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timama [110]
3 years ago
6

Giving brainliest to whoever can solve!! Light of wavelength 675 nm passes through a double slit , and makes its first minimum (

m = 1) at an angle of 0.111 deg. What is the separation of the slits , d , in MILLIMETERS ?

Physics
1 answer:
Aloiza [94]3 years ago
5 0

Answer:

0.174

Explanation:

Using the equation with both the minimum and where you're solving for d, (1-1/2) 675/sin(0.111), you get 174210.2 so you convert that into millimeters by moving the decimal point up 6 (the difference of nm 10^-9 and millimeters 10^-3),leaving you with 0.174.

I hope this helped!

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If the radio waves transmitted by a radio station have a frequency of 76.5 mhz, what is the wavelength of the waves, in meters?
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Given: Velocity of light    c = 3.00 x 10⁸ m/s

Frequency f = 7.65 x 10⁷/s

Required: Wavelength λ = ?

Formula:  λ = c/f

λ = 3.00 x 10⁸ m/s/7.65 x 10⁷/s

λ = 3.92 m
3 0
3 years ago
Please help me find the equations guys
Alexeev081 [22]

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4 0
3 years ago
The mass of the Sun is 2multiply1030 kg, and the mass of the Earth is 6multiply1024 kg. The distance from the Sun to the Earth i
spayn [35]

Answer:

<em>a) 3.56 x 10^22 N</em>

<em>b) 3.56 x 10^22 N</em>

<em></em>

Explanation:

Mass of the sun M = 2 x 10^30 kg

mass of the Earth m = 6 x 10^24 kg

Distance between the sun and the Earth R = 1.5 x 10^11 m

From Newton's law,

F = \frac{GMm}{R^2}

where F is the gravitational force between the sun and the Earth

G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

m is the mass of the Earth

M is the mass of the sun

R is the distance between the sun and the Earth.

Substituting values, we have

F = \frac{6.67*10^{-11}*2*10^{30}*6*10^{24}}{(1.5*10^{11})^2} = <em>3.56 x 10^22 N</em>

<em></em>

A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to <em>3.56 x 10^22 N</em>

b) The force exerted by the Earth on the Sun = <em>3.56 x 10^22 N</em>

7 0
3 years ago
For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par
Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

  • <em>initial temperature of metals, =  </em>T_m<em />
  • <em>initial temperature of water, = </em>T_i<em> </em>
  • <em>specific heat capacity of copper, </em>C_p<em> = 0.385 J/g.K</em>
  • <em>specific heat capacity of aluminum, </em>C_A = 0.9 J/g.K
  • <em>both metals have equal mass = m</em>

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w )  = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w)  = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0
3 years ago
I NEED ANSWER ASAP
umka21 [38]

Answer:

97% will sink below the water

Explanation:

Waters density is 1 g/cm3

If an object's density is greater than 1g/cm3, it will sink. If it's less then it would float. 1-0.97 = 0.03. Only 3% of the ice would show while 97 will be under.

7 0
3 years ago
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