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2 years ago

Which formula can be used to find the angle of the resultant vector has

1 answer:

5 0

The diagram shows components that have been added together to form Rx and Ry. Rx and Ry are the components of the resultant vector.
Which formula can be used to find the angle of the resultant vector?
the answer is C
C. tan0= Ry/Rx

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Calculate the wavelength (in nm) of the blue light emitted by a mercury lamp with a frequency of 6.88 × 1014 Hz. The speed of li

Nezavi [6.7K]

Answer: 430 nm.

Explanation:

The relation of wavelength and frequency is:

Formula used : $\nu=\frac{c}{\lambda}$

where,

$\nu$ = frequency = $6.88\times 10^{14}Hz$

$\lambda$ = wavelength = ?

c = speed of light = $3.00\times 10^{8}m/s$

Now put all the given values in this formula, we get

$6.88\times 10^{14}=\frac{3.00\times 10^{8}m/s}{\lambda}$

$\lambda=\frac{3.00\times 10^{8}m/s}{6.88\times 10^{14}}=0.43\times 10^{-6}m=430m$ $(1nm=10^{-9}m)$

Thus the wavelength (in nm) of the blue light emitted by a mercury lamp is 430 nm.

6 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;

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

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

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

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

The change in heat energy for copper metal;

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

For aluminum metal, the quantity of heat transferred is calculated as;

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

The change in heat energy for aluminum metal ;

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

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

How many grams of ice at 0°c will melt if 1.50 kj of heat are added? (the molar heat of fusion of water is 6.01 kj/mol.) 4.49 g

Nata [24]

The computation would be:moles = mass/ Molar Mass, but we are looking for the mass, so rearranging, will give us: mass = moles x MM
Q = moles x Hf
Q = (mass/MM) x Hf
mass = (Q x MM) / Hf
= (1.50-kJ x 18.0-g/mol) / 6.01-kJ/mol
=4.49 g H20 is the answer

7 0

3 years ago

Read 2 more answers

What is observed as the distance between the rest position and the crest of a light wave decreases?

Alex_Xolod [135]

Answer: answers B

Explanation: it gets dimmer becuase less light......

8 0

2 years ago

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According to the first law of thermodynamics, the amount of work done by a heat engine equals the amount of

AleksandrR [38]