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3 years ago

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PLEASE ANSWER THIS I AM SO STUCK The increasing trend of energy requirements in the U.S.: a) is directly proportional to populat

ion growth b) is inversely proportional to population growth c) exceeds the rate of population growth

1 answer:

stiks02 [169]3 years ago

7 0

Answer:

It should be A

Explanation:

the amount of energy a population has should be based on how many people there are. if more people come into the population, the more energy it should have. if people leave the population, the less energy it should have.

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A train has a constant velocity of 2 m/s east. What is the magnitude of the horizontal acceleration of the train?

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"Constant velocity" is practically a definition for zero acceleration.

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4 years ago

Hydrogen gas is generated when acids come into contact with certain metals. When excess hydrochloric acid reacts with 2.2 g of (

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To solve the problem it is necessary to apply the concepts related to Byle's Law and Avogadro's Law.

The ideal gas equation would help us find the final solution to the problem, defined by

$PV = nRT$

Where,

T= Temperature of the gas

R = Universal as constant

n = number of moles

V = Volume

P = Pressure

For our case we have that the mass of Zn is 2.2g in moles would be

$[tex]Zn = \frac{2.2}{65}$ [/tex]

$Zn = 0.0338$

We know that 1 mole of hydrogen gas is proceed by 1 mole of zinc and the result is $Zn^{2+}$ , then Hydrogen can produce the same quantity,

$H_2 = 0.0338$

Applying the previous equation we have that

$V= \frac{nRT}{P}$

$V = \frac{0.0338*0.08206*293.15}{0.98}$

$V = 0.829L$

Therefore the volume of hydrogen gas is collected is 0.829L

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3 years ago

The cross section of a copper strip is 1.2 mmthick and 20 mm wide. There is a 25-A current through this cross section, with the

Naily [24]

To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.

The drift velocity is given by the equation:

$V_d = \frac{I}{nAq}$

Where

I = current

n = Number of free electrons

A = Cross-Section Area

q = charge of proton

Our values are given by,

$I = 25 A$

$A= 1.2*20 *10^{-6} m^2$

$q= 1.6*10^{-19}C$

$N = 8.47*10^{19} mm^{-3}$

$V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}$

$V_d = 7.68*10^{-5}m/s$

The hall voltage is given by

$V=\frac{IB}{ned}$

Where

B= Magnetic field

n = number of free electrons

d = distance

e = charge of electron

Then using the formula and replacing,

$V=\frac{(2.5)(25)}{(8.47*10^{28})(1.6*10^{-19})(1.2*10^{-3})}$

$V = 3.84*10^{-6}V$

5 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

RUDIKE [14]

The cup is acted upon by an unbalanced force which is the cars acceleration, but before it was an object at rest that stayed at rest. This jet propels their body forward.

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3 years ago

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