation:

Given:

Total length of the tape is, $d= 297$ m

Total time of run is, $t = 2.1$ hours

We know, 1 hour = 3600 s

So, 2.1 hours = 2.1 × 3600 = 7560 s

So, total time of run is, $t= 7560$ s

Inner radius is, $r = 10\ mm = 0.01\ m
$

Outer radius is, $R = 47\ mm = 0.047\ m
$

Now, linear speed of the tape is, $v =\frac{d}{t}=\frac{297}{7560}=0.039\ m/s
$

Let the same angular speed be $\omega$ .

Now, average radius of the reel is given as the sum of the two radii divided by 2.

So, average radius is, $R_{avg}=\frac{R+r}{2}=\frac{0.047+0.01}{2}=\frac{0.057}{2}=0.0285\ m
$

Now, common angular speed is given as the ratio of linear speed and average radius of the tape. So,

$\omega=\dfrac{v}{R_{avg}}\\\\\\\omega=\dfrac{0.039}{0.0285}\\\\\\\omega=1.37\ rad/s
$

Therefore, the common angular speed of the reels is 1.37 rad/s.

5 0

3 years ago

The muscular system brings what to the body?

kow [346]

The muscular system brings strength and endurance to the body. It helps us perform everyday activities. As well as soaks up water to keep us hydrated longer.

6 0

3 years ago

Feng and Isaac are riding on a merry-ground. Feng rides on a horse at the outer rim of the circular platform, twice as far from

kogti [31]

Answer: The question is incomplete or missing details. here is the remaining part of the question ;

1. impossible to determine

2. half of Isaac’s

3. the same as Isaac’s

4. twice Isaac’s

The angular speed of feng will be the same as that of Isaac. Hence the answer is option 3

Explanation:

Since we have been told that both feng and isaac are riding on a merry go round i.e in a circular motion, irrespective of how fast one ride above the other, the angular speed will be constant since they are riding on a merry go round, as such both feng and isaac will maintain equal angular speed, hence the angular speed of feng will be the same as that of Isaac.

4 0

3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$