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3 years ago

How does the water cycle influence weather in your area? Use complete sentences and a specific example from your experience.

2 answers:

7 0

The water cycle increases precipitation which causes more clouds, severe weather, and changes the temperature.I live by a lake so it causes rising sea level which could sometimes result to flooding.

I hope this helps ☺ Let me know if you have other questions

Charra [1.4K]3 years ago

5 0

Evaporation is the process by which water changes from a liquid to a gas or vapor. Evaporation is the primary pathway that water moves from the liquid state back into the water cycle as atmospheric water vapor. Studies have shown that the oceans, seas, lakes, and rivers provide nearly 90 percent of the moisture in the atmosphere via evaporation, with the remaining 10 percent being contributed by plant transpiration.

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A light-year measures the _______ that light travels in 1 year.

o-na [289]

A light-year measures the distance that light travels in 1 year.

Answer : B ) Distance

-Hope this helps.

4 0

3 years ago

Read 2 more answers

Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly becau

Tamiku [17]

Answer:

$7.0\cdot 10^{-13}C$

Explanation:

The magnitude of the electrostatic force between two charged objects is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

The force is attractive if the charges have opposite sign and repulsive if the charges have same sign.

In this problem, we have:

$r=0.070 cm =7\cdot 10^{-4} m$ is the distance between the charges

$q_1=q_2=q$ since the charges are identical

$F=9.0\cdot 10^{-9}N$ is the force between the charges

Re-arranging the equation and solving for q, we find the charge on each drop:

$F=\frac{kq^2}{r^2}\\q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(9.0\cdot 10^{-9})(7\cdot 10^{-4})^2}{8.99\cdot 10^9}}=7.0\cdot 10^{-13}C$

8 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;

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