How do you un rewrap something?

aev [14]

Answer:

the answer is UV Radiation

6 0

2 years ago

How does the direction of the electric current, moving across a battery powered device, differ from the direction of travel in t

svetoff [14.1K]

They both have different wave traction's

8 0

3 years ago

Read 2 more answers

The forces on a car are balanced.state and explain the resultant force on the car

Vanyuwa [196]

there will no resultant force

Explanation:

this is because if the forces are balanced on opposite direction. then they cancel each other out

5 newton's ---------> <--------- 5 newton's

then both forces will cancel each other out as a result there is no resultant force and the newton's laws states that if there is no resultant the object will continue in its state of rest (remains there) or it will in continue in its uniform motion in a straight line.

I hope you understand,

5 0

2 years ago

A projectile enters a resisting medium at x = 0 with an initial velocity v0 = 910 ft/s and travels 5 in. before coming to rest.

evablogger [386]

Answer:

a = - 1.987 × 10⁶ ft/s²

t = 6.84 × 10⁻⁴ s

Explanation:

v₀ = 910 ft/s

x = 5 in.

relation v = v₀ - k x

v = 0 as body comes to rest

0 = 900 - 5k/12

k = 2184 s⁻¹

acceleration

$\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}$

where

(A) a = -k × v

at v= 910 ft/s

a = - 1.987 × 10⁶ ft/s²

(B) at x = 3.9 in.

v = 910 - 3.9(2184)/12

v = 200.2 m/s

$\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{dv}{v} = -kdt$

$\int\limits^{200.2}_{900} {\frac{1}{v} }dv = -k\int\limits^t_0 dt$

$ln(200.2)-ln(900) = -kt$

t = 6.84 × 10⁻⁴ s

3 0

2 years ago

what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three

grin007 [14]

Answer:

Approximately $5.11 \times 10^{-19}\; {\rm J}$ .

Explanation:

Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

Look up the Rydberg constant for hydrogen: $R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}$ .

Look up the speed of light in vacuum: $c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}$ .

Look up Planck's constant: $h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}$ .

Apply the Rydberg formula to find the wavelength $\lambda$ (in vacuum) of the photon in question:

$\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}$ .

The frequency of that photon would be:

$\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}$ .

Combine this expression with the Rydberg formula to find the frequency of this photon:

$\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}$ .

Apply the Einstein-Planck equation to find the energy of this photon:

$\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}$ .

(Rounded to three significant figures.)

6 0

2 years ago