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Verizon [17]
2 years ago
12

Why would an object have a much lower maximum speed falling through water than falling through air?

Physics
2 answers:
Zinaida [17]2 years ago
8 0
The exact same principles of physics are involved regardless of the medium through which an object is falling.

That said, which particular principles are having the most effect may change. In air, for example, drag matters more than buoyancy for most solid objects, but in water this can be reversed, at least for objects that have any significant buoyancy to begin with.

No matter how you slice it, however, the same principles apply in all cases where terminal velocity applies, i.e. not in empty space.
Nonamiya [84]2 years ago
4 0

Answer:

Explanation:

when objects move through a fluid such as air or water the fluid exerts a frictional force on the moving object. the frictional force from fluid is called a drag force friction drag force causes objects to slow down as they move through a fluid such as air or water.

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uppose a wheel is initially rotating at 10.0 rad/s while undergoing constant angular acceleration reaching a speed of 30.0 rad/s
Paul [167]

Answer:

t = 22.2 s

Explanation:

angular distance covered in the 36.0 s is

θ = ω(avg)t = ½(10.0 + 30.0)36 = 720 radians

720/2 = 360 radians

α = Δω/t = (30 - 10)/36 = 5/9 rad/s²

θ = ω₀t + ½αt²

360 = 10.0t + ½(5/9)t²

   0 = (5/18)t² + 10.0t - 360

   0 = t² + 36t - 1296

t = (-36 ±√(36² - 4(1)(-1296))) / 2

t = (-36 ±√(6480)) / 2

t = -18 ±√1620

we ignore the negative time result as it occurs before we care.

t = -18 + √1620 = 22.249223... s

8 0
3 years ago
Otion
Scorpion4ik [409]

Answer:

SKID

Explanation:

In general, airplane tracks are flat, they do not have cant, consequently the friction force is what keeps the bicycle in the circle.

Let's use Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis.

Y axis y

     N- W = 0

     N = W

X axis (radial)

        fr = m a

the acceleration in the curve is centripetal

         a = \frac{v^2}{r}

the friction force has the expression

        fr = μ N

we substitute

       μ mg = m v²/r

       v = \sqrt{\mu g r}

we calculate

      v = \sqrt{0.1 \ 9.8 \ 3}

      v = 1,715 m / s

to compare with the cyclist's speed let's reduce to the SI system

        v₀ = 18 km / h (1000 m / 1 km) (1 h / 3600 s) = 5 m / s

We can see that the speed that the cyclist is carrying is greater than the speed that the curve can take, therefore the cyclist will SKID

5 0
3 years ago
The wavelength of an electromagnetic waves is __________ also known as the period. the number of waves that pass a given point i
Andreas93 [3]

The sentence can be completed as follows:

The wavelength of an electromagnetic waves is the spatial distance between two successive troughs.


Note that the wavelength of a wave can also measured as the spatial distance between two successive crests of the wave. Also note that the second part of the sentence ("also known as the period") is not true, because period is another thing (in fact, the period is the time interval between two successive troughs).



6 0
3 years ago
Just need help with 1 and 2 please :D i’m having a bit of trouble :/
dexar [7]
1. Traveling by car means you have specific roads to follow. You won’t be able to go straight to Banning high from POLAHS. The 8.4km will be defined as distance. Traveling by helicopter you don’t have roads to follow that means you can fly directly to banning high. 6.8km will be defined as displacement.

2. A) 400m
B)0m
C)d=1/2(vi+vf)t
400=1/2(0+vf)92
8.7m/s
D) 0m/s
E) Not sure but instantaneous velocity refer to velocity at a given point. Average velocity is just the average. Usually instantaneous velocity won’t be same as the average velocity.
Plz like if it helped.
7 0
3 years ago
In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le
expeople1 [14]

Answer:

The  value  is   m \approx   310

Explanation:

From the question we are told that

     The  focal length of the objective is  f_o =  1.0 \ cm

    The  focal length of the eyepiece is  f_e  =  2.0 \  cm

    The  tube length is  L  =  25 \  cm

Generally the magnitude of the overall magnification is mathematically represented as

            m =  m_o  *  m_e

Where  m_o is the objective magnification which is mathematically represented as

        m_o  =  \frac{L}{f_o }

=>      m_o  =  \frac{25}{1 }

=>      m_o  =  25

m_e is the eyepiece magnification which is mathematically evaluated as

     m_e  =  \frac{L }{f_e }

     m_e  =  \frac{25 }{ 2}

      m_e  =  12.5 \  cm

So

    m =  25 * 12.5

     m \approx   310

6 0
3 years ago
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