Questions 6-10? Please

4. How much mass is required to exert a force of 25 Newtons, accelerating at 5 m/s?

lions [1.4K]

F = ma
F/a = m
(25 N) / (5 m/s2) = m
4 kg = m

8 0

3 years ago

Example forces that can make objects move without touching them?

jek_recluse [69]

Answer:

Gravity such as magnets

Explanation:

Hope this helped

3 0

3 years ago

A solar eclipse occurs when the Moon moves directly between the Earth and the Sun. When Venus moves directly between the Earth a

Serga [27]

Answer:

it's because the eclipse hides the moon and Venus and the sun

Explanation:

I found it

7 0

2 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

A submarine deep underwater releases a bubble of air. The total pressure at this depth is P. The bubble has diameter, D, and abs

Sunny_sXe [5.5K]

Answer:

P V = n R T ideal gas equation

P2 V2 / P1 V1 = T2 / T1

V2 / V1 = T2 / T1 * P1 / P2 = T2 P1 / (T1 P2)

V2 / V1 = (1.17 T1) / T1 * (P1 / .22 P1) assuming absolute temp as 1.17 P1

V2 / V1 = 1.17 / .22 = 5.32

V = 4/3 pi R^3 = 4/3 pi (D/2)^3 = 4/3 pi D^3 / 8 = pi D^3 / 6

V2 / V1 = D2^3 / D1^3

D2 = (V2 / V1 * D1^3)^1/3

D2 = 5.32^1/3 * D = 1.75 D (D1 = D)

7 0

3 years ago