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luda_lava [24]
3 years ago
14

Accurate forecasts require careful study of the location, size, movement, and characteristics of large _____.

Physics
1 answer:
tankabanditka [31]3 years ago
3 0

Answer: B air masses

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If a freely falling object were equipped with a speedometer on a planet where the acceleration due to gravity is 20 m/s², then i
babymother [125]

Answer:

20 m/s

Explanation:

The acceleration of an object determines the increase in velocity per second.

In this case the acceleration is 20m/s^2.

a=\frac{v}{t}=\frac{\frac{20m}{s}}{s}

That is, the increase is 20m/s each second.

I hope this is useful for you

Regards

7 0
4 years ago
Read 2 more answers
Whitch family has full valence energy level
levacccp [35]
Group 0 - they are called the NOBLE GASES
(as they're very unreactive) because...

ALL of their atoms are full.
Example : Helium contains 2 electrons so it's outer shell is full
7 0
4 years ago
Calculate the force, when a object acelerattes to 30 m/s^2 and have a mass of 40 kg
faust18 [17]

Hello!

Use the <u>second law of Newton:</u>

F = ma

Replacing:

F = 40 kg * 30 m/s^2

Resolving:

F = 1200 N

The force is <u>1200 Newtons.</u>

3 0
3 years ago
Read 2 more answers
Please help me do this. Study the scenario.
dem82 [27]

The mechanical advantage of the lever is 3 and the right option is A.) 3.

<h3>What is mechanical advantage?</h3>

Mechanical advantage can be defined as the ratio of load to effort in a machine.

To calculate the mechanical advantage of the lever, we use the formula below.

Formula

  • M.A = y/x................ Equation 1

Where:

  • M.A = mechanical advantage of the lever
  • y = distance moved by effort (input arm)
  • x = distance moved by load (output arm)

From the question,

Given:

  • y = 3 m
  • x = 1 m

Substitute these values into equation 1

  • M.A = 3/1
  • M.A = 3.

Hence, The mechanical advantage of the lever is 3 and the right option is A.) 3.

Learn more about mechanical advantage here: brainly.com/question/19038908

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;

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

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

Q = mcΔt

<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

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

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

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

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

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

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

\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
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