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3 years ago

How does a fuse work?

Physics

1 answer:

Vilka [71]3 years ago

5 0

Answer:

A. a material burns out when current is excessive

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A vector must always have both size and ..

DochEvi [55]

Both magnitude and DIRECTION
For example,
• 12m East
• -2 miles
•9 meter north
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3 years ago

What is true of the moon's orbital and rotational periods?

Eddi Din [679]

The moon's orbital and rotational periods are identical or the same, Its rate of spin is done in unison with its rate of revolution (the time that is needed to complete one orbit). Thus, the moon rotates exactly once every time it circles the Earth.

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3 years ago

8. While taking a measurement, Ajay put the 2nd mark of the scale to the edge of the line and the mark that pointed to the end o

Leni [432]

Answer:

The length of line is 78 cm or 0.78 m.

Explanation:

initial reading 2 mark

final reading 80 cm

The length of the line

= final reading - initial reading

= 80 - 2

= 78 cm

1 cm = 0.01 m

So, 78 cm = 0.78 m

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2 years ago

[BWS.02]If the same experiment is repeated in different parts of the world by different scientists,

guajiro [1.7K]

Answer:

the results will be the same.it may be

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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