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1 year ago

5

Classify the following as alkali metals, alkaline earth metals, transition elements, or inner transitional elements: calcium, go

ld, iron, magnesium, plutonium, potassium, sodium and uranium

2 answers:

Softa [21]1 year ago

8 0

1.) Alkali Metals: Sodium, Potassium

2.) Alkaline Earth metals: Magnesium, Calcium

3.) Transitional elements: Iron, Gold

4.) Inner-Transitional Elements: Uranium, Plutonium

Hope this helps!

Digiron [165]1 year ago

6 0

Alkali metals : sodium , potassium
alkaline earth : magnesium , calcium
the rest are transition elements... i don't know about "inner transition"

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g A projectile is launched with speed v0 from point A. Determine the launch angle ! which results in the maximum range R up the

Svetlanka [38]

Answer:

The range is maximum when the angle of projection is 45 degree.

Explanation:

The formula for the horizontal range of the projectile is given by

$R = \frac{u^{2}Sin2\theta }{g}$

The range should be maximum if the value of Sin2θ is maximum.

The maximum value of Sin2θ is 1.

It means 2θ = 90

θ = 45

Thus, the range is maximum when the angle of projection is 45 degree.

If the angle of projection is 0 degree

R = 0

It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.

If the angle of projection is 30 degree.

$R = \frac{u^{2}Sin60 }{9.8}$

R = 0.088u^2

If the angle of projection is 45 degree.

$R = \frac{u^{2}Sin90 }{g}$

R = u^2 / g

5 0

2 years ago

A 4 kg box is on a frictionless 35° slope and is connected via a massless string over a massless, frictionless pulley to a hangi

Anarel [89]

Answer:

(a) 19.62 N

(b) Box moves down the slope

(c) 24.43 N

Explanation:

(a)

2 Kg box causes tension

$T=mgwhere m is mass, g is gravitational force taken as 9.81T=2*9.81 =19.62 N (b) Block mass of 4 Kg [tex]T'-mg sin \theta=0$ hence $T'=mg sin \theta$ where m is mass and g is gravitational force

T'=4*9.81 sin 35= 22.5071 N

Since T' is greater than $mg sin\theta$ , then the box moves down the slope

(c)

Acceleration a= $\frac {forward force-backward force}{Total mass}= \frac {mg sin \theta -mg}{m1 + m2}$

$a= \frac {22.51-19.62}{2+4}=0.48$

When moving, the box will exert force T"= $mgsin \theta + ma$

T"= 4*9.81 sin 35 +(4*0.48)= 24.43 N

7 0

2 years ago

Read 2 more answers

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

2 years ago

the millersburg ferry (m=13000.0 kg loaded) puts its engines in full reverse and stops in 65 seconds. if the speed before brakin

kenny6666 [7]

The braking force is -400 N

Explanation:

We can solve this problem by using the impulse theorem, which states that the impulse applied on the ferry (the product of force and time) is equal to its change in momentum:

$F \Delta t = m(v-u)$

where in this problem, we have:

F is the force applied by the brakes

$\Delta t = 65 s$ is the time interval

m = 13,000 kg is the mass of the ferry

u = 2.0 m/s is the initial velocity

v = 0 is the final velocity

And solving for F, we find the force applied by the brakes:

$F=\frac{m(v-u)}{\Delta t}=\frac{(13000)(0-2.0)}{65}=-400 N$

where the negative sign indicates that the direction is backward.

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

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2 years ago

A transition metal complex in solution has an absorption peak at 450 nm, in the blue region of the visible spectrum. What color

Ivan

Answer:

In the case of a solution transition metal complex that has an absorption peak at 450 nm in the blue region of the visible spectrum, the (complementary) color of this solution is orange (option B).

Explanation:

The portion of UV-visible radiation that is absorbed implies that a portion of electromagnetic radiation is not absorbed by the sample and is therefore transmitted through it and can be captured by the human eye. That is, in the visible region of a complex, the visible color of a solution can be seen and that corresponds to the wavelengths of light it transmits, not absorbs. The absorbing color is complementary to the color it transmits.

So, in the attached image you can see the approximate wavelengths with the colors, where they locate the wavelength with the absorbed color, you will be able to observe the complementary color that is seen or reflected.

<u><em> In the case of a solution transition metal complex that has an absorption peak at 450 nm in the blue region of the visible spectrum, the (complementary) color of this solution is orange (option B).</em></u>

7 0

2 years ago