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3 years ago

Your code returns a number of 99.123456789 +0.00455679 for your calculation. How should you report it in your lab write-up?

Physics

1 answer:

Aneli [31]3 years ago

6 0

Answer: Your code returns a number of 99.123456789 +0.00455679

Ok, you must see where the error starts to affect your number.

In this case, is in the third decimal.

So you will write 99.123 +- 0.004 da da da.

But you must round your results. In the number you can see that after the 3 comes a 4, so the 3 stays as it is.

in the error, after the 4 comes a 5, so it rounds up.

So the final presentation will be 99.123 +- 0.005

you are discarding all the other decimals because the error "domains" them.

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3 years ago

Read 2 more answers

Sam is recklessly driving 60 mph in a 30 mph speed zone when he suddenly sees the police. he steps on the brakes and slows to 30

barxatty [35]

For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:

a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time

The solution is as follows;

a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²

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2 years ago

How much heat is lost by 2.0 grams of water if the temperature drops from 31 °C to 29 °C? The specific heat of water is 4.184 J/

Elanso [62]

Given :

Mass of water, m = 2 grams.

The temperature of water drops from 31 °C to 29 °C .

The specific heat of water is 4.184 J/(g • °C).

To Find :

Amount of heat lost in this process.

Solution :

We know, heat lost is given by :

$Heat\ lost,H = ms( T_f - T_i)\\\\H = 2\times 4.184 \times ( 31 - 29 )\ J\\\\H = 16.736\ J$

Therefore, amount of heat lost in this process is 16.736 J.

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3 years ago

a 15 kg television sits on a shelf at a height of 0.3 m how much graviitional potential energy is added to the television when i

Readme [11.4K]

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3 years ago

Very lost please help.

V125BC [204]

A) The acceleration is due to gravity at any given point if you look at it vertically, so $-10 m/s^2$ .

b) $sin(25) = V_y/V$ , so $V_y = V*sin(25)$ . We use $V = V_0 + a t$ and then the final speed must be 0 because it stops at the highest point. So $0 = V_y - 10t$ . Solve for $t$ and you get $t = 32sin(25)/10 = 16sin(25)/5$

c) $Y = Y_0 + V_0t + (1/2)at^2$ , and then we plug the values: $Y_m_a_x = 32sin(25)*t - (1/2)*10*t^2$ and we already have the time from "b)", so $Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2$ ; then we just rearrange it $Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100]$ and finally $Y_m_a_x = 5[(32sin(25))^2/100] = (32sin(25))^2/20$

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2 years ago