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3 years ago

6

The total amount of a solute that can dissolve into a particular solvent is called __________.A.dissolving point B.evaporation C

.solubility D.distillation

1 answer:

diamong [38]3 years ago

5 0

The answer is C BeacUse solubility means the total of what one substance can dissolve

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What is the energy (in evev) of a photon of visible light that has a wavelength of 500 nmnm?

lisabon 2012 [21]

<h3>Answer:</h3>

E≈2,5 eV

<h3>Explanation:</h3>

_______________

λ=500 nm = 500·10⁻⁹ m

c=3·10⁸ m/s

h=6,63·10⁻³⁴ J·s = 4,14·10⁻¹⁵ eV·s

_______________

E - ?

_______________

$\displaystyle \boldsymbol{E}=h\nu =h \frac{c}{\lambda} =4,14\cdot 10^{-15} \; eV\cdot s\cdot \; \frac{3\cdot 10^8\; m/s}{500\cdot 10^{-9}\; m} =2,484\; eV\approx \boldsymbol{2,5\; eV}$

6 0

2 years ago

Let's apply Snell's law to the refraction of light across a water–air interface. Suppose you kneel beside the fishpond in your b

Nataliya [291]

Answer:

The angle of refraction is 41.68°.

Explanation:

The refractive index for water is $n_2 = 1.333$ , and for air $n_1 = 1.00$ : the angle of light with the normal is $90^o-60^o = 30^o$ ; therefore Snell's law gives

$n_1sin(\theta_1)= n_2sin(\theta_2)$

$1.00*sin(\theta_1) = 1.33 sin(30^o)$

$sin (\theta_1) = \dfrac{1.33sin(30^o)}{1.00}$

$sin (\theta_1) = 0.665$

$\theta _1 = sin^{-1}(0.665)$

$\boxed{\theta_1 = 41.68^o}$

4 0

3 years ago

. You have completed your design of a new surgical headlamp. You have decided to power the headlamp using four AA rechargeable b

SSSSS [86.1K]

Answer:

LED run before the batteries are depleted is 3.87 hours

Explanation:

given data

battery = 4 AA

battery is rated = 1.2V, 2550 mAh

power = 3 W

efficient = 95%

to find out

How long will the LED run before the batteries are depleted

solution

we consider here power delivered by battery is = x

so power = x × efficient

3 = x 95%

x = 3.15 W

and

voltage by 4 battery is = 4 × 1.2 = 4.8 V

so current will be = $\frac{x}{volatge}$

current = $\frac{3.15}{4.8}$

current = 0.6577 A

so total discharge hours = $\frac{2550}{current}$

total discharge hours = $\frac{2550}{657.7}$

time = 3.87 hours

so LED run before the batteries are depleted is 3.87 hours

5 0

3 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

3 years ago

A car drives 16 miles south and then 12 miles west. What is the magnitude of the car’s displacement?

Lorico [155]

20miles (16^2+12^2)^1/2=20

7 0

4 years ago