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3 years ago

A certain shade of blue has a frequency of 7.24 × 1014 Hz. What is the energy of exactly one photon of this light?

Physics

2 answers:

ANEK [815]3 years ago

4 0

E = hf

E = 6.63* 10 ⁻³⁴ * 7.24* 10¹⁴

E = 4.80012 × 10⁻¹⁹ J

Kitty [74]3 years ago

4 0

The energy of the photon is proportional to frequency of the photon. The constant of proportionality here is called Planck's constant with a value of 6.62607004 × 10-34 m2 kg / s. In this case, substituting to the formula E = h v where h is the constant and v is the frequency, the answer is 4.80 × 10^-19 Joules

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To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?

Nimfa-mama [501]

The energy stored in a capacitor is given by:
$U= \frac{1}{2}CV^2$
where
U is the energy
C is the capacitance
V is the potential difference

The capacitor in this problem has capacitance
$C=3.0 \mu F = 3.0 \cdot 10^{-6} F$
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
$V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V$

8 0

3 years ago

A rectangular block of steel measures 4cm*2cm*1.5cm*. its mass is 93.6g

nata0808 [166]

Answer:

Explanation:

5 0

3 years ago

The diffusion rate for a solute is 4.0 x 10^-11 kg/s in a solvent- filled channel that has a cross-sectional area of 0.50 cm^2 a

zlopas [31]

Answer:

$s = 9.6\times 10^{-12}kg/s$

Explanation:

Given:

Solute Diffusion rate = 4.0 × 10⁻¹¹ kg/s

Area of cross-section = 0.50 cm²

Length of channel =0.25 cm

Now for the new channel

Area of cross-section = 0.30 cm²

Length of channel =0.10 cm

let the Solute Diffusion rate of new channel = s

now equating the diffusion rate per unit volume for both the channels

$\frac{4\times 10^{-11}}{0.50\times 0.25}=\frac{s}{0.30\times 0.10}$

thus,

$s = 9.6\times 10^{-12}kg/s$

7 0

3 years ago

Select the situation for which the torque is the smallest.

alina1380 [7]

Answer:

e. The torque is the same for all cases.

Explanation:

The formula for torque is:

τ = Fr

where,

τ = Torque

F = Force = Weight (in this case) = mg

r = perpendicular distance between force an axis of rotation

Therefore,

τ = mgr

Here,

m = 200 kg

r = 2.5 m

Therefore,

τ = (200 kg)(9.8 m/s²)(2.5 m)

τ = 4900 N.m

Here,

m = 20 kg

r = 25 m

Therefore,

τ = (20 kg)(9.8 m/s²)(25 m)

τ = 4900 N.m

Here,

m = 8 kg

r = 62.5 m

Therefore,

τ = (8 kg)(9.8 m/s²)(62.5 m)

τ = 4900 N.m

Hence, the correct answer will be:

e. The torque is the same for all cases.

6 0

3 years ago

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle

ad-work [718]

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E = 5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

7 0

3 years ago