Ask question

Ask question

All categories

3 years ago

10

A typical running track is an oval with 74-mm-diameter half circles at each end. A runner going once around the track covers a d

istance of 400 mm . Suppose a runner, moving at a constant speed, goes once around the track in 1 minmin 40 s.What is her centripetal acceleration during the turn at each end of the track?

1 answer:

lisabon 2012 [21]3 years ago

6 0

The centripetal acceleration a is 4.32 $\times$ 10^-4 m/s^2.

<u>Explanation:</u>

The speed is constant and computing the speed from the distance and time for one full lap.

Given, distance = 400 mm = 0.4 m, Time = 100 s.

Computing the v = 0.4 m / 100 s

v = 4 $\times$ 10^-3 m/s.

radius of the circular end r = 37 mm = 0.037 m.

centripetal acceleration a = v^2 / r

= (4 $\times$ 10^-3)^2 / 0.037

a = 4.32 $\times$ 10^-4 m/s^2.

You might be interested in

If a longitudinal wave passes a specific point seven times per second, and the distance between wave rarefaction points is 2 met

zavuch27 [327]

Answer: 14 m/s

Explanation:

The speed $S$ of a sound wave is given by the following equation:

$S=\lambda f$

Where:

$\lambda=2 m$ is the wavelength of the sound wave

$f=7 times/s=7 Hz$

$S=(2 m)(7 Hz)$

Hence:

$S=14 m/s$

4 0

3 years ago

Describe an experiment to show that pressure increases with the decrease in the area of surface

cluponka [151]

Answer:
press a baloon against one pin it bursts
but now arrange lot of pins parallel closely to each other if u press a baloon against them it does not burst hope this helps u

7 0

2 years ago

I've got an energy and work problem. The premise of the problem is:

Alenkasestr [34]

Refer to the diagram shown below.

μ = the coefficient of dynamic friction between the crate and the ramp.

1. The applied force of F acts over a distance, d.
The work done is F*d.

2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.

3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.

4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.

5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)

6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)

7 0

4 years ago

Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of

marusya05 [52]

Answer:

a) $\dot m = 6.878\,\frac{kg}{s}$ , b) $T = 104.3^{\textdegree}C$ , c) $\dot S_{gen} = 11.8\,\frac{kW}{K}$

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

$-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The mass flow rate is:

$\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}$

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

$P = 7000\,kPa$

$T = 500^{\textdegree}C$

$h = 3411.4\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

State 2s - Liquid-Vapor Mixture

$P = 100\,kPa$

$h = 2467.32\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

$x = 0.908$

The isentropic efficiency is given by the following expression:

$\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}$

The real specific enthalpy at outlet is:

$h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})$

$h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )$

$h_{2} = 2684.46\,\frac{kJ}{kg}$

State 2 - Superheated Vapor

$P = 100\,kPa$

$T = 104.3^{\textdegree}C$

$h = 2684.46\,\frac{kJ}{kg}$

$s = 7.3829\,\frac{kJ}{kg\cdot K}$

The mass flow rate is:

$\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}$

$\dot m = 6.878\,\frac{kg}{s}$

b) The temperature at the turbine exit is:

$T = 104.3^{\textdegree}C$

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

$\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0$

$\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})$

$\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )$

$\dot S_{gen} = 11.8\,\frac{kW}{K}$

4 0

3 years ago

A train is approaching the station. Explain what happens to the frequency of the sound as the train draws closer. What do you he

Hitman42 [59]

The frequency of sound waves goes up, so the pitch goes up as well. More frequent=higher pitched

3 0

4 years ago

Read 2 more answers