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Marrrta [24]
2 years ago
8

3. How do you record your score in the 3-Minute Step Test? A record the 60-second heart rate before activity B. record the 30-se

cond heart rate after the activity C. record the 60-second heart rate after the activity D. record the 60-second heart rate during7 the activity ?? plss po nid ko ngayon​ plsss free piont makakasagot neto

Physics
1 answer:
Lisa [10]2 years ago
5 0

Answer:

d

Explanation:

you record it during the test

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Does wind have atoms and molecules in it?
Likurg_2 [28]
<span>
Of course.  Wind is air in motion, and the gases in air are composed of
all the usual familiar stuff ... atoms, molecules, mass, etc.  That's how
the wind moves things ... it has momentum and kinetic energy, which
get transferred to the things that move in the wind.</span>


4 0
3 years ago
Read 2 more answers
2. Four resistors of 2 ohm each are connected first in series and then in parallel with a battery.Find the ratio of electric cur
Serjik [45]

Answer:

Ratio of series current to parallel

= 1 : 8

Explanation:

Total resistance Rt

For series, Rt = 2+2+2+2 = 4ohms

For parallel, 1/Rt = 1/2 + 1/2 + 1/2 + 1/2

1/Rt = 4/2, Rt = 2/4 ohms.

If we use a 1V battery, then,

I = V/Rt

I = 1/4 = 0.25 ampere for series arrangement.

I = 1/0.5 = 2 ohms.

Ratio of current of series to parallel = 0.25 : 2

= 1 : 8

4 0
3 years ago
Read 2 more answers
If there is an object in top of another object, why does the upper object exert a downward normal force?​
rusak2 [61]

Answer:

This is because normal force is exerted perpendicularly to the point of contact between the upper and lower objects.  

Explanation:

This is because the upper object is still subject to gravitational pull. Therefore, the amount of force it exerts on the lower object due to gravity will be equal to the normal force that acts in the negative direction of gravitational force. Additionally, normal force is evident because the upper object will not go into the lower object.

4 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
At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci
Klio2033 [76]

Answer:

The radius of a chamber is 2.36 meters.

Explanation:

Given that,

The outer wall moves at a speed of 2.72 m/s.

Mass of the person, m = 75.1 kg

The person feels a force of 235 N force pressing against his back. The force acting on the person is centripetal force. It is given by the below formula :

F=\dfrac{mv^2}{r}

r is the radius of a chamber

r=\dfrac{mv^2}{F}

r=\dfrac{75.1\times (2.72)^2}{235}

r = 2.36 meters

So, the radius of a chamber is 2.36 meters. Hence, this is the required solution.

7 0
3 years ago
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