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trapecia [35]
3 years ago
13

An athlete is running a 400m race around a 400m track. On the backstretch the athlete's velocity is 8m/s but he is running into

a 2m/s headwind. How large is the drag force that acts on him? Assume that the density of the air is 1.2kg/m^3, his cross sectional area is 0.5m^2, and the coefficient of drag is 1.1.
Physics
1 answer:
Aleksandr-060686 [28]3 years ago
3 0

Answer:

33 N

Explanation:

v = Velocity of fluid = 8+2 = 10 m/s

\rho = Density of fluid = 1.2 kg/m³

C = Coefficient of drag = 1.1

A = Cross sectional area = 0.5 m²

Drag force is given by

F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2\times 1.1\times 0.5\times (8+2)^2\\\Rightarrow F=33\ N

The drag force on the athlete is 33 N

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Consider an ideal monatomic gas of N particles with mass m in thermal equilibrium at a temperature T. The gas is contained in a
harina [27]

Answer:

K.E.=\dfrac{3}{2}KT

Explanation:

Given that

Number of particle =N

Equilibrium temperature= T

Side of cube = L

Gravitational acceleration =g

The kinetic energy of an atom  given as

K.E.=\dfrac{3}{2}KT

Where

Equilibrium temperature= T

Boltzmann constant =K

        K =1.380649×10−23 J/K

3 0
3 years ago
Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendu
erica [24]

Answer:

a)  v = 16.57 m / s, b)  a = 19.6 m / s², d)    N = 1.76 10³ N,     N / W = 3

Explanation:

This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.

Let's answer the questions.

a) For this part we can use energy considerations.

Starting point. The upper part of the trajectory indicates that the arm is horizontally

          Em₀ = U = m g h

in this case h = r

Final point. For lower of the trajectory

          Em_f = K = ½ m v²

as they indicate that there is no friction

         Em₀ = em_f

         mgh = ½ m v²

         v = \sqrt{2gh}

let's calculate

        v = \sqrt{2 \ 9.8 \ 14.0}

         v = 16.57 m / s

b) the centripetal acceleration has the formula

           a = v² / r

           a = 16.57² / 14.0

           a = 19.6 m / s²

c) see attached where the diagram is

where N is the normal and w the weight

d) let's use Newton's second law

               N-W = m a

               N - mg = m ar

               N = m (g + a)

let's calculate

               N = 60.0 (9.8 + 19.6)

               N = 1.76 10³ N

the relationship with weight is

              N / W = 1.76 10³/( 60 9.8)

              N / W = 3

normal is three times greater than body weight

e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it

6 0
2 years ago
Understand that the acceleration vector is in the direction of the change of the velocity vector. In one dimensional (straight l
MaRussiya [10]

Answer:

 a = √ (a_t² + a_c²)

a_t = dv / dt ,    a_c = v² / r  

Explanation:

In a two-dimensional movement, the acceleration can have two components, one in each axis of the movement, so the acceleration can be written as the components of the acceleration in each axis.

            a = aₓ i ^ + a_y j ^

Another very common way of expressing acceleration is by creating a reference system with a parallel axis and a perpendicular axis. The axis called parallel is in the radial direction and the perpendicular axis is perpendicular to the movement, therefore the acceleration remains

         a = √ (a_t² + a_c²)

where the tangential acceleration is

           a_t = dv / dt

the centripetal acceleration is

          a_c = v² / r

8 0
3 years ago
The air sealed between two panes of glass in some windows helps to _____.
olga_2 [115]
<span>The air sealed between two panes of glass in some windows helps to </span><span>insulate and helps it to keep the warmth or coolness inside.

Hope that helps!
-Chris</span>
3 0
3 years ago
Read 2 more answers
Bottlenose dolphins use echolocation pulses with a frequency of about 100 kHz, higher than the frequencies used by most bats. Wh
Alinara [238K]

Answer:

1. greater

2. direct

3. smaller

4. inverse

Explanation:

The speed of sound in water is greater than in air; hence for the same frequency the sound wavelength in water is <u>greater </u>than in air (for the given frequency the wavelength is in the <u>direct </u>proportion with the speed of sound).

To "see" an object via the echolocation creature needs to use sound with the wavelength <u>smaller </u>than the size of an object viewed.

That means to "see" objects of the same size dolphin and bat need to use ultrasound of the same wavelength, hence dolphin needs to use higher frequency (for the given speed of sound the wavelength is in <u>inverse </u>proportion with the frequency).

8 0
3 years ago
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