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2 years ago

The measured value of mass M in an experiment is M = 0.743 ± 0.005kg. The error in 2M is

Physics

1 answer:

omeli [17]2 years ago

8 0

Given the measured value of mass M in the experiment the error in 2M is ; dM

<u>Given data :</u>

Measured value of mass M = ( 0.743 ± 0.005 ) kg

<h3>The error in 2M in the measured value </h3>

The error in 2M in the measured value M = 0.743 ± 0.005 kg is dM while the error in M² will be 2dM

Hence we can conclude that the measured value of mass M in the experiment the error in 2M is ; dM.

Learn more about error in 2M : brainly.com/question/6650225

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3 years ago

A) One Strategy in a snowball fight the snowball at a hangover level ground. While your opponent is watching this first snowfall

Alexandra [31]

Answers:

a) $\theta_{2}=38\°$

b) $t=0.495 s$

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: <u>x-component</u> and <u>y-component</u>. Being their main equations as follows for both snowballs:

<h3><u>Snowball 1:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{1} t_{1}$ (1)

Where:

$V_{o}=14.1 m/s$ is the initial speed of snowball 1 (and snowball 2, as well)

$\theta_{1}=52\°$ is the angle for snowball 1

$t_{1}$ is the time since the snowball 1 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)

$y=0$ is the final height of the snowball 1

$g=-9.8m/s^{2}$ is the acceleration due gravity (always directed downwards)

<h3><u>Snowball 2:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{2} t_{2}$ (3)

Where:

$\theta_{2}$ is the angle for snowball 2

$t_{2}$ is the time since the snowball 2 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}$ (4)

Having this clear, let's begin with the answers:

<h2>a) Angle for snowball 2</h2>

Firstly, we have to isolate $t_{1}$ from (2):

$0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (5)

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$ (6)

Substituting (6) in (1):

$x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})$ (7)

Rewritting (7) and knowing $sin(2\theta)=sen\theta cos\theta$ :

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})$ (8)

$x=-\frac{(14.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(52\°))$ (9)

$x=19.684 m$ (10) This is the point at which snowball 1 hits and snowball 2 should hit, too.

With this in mind, we have to isolate $t_{2}$ from (4) and substitute it on (3):

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$ (11)

$x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})$ (12)

Rewritting (12):

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})$ (13)

Finding $\theta_{2}$ :

$2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})$ (14)

$2\theta_{2}=75.99\°$

$\theta_{2}=37.99\° \approx 38\°$ (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle $(\theta_{2} < \theta_{1})$ .

<h2>b) Time difference between both snowballs</h2>

Now we will find the value of $t_{1}$ and $t_{2}$ from (6) and (11), respectively:

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$

$t_{1}=-\frac{2(14.1 m/s)sin(52\°)}{-9.8m/s^{2}}$ (16)

$t_{1}=2.267 s$ (17)

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$

$t_{2}=-\frac{2(14.1 m/s)sin(38\°)}{-9.8m/s^{2}}$ (18)

$t_{2}=1.771 s$ (19)

Since snowball 1 was thrown before snowball 2, we have:

$t_{1}-t=t_{2}$ (20)

Finding the time difference $t$ between both:

$t=t_{1}-t_{2}$ (21)

$t=2.267 s - 1.771 s$

Finally:

$t=0.495 s$

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4 years ago

The heating curve of an ice-water mixture that is slowly heated to 125°C contains three sloped and two level portions. What do t

marin [14]

Answer:

Explanation:

Hi there,

To get started, please recall what heating curves are, latent heat, and enthalpy of phase changes (vaporization, fusion).

In a heating curve for ice water, temperature is the dependent variable (y-axis) and heat or time elapsed is independent variable (x-axis). There are three sloped portions. These sloped portions are showing where the <u>mixture is not changing phase, and is in a specific phase.</u> (solid first, then liquid then gas).

The reason is because the addition of more heat in time is causing the temperature to rise, so the mixture is getting "hotter." The temperature is still rising, yet the mixture is still in the current phase (depending on where you are referencing on the curve).

The two leveled portions are the process of melting (first leveled) and boiling (second leveled). The reason they are horizontal is because the addition of heat is not causing temperature to go up. The molecules are in the process of distributing heat in a different form by changing the mixtures entire structure (AKA phase), and so they are flat because the average kinetic energy from heat is being used change phase.

If you liked this solution, hit Thanks or leave a Rating!

thanks,

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3 years ago

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Answer:

If a material has high thermal conductivity, it is a conductor.

If a material has low thermal conductivity, it is an insulator.

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4 years ago

Thanks+BRAINLIST

Fynjy0 [20]

Answer:

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3 years ago