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3 years ago

Which statement describes a unit rate?

Physics

2 answers:

Nostrana [21]3 years ago

7 0

Answer:

katherine paid 1.25 for 1 bottle of water

Explanation:

Gwar [14]3 years ago

6 0

Katherine paid $1.25 for a bottle of water

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Cohort studies allow the investigator to examine multiple outcomes and multiple exposures. Consider the following three exposure

mojhsa [17]

Answer:

See the explanation for the answer.

Explanation:

Cohort study for smoking

Outcomes that were examined

1. People at risk of lung cancer due to smoking

2. Influence of outdoor air pollution in lung cancer

3. Variation in tobacco smoking and heavy tobacco smoking was observed.

4. Occupational factors and housing factor were observed.

Cohort study for low vitamin D

Outcomes that were examined

1. People at risk of vitamin D deficiency.

2. Observing value in summer.

3. Correlating vitamin D value with weather and age

4. 25-hydroxy vitamin D was observed.

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Outcomes that were examined

1. Allergic and asthma risked people were observed.

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3. Observation on the most harsh weather day.

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4 years ago

Some fuel cells are powered by hydrogen. Scientists are looking into the decomposition of water (H2O) to make hydrogen fuel with

mr_godi [17]

A challenge scientists face with this process is the use of ultrathin iron oxide, to pull protons off water and produce hydrogen gas, which itself is a poor electrical conductor.

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3 years ago

Read 2 more answers

Three point charges are on the x axis: −1 µC

Nina [5.8K]

Answer:

0.0078 N

Explanation:

The electrostatic force between two charges is given by Coulomb's law:

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

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

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

In this problem, we have:

$q_1 = -1 \mu C = -1 \cdot 10^{-6}C$ located at $x_1=-3 m$

$q_2=+9 \mu C = +9\cdot 10^{-6}C$ located at $x_2=0 m$

$q_3 = -5 \mu C = -5\cdot 10^{-6} C$ located at $x_3=+3 m$

The force between charge 1 and charge 2 is:

$F_{12}=k\frac{q_1 q_2}{(x_2-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(+9\cdot 10^{-6})}{3^2}=0.0090 N$

And since the two charges have opposite sign, the force is attractive, so the force on charge 1 is towards the right.

The force between charge 1 and charge 3 is:

$F_{13}=k\frac{q_1 q_3}{(x_3-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(5\cdot 10^{-6})}{6^2}=0.0012 N$

And since the two charges have same sign, the force is repulsive, so the force on charge 1 is towards the left.

Therefore, the net force on charge 1 is:

$F=F_{12}-F_{13}=0.0090-0.0012 = 0.0078 N$

towards the right.

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3 years ago

You observe a distant galaxy. you find that a spectral line normally found in the visible part of the spectrum is shifted toward

MrRissso [65]

The galaxy is moving away from you.

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3 years ago

A 100 kg mass is pulled along a frictionless surface by a horizontal force F such that its acceleration is 8.0 m/s2. A 20 kg mas

nadya68 [22]

Answer:

a) 60 N

b) 860 N

Explanation:

Given that,

$m_1$ = 100 kg

$m_2$ = 20 kg

$a_{1}$ = 8.0 $\frac{m}{s^{2}}$

$a_{2}$ = 3.0 $\frac{m}{s^{2}}$

a) By Newton's Law,

∑ $F_{m_2,x} = f_k = m_{2} * a_{2}$

∑ $F_{m_2,y} = F_{N} - m_{2} * g = 0$

Hence,

$f_k = m_2 * a_2 = 20 * 3 = 60 N$

b) By Newton's Law

∑ $F_{m_1,x} = F = m_{1} * a_{1}$

Hence,

$F = m_{1} * a_{1} = 100 * 8 = 800 N$

Net force acting on 100 kg mass,

$F_{net} = F + f_k = 800 + 60 = 860 N$

6 0

3 years ago