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Nutka1998 [239]
3 years ago
7

Think about lifestyle choices that you make on a daily basis. Choose at least four lifestyle choices which you could change to i

mprove your overall health. Explain how making those changes would improve your health. Read More >>
Physics
1 answer:
eimsori [14]3 years ago
4 0
A.The four lifestyle choices which I could change are:
1. The amount of physical exercise I engage in on a daily basis.
2, The number of hours I sleep.
3. The quantity of junk foods I consume daily and 
4. Cultivation of the habit of spending time alone daily.
B. My days are always very busy, so I do not engage at all in physical exercise. Adding at least a thirty minute of physical exercise to my daily tasks will improve my health considerable. Due to the nature of my day, I usually eat on the move, thus I eat junk foods everyday. Eating healthy and balanced diet will go a long way in improving my health. I sleep for three hours only everyday, from 2 am to 5 am.This is because I have to leave my house very early in the morning and I don't usually get to sleep until 2 am in the morning. Spending more hours sleeping will improve my overall health. I want to start spending at least 30 minutes alone on a daily basis in order to meditate and to reflect over the events of the day. This will be a great boost to my mental health.
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A total charge of 7.5 mC passes through a cross-sectional area of a wire in 0.9 s. What is the current in the wire
Elan Coil [88]

Answer:

the  current in the wire is  0.008333333  A

Explanation:

The computation of the current in the wire is as follows;

Current in the wire is

= Total charge ÷ cross sectional area of the wire

= 7.5 × 10^-3C ÷ 0.9s

= 0.008333333  A

Hence, the  current in the wire is  0.008333333  A

We simply applied the above formula so that the correct current of the wire could come

And, the same is considerd and relevant too

4 0
3 years ago
A source vibrating at constant frequency generates a sinusoidal wave on a string under constant tension. If the power delivered
just olya [345]

Answer:

Amplitude changes by factor of 2 means double.

Explanation:

Given:

If the power delivered to the string is quadrupled.

From the formula of power transmitted by a sinusoidal wave on a stretched string is,

  P = \frac{1}{2} \mu  \omega ^{2} A^{2}   v

Where \mu = mass per unit length, \omega  = angular speed, A = amplitude, v = wave speed, P = power delivered.

Here we need only two terms power and amplitude.

       P∝ A^{2}

All other quantities are constant for our problem,

     A = k \sqrt{P}

Where k = constant

Here Power become quadruple means four times,

     A = \sqrt{4P}

     A = 2\sqrt{P}

So amplitude changes by factor of 2 means double

5 0
3 years ago
The force of gravity pulls down on your school with a total force of 400,000 newtons. The force of gravity pulling down on your
gizmo_the_mogwai [7]

Answer: a Had twice as much mass

Explanation:

The data that we have is:

"The force of gravity pulls down on your school with a total force of 400,000 newtons. "

First, remember that, by the second Newton's law that:

F = a*m

F = force

a = acceleration

m = mass

In the case of the gravitational force, the gravitational acceleration is a constant: a = 9.8m/s^2

Then, if we want to have twice as much force the only thing that we can change in the equation is the mass:

Then if the initial force is written as:

F = a*m

twice as much that force is written as:

2*F = a*x

x is a variable that represents the new mass.

We know that F = a*m

2*F = 2*a*m

2*a*m  = a*x

2*m = x

Then, if we want to have twice as much force, we should have twice as much mass.

7 0
3 years ago
Indirect economic value to Steven and his family,
Eduardwww [97]

Answer:

indirect economic value

7 0
3 years ago
Read 2 more answers
(a) What is the angular speed ω about the polar axis of a point on Earth's surface at a latitude of 55° N? (Earth rotates about
Vaselesa [24]

Answer

given,

radius of earth = 6370 km

earths latitude = 55° N

time  = 24 hour = 86400 s

angular speed = ω

a) \omega = \dfrac{2\pi}{T}

\omega = \dfrac{2\pi}{86400}

\omega = 7.27 \times 10^{-5}rad/s

b) v = r\omega cos\theta

v = 6370\times 7.27 \times 10^{-5}\times cos 55^0

v = 266 m/s

c) ω at equator

\omega = \dfrac{2\pi}{T}

\omega = \dfrac{2\pi}{86400}

\omega = 7.27 \times 10^{-5}rad/s

d) v = r\omega cos\theta

v = 6370\times 7.27 \times 10^{-5}\times cos 0^0

v = 463 m/s

6 0
3 years ago
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