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3 years ago

Scientific way of thinking

Physics

2 answers:

Sholpan [36]3 years ago

8 0

What about it, explain more

noname [10]3 years ago

7 0

Answer:

huh,? can you explain the question more please

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In a region of space where gravitational forces can be neglected, a sphere is accelerated by a uniform light beam of intensity 8

BlackZzzverrR [31]

Answer:

The correct answer is B

Explanation:

To calculate the acceleration we must use Newton's second law

F = m a

a = F / m

To calculate the force we use the defined pressure and the radiation pressure for an absorbent surface

P = I / c absorbent surface

P = F / A

F / A = I / c

F = I A / c

The area of area of a circle is

A = π r²

We replace

F = I π r² / c

Let's calculate

F = 8.0 10⁻³ π (1.0 10⁻⁶)²/3 10⁸

F = 8.375 10⁻²³ N

Density is

ρ = m / V

m = ρ V

m = ρ (4/3 π r³)

m = 4500 (4/3 π (1 10⁻⁶)³)

m = 1,885 10⁻¹⁴ kg

Let's calculate the acceleration

a = 8.375 10⁻²³ / 1.885 10⁻¹⁴

a = 4.44 10⁻⁹ m/s² absorbent surface

The correct answer is B

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2 years ago

Permanent magnets consist of multiple _____

julsineya [31]

Answer: Permanent magnets consist of multiple "ferromagnetic materials" I think that might be the answer, there weren't really any choices for me to choose from.

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3 years ago

How can I become a good science student ?

iris [78.8K]

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Study hard , focus on your studies and alyways ask questions .

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2 years ago

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What type of specialized training is necessary to learn the skills needed to work in this career field

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5 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

2 years ago