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-Dominant- [34]

3 years ago

15

Who reported four “element” classifications, but included some substances that were combinations of elements rather than true el

ements in his listing?

1 answer:

notsponge [240]3 years ago

7 0

Antoine-Laurent Lavoisier was the first person to report the four element classification system but also ended up including some compounds rather than elements.

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How does a vacuum flask keep drinks hot? Explain in terms of conduction, convection and radiation.

Flura [38]

Vacum flask has inner layer of glass and outer body of plastic.the glass has a reflective metal layer..when a hot drink is poured into the flask it prevents conduction.tight lid stops the convection.when radiation tries to leave the hot drink the reflective layer of the glass reflects it back so no heat can escape.

4 0

3 years ago

If a bullet travels at 593.0 m/s, what is its speed in miles per hour?

Ksenya-84 [330]

We have $1 \; mile = 1609.34\; meters$ . So, $1 \; meter = \frac{1}{1609.34} \;mile$ .

$1 \; hour = 3600 \; s$ . So $1 \; s = \frac{1}{3600} \; hour$ .

Thus we can convert the units of the given quantity.

That is,

$593\;m/s=593\;\frac{1/1609.34}{1/3600} \;miles/hour\\
593\;m/s=1,326.51\;miles/hour$ .

The quantity is converted to the required units.

7 0

2 years ago

Use the following half-life graph to answer the following question:

Temka [501]

Answer:

A 1.0 min

Explanation:

The half-life of a radioisotope is defined as the time it takes for the mass of the isotope to halve compared to the initial value.

From the graph in the problem, we see that the initial mass of the isotope at time t=0 is

$m_0 = 50.0 g$

The half-life of the isotope is the time it takes for half the mass of the sample to decay, so it is the time t at which the mass will be halved:

$m'=\frac{50.0 g}{2}=25.0 g$

We see that this occurs at t = 1.0 min, so the half-life of the isotope is exactly 1.0 min.

3 0

3 years ago

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

1)

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$

Which has 2 solutions:

x = 0 (origin)

and

$tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft$

So, the distance d down the hill at which the shell strikes the hill is

$d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m$

2)

In order to find how long the mortar shell remain in the air, we can use the equation:

$t=\frac{x}{u cos \theta}$

where:

x = 1322 ft is the final position of the shell when it strikes the hill

$u=156 ft/s$ is the initial velocity of the shell

$\theta=49.0^{\circ}$ is the angle of projection of the shell

Substituting these values into the equation, we find the time of flight of the shell:

$t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s$

3)

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

$v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s$

Instead, the vertical component of the velocity is given by

$v_y=usin \theta -gt$

And substituting at t = 12.9 s (time at which the shell strikes the hill),

$v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s$

Therefore, the final speed of the shell is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0

3 years ago

In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur

Margaret [11]

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

$3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)$

First, we isolate C from equation (2):

$2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)$

using this value of C from equation (3) in equation (1):

$3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}$

<u>D = - 0.57</u>

Put this value in equation (3), we get:

$C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C = \frac{17}{7}\\$

<u>C = 2.43</u>

5 0

3 years ago