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jasenka [17]
3 years ago
8

Andy has been consuming numbers of calories and recently added and additional 100 calories in the form of a daily snack. Althoug

h he is not trying to lose weight , he has lost one pound per week over the last two months l. He wants to increase his level of activity and exercise, and he wants to return to his weight from two months ago. Which strategy should he follow to prevent becoming underweight?
Physics
1 answer:
kipiarov [429]3 years ago
8 0
He needs to make sure he is consuming more than he is burning off
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Help me with this one aswell, too confused and tired for this
gogolik [260]
D all of the above applies to the functions of the nervous system.
7 0
2 years ago
Un vas plin cu lichid cântăreşte 175kg. Ceea ce reprezintă de 5 ori masa vasului gol. Ştiind că volumul interior al vasului este
Mariana [72]

a) Density of the liquid: 823.5kg/m^3

b) Weight of the liquid: 1372 N

Explanation:

Translation of the text:

<em>"A full tank with liquid weighs 175kg. Which is 5 times the mass of the empty vessel. Knowing that the inside volume of the vessel is 0.17kl, calculate: </em>

<em>a) the density of the liquid; </em>

<em>b) the weight of the liquid."</em>

a)

We know that the full tank with liquid has a total mass of M = 175 kg. We can write the total mass as

M=m_L + m_V (1)

where

m_L is the mass of the liquid

m_V is the mass of the vessel

We also know that the total mass M is 5 times the mass of the empty vessel, so we have:

M=5m_V\\m_V=\frac{M}{5}=\frac{175}{5}=35 kg

which is the mass of the empty vessel.

Therefore, we can find the mass of the liquid only using (1):

m_L=M-m_V=175-35=140 kg

The density of the liquid is given by

d=\frac{m}{V}

where

m = 140 kg (mass of the liquid)

V = 0.170 kL = 170 L = 0.170 m^3 (volume of the liquid, which is equal to the volume of the vessel)

So we get

d=\frac{140}{0.170}=823.5kg/m^3

b)

The weight of a body is given by

F=mg

where

m is its mass

g is the acceleration due to gravity

For the liquid in this problem, we have

m = 140 kg (mass)

g=9.8 m/s^2 (acceleration due to gravity)

Therefore, its weight is

F=(140)(9.8)=1372 N

Learn more about density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

6 0
3 years ago
A solar eclipse that occurs when the new moon is too far from earth to completely cover the sun can be either a partial solar ec
Sonja [21]
1) <span>A solar eclipse that occurs when the new moon is too far from earth to completely cover the sun can be either a partial solar eclipse or an -->
Answer: ANULAR ECLIPSE. Since the moon is too far, it will cover only a part of the sun, and only the external ring of the moon will be visible; this is called anular eclipse.

2) </span><span>anyone looking from the night side of earth can, in principle, see a -->
Answer: LUNAR ECLIPSE. If the moon is the right position, and the Earth's shadow covers partially or totally the moon, then a lunar eclipse occurs.

3) </span><span>during some lunar eclipses, the moon's appearance changes only slightly, because it passes only through the part of earth's shadow called the -->
Answer: PENUMBRA. 

4) </span><span>a ... can occur only when the moon is new and has an angular size larger than the sun in the sky -->
Answer: TOTAL SOLAR ECLIPSE. When the moon is new, it means it is between the sun and the Earth, and its dark side faces the Earth. If the moon's angular size is also larger than the sun angular size, than it will completely cover the sun, and a total solar eclipse occurs.

5) </span><span>a partial lunar eclipse begins when the moon first touches earth's --> 
Answer: SHADOW. The Earth's shadow will start to cover the moon, and partial lunar eclipse will start.

6) </span><span> a point at which the moon crosses earth's orbital plane is called a(n)  --> 
Answer: NODE. Eclipses occur only when the Moon is at or close to a node, otherwise sun, earth and moon are not "aligned".</span>
5 0
3 years ago
Electromagnetic waves travel through space at a speed of _____.
Veseljchak [2.6K]
The answer is C 300,000 kilometers per second

6 0
3 years ago
Read 2 more answers
(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
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