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2 years ago

describe three small changes that you can make to your current eating habits to support a healthier lifestyle.

1 answer:

7 0

Answer: Eat only when you're truly hungry instead of when you are tired, anxious, or feeling an emotion besides hunger.

Plan meals ahead of time to ensure that you eat a healthy well-balanced meal.

Keep more fruits, low-fat dairy products (low-fat milk and low-fat yogurt), vegetables, and whole-grain foods at home and at work.

Explanation:

You might be interested in

What Energy transformations when climbing a hill?

elixir [45]

Your kinetic energy goes down and your potential energy rises. This happens till you reach the top or start falling, in which the opposite happens. Hope this helps!

4 0

2 years ago

a bus takes to reach from station A to station B and then 3 hour to return from station B to station A find the average velocity

Anton [14]

Answer:

you have probably missed some details in the question.

3 0

2 years ago

In 3 meters a person running 0.5 m/s accelerates 1.2 m/s 2. How fast were they going afterward? Choose the right equation for th

Feliz [49]

Answer:

2.72 m/s

Explanation:

In 3 meters a person running 0.5 m/s accelerates 1.2 m/s².

It means,

Distance, s = 3 m

Initial velocity, u = 0.5 m/s

Acceleration, a = 1.2 m/s²

We need to find the final velocity of the person. Using equation of motion to find it as follows :

$v^2-u^2=2as\\\\v^2=2as+u^2\\\\v^2=2\times 1.2\times 3+(0.5)^2\\\\v=2.72\ m/s$

So, the final velocity of the person is 2.72 m/s.

7 0

2 years ago

Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

Now, gravitational force on particle 3 due to particle 1:

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

gravitational force on particle 3 due to particle 2:

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

Now the net force

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

For angle in counterclockwise direction from the +x-axis

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

4 0

3 years ago

say I have two rocks of different sizes but are taken from the same rock. Will their density be equalvilent

poizon [28]

yes their density are equal

8 0

2 years ago