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3 years ago

Determine tge energy in joules of a photon whose frequency is 3.55x10

1 answer:

6 0

E = hf
E = 6.63×10^-34 × 3.55×10 eV
1 eV = 1.60×10^-19 J
E = 6.63×10^-34 × 3.55×10 × 1.60×10^-19
E = 3.77×10^-51 J

Hope it helped!

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An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe

kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

3 0

3 years ago

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?

Elan Coil [88]

Answer:

Time period, $T=3.05\times 10^{-5}\ s$

Explanation:

Given that,

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz

We need to find the period of the crystal's motion. The relationship between the frequency and the time period is given by :

$T=\dfrac{1}{f}$

T is the time period of the crystal's motion.

Time period is given by :

$T=\dfrac{1}{32768}$

$T=3.05\times 10^{-5}\ s$

So, the time period of the crystal's motion is $3.05\times 10^{-5}\ s$ . Hence, this is the required solution.

8 0

3 years ago

A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

$\sum F_x = 0$

$f-N_w = 0$

$N_w = f$

The forces in the horizontal direction would be,

$\sum F_y = 0$

$N_f -W =0$

$N_f = W$

The sum of Torques at equilibrium,

$\sum \tau = 0$

$Wdcos\theta - N_wLsin\theta = 0$

$WdCos\theta = fLSin\theta$

$f = \frac{Wd}{Ltan\theta}$

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

$f_{max} = \mu W=\frac{Wd}{Ltan\theta}$