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koban [17]
3 years ago
8

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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If you place a magnet under a clear dish, and sprinkle iron fillings over it, above what part of the magnet will most of the fil
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round the corners of the magnet that where it is stronger

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3 years ago
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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
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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)
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5 0
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 :

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

4 0
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
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15x 9.8 x .3 = 44.1
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5 0
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
6 0
2 years ago
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