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deff fn [24]
3 years ago
15

Whenever you work a physics problem you should get into the habit of thinking about whether the answer is physically realistic.

Think about how far off the ground a typical small truck is. Is the answer to Part B physically realistic?
Physics
2 answers:
Marta_Voda [28]3 years ago
4 0

Answer:

Yes, the answer on <em>how far off the ground a typical truck is happens to be</em> physically realistic.

Explanation:

This concept could be explained with when someone is viewing a picture from a particular distance. The farther the person is from the picture, the smaller the details of the picture and the higher they picture would be unappriciated and vice versa.

Likewise the small truck on ground. <em>The distance off ground the truck is being viewed, the smaller the truck would become.</em>

Alisiya [41]3 years ago
4 0

Answer:

the is how far off the ground a typical small truck is it is happen to be physically realistic

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After observing a moth that is camouflaged against dark-colored bark, a scientist asks a question. The scientist discovers that
Rasek [7]
Because of how it's worded the answer would most likely be number four                                                                                       

6 0
3 years ago
Read 2 more answers
A 1500 kg car skids to a halt on a wet road where μk = 0.47. You may want to review (Pages 141 - 145) . Part A How fast was the
shusha [124]

The car travels at a speed of 25m/s.

<u>Explanation:</u>

Given-

Mass, m = 1500kg

Coefficient of friction, μk = 0.47

Distance, x = 68m

Speed, s = ?

We know,

Force, F = ma

and

F = μ X m X g

Therefore,

μ * m * g = m * a

μ * g = a

Let, g = 9.8m/s²

So,

a = 0.47 * 9.8 m/s^2

a = 4.606m/s^2

We know,

v^2 - u^2 = 2as

where, v is the final velocity

           u is the initial velocity

           a is the acceleration

           s is the distance

If the car comes to rest, the final velocity, v becomes 0.

So,

u^2 = 2 * 4.606 * 68\\\\u^2 = 626.416m/s\\\\u = 25m/s

The car travels at a speed of 25m/s.

6 0
3 years ago
Suppose I have a vector that is 7 units long and that makes an angle of +30 degrees from the positive x-axis. I want to add to t
Vinil7 [7]

Answer:

sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

Explanation:

We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis

So x component of the vector =7cos30^{\circ}=7\times 0.866=6.06

y component of the vector =7sin30^{\circ}=7\times 0.5=3.5

So vector will be 6.06i+3.5j

Now other vector of length of 7 units and makes an angle of 120° with positive x-axis

So x component of vector  =7cos120^{\circ}=7\times -0.5=-3.5i

y component of the vector =7sin120^{\circ}=7\times 0.866=6.06j

Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

5 0
3 years ago
Listed following is a set of statements describing individual stars or characteristics of stars. Match these to the appropriate
Oksana_A [137]

Answer:

Red giant or super giant → very cool but very luminous

                                      → found in the upper right of the H-R diagram.

Main sequence →The majority of stars in our galaxy

                        → Sun, for example

                        → a very hot and very luminous star

White dwarfs → very hot but very dim

                     → not much larger in radius than earth  

Explanation:

Giant:

When the stars run out of their fuel that is hydrogen for the nuclear fusion reactions then they convert into Giant stars.That's why they are very cool. Giant stars have the larger radius and luminosity then the main sequence stars.

Main Sequence:

Stars are called main sequence stars when their core temperature reaches up to 10 million kelvin and their start the nuclear fusion reactions of hydrogen into helium in the core of the star. That is why they are very hot and luminous. For example sun is known as to be in the stage of main sequence as the nuclear fusion reactions are happening in its core.

White dwarfs:

When the stars run out of their fuel then they shed the outer layer planetary nebula, the remaining core part that left behind is called as white dwarf. It's the most dense part as the most of the mass is concentrated in this part.

6 0
3 years ago
Calculate the mass of a liquid with a density of 2.5 g/ml and a volume of 15ml
Nana76 [90]
Using the density equation and clearing mass:

\rho = \frac{m}{V}\ \to\ m = \rho\cdot V = 2.5\frac{g}{mL}\cdot 15\ mL = \bf 37.5\ g
3 0
3 years ago
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