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Pavlova-9 [17]
3 years ago
14

Experiment: Group 1 drinks 500 mL of coffee a day.

Physics
1 answer:
slega [8]3 years ago
3 0

Answer:

How does the drink content affect an individual's blood pressure?

Explanation:

In every experiment using the scientific method, an observation lays the foundation of that experiment. A problem must be observed, which then leads to asking a SCIENTIFIC QUESTION in order to investigate. A scientific question must include the variable being changed called INDEPENDENT VARIABLE and the variable being measured called DEPENDENT VARIABLE.

In this experimental procedure or set up,

- Group 1 drinks 500 mL of coffee a day.

- Group 2 drink 500 mL of tea a day,

- Group 3 is a control group i.e no drink

At the end of 60 days all participants

blood pressure is tested.

This set up indicates that the variable being changed (independent) is the DRINK CONTENT while the variable being measured (dependent) is the BLOOD PRESSURE. Hence, these variables serve as the template to ask a scientific question which goes thus:

HOW DOES THE DRINK CONTENT AFFECT AN INDIVIDUAL'S BLOOD PRESSURE?

This scientific question relates how the independent variable (drink) causes the dependent variable to respond (blood pressure).

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Which unit of measurement is best to estimate the volume of a juice box
stepan [7]

The unit of measurement is best to estimate the volume of a juice box will be kilogram.

<h3>What is volume?</h3>

The capacity of a container like how much quantity of liquid can be filled in that container is called the volume.

The juice bottle is estimated by liter as it contains the liquid. The juice box is a solid cardboard with some volume into which juice bottles are kept. So, the weight will be shown in kilograms.

Thus, the unit of measurement is best to estimate the volume of a juice box will be kilogram.

Learn more about volume.

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8 0
2 years ago
Why isn't Coulomb's law valid for dielectric objects, even if they are spherically symmetrical?
marshall27 [118]

Answer:

Explanation:

The "traditional" form of Coulomb's law, explicitly the force between two point charges. To establish a similar relationship, you can use the integral form for a continuous charge distribution and calculate the field strength at a given point.

In the case of moving charges, we are in presence of a current, which generates magnetic effects that in turn exert force on moving charges, therefore, no longer can consider only the electrostatic force.

4 0
3 years ago
The following equation is an example of decay?<br><br> 232/90 TH---4/2 HE +228/88 RA?
DaniilM [7]

Answer:

Alpha decay

Explanation:

  • Alpha decay is one of the three major types of decays, others being, beta decay and gamma decay.
  • <em><u>When a radioactive isotope undergoes alpha decay it emits alpha particles. An alpha particle is equivalent to the nucleus of Helium atom.</u></em>
  • <em><u>Therefore, an atom undergoing decay, its atomic mass is decreased by 4 and its atomic number is decreased by 2. </u></em>
  • Thus, since 232/90 Th, has undergone alpha decay its mass number is reduced by 4 to 228 and its atomic number by 2 to 88, and becomes 228/88 Ra.
5 0
2 years ago
Right Hand Rule 1 requires you to put the thumb of your right hand in the direction of current and your curled fingers will indi
Jet001 [13]
Answer: false
explanation: magnetic field does that
3 0
2 years ago
Read 2 more answers
determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund
wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is a_G= 6.78 m/s^2

<u>Explanation</u>:

N_A+N_B-Mg = 0

-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0

0.8N_B +0.8N_A = 975a_G

N_A+N_B = 9564.75 -------------(1)

-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0

-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0

-2.26N_A -0.06N_B= 0 ----------------(2)

Solving the equation (1) and(2)

N_A + N_B = 9564.75

-2.26N_A-0.06N_B=0

N_A = -260.85N

N_B = 9825.60N

\mu_s N_B + \mu_s N_A = 975a_G

0.8(9825.60)+0.8(-260.85) = 975a_Ga_G=\frac{7651.8}{975}a_G_1=7.4848m/s^2

Next lets assume that the front wheels contact with the ground N_A = 0

F_B = Ma_G

N_B = M_g

N_B - M_g = 0

N_B(b-a) –F_Bh = 0

F_B = 975a_G

N_B-975(9.8) = 0

N_B=9564.75N

9564.75(2.20 -1.82) -F_B(0.55)=0

\frac{3634.605}{0.55}=F_B

F_B = 6608.3

F_B = Ma_G

6608.3 = 975a_G

a_G = 6.7778 m/s^2

a_G_2 = 6.78m/s^2

Choosing the critical case

a_G = min(a_G_1 ,a_G_2)

a_G = min(7.848, 6.78)

a_G= 6.78 m/s^2

3 0
3 years ago
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