Ask question

Ask question

All categories

3 years ago

9

The radial component of acceleration of a particle moving in a circular path is always:________ a. negative. b. directed towards

the center of the path. c. perpendicular to the transverse component of acceleration d. all of the above

1 answer:

lesya [120]3 years ago

7 0

Answer:

d. all of the above

Explanation:

There are two components of acceleration for a particle moving in a circular path, radial and tangential acceleration.

The radial acceleration is given by;

$a_r = \frac{V^2}{R}$

Where;

V is the velocity of the particle

R is the radius of the circular path

This radial acceleration is always directed towards the center of the path, perpendicular to the tangential acceleration and negative.

Therefore, from the given options in the question, all the options are correct.

d. all of the above

You might be interested in

A rotating cup viscometer has an inner cylinder diameter of 2.00 in., and the gap between cups is 0.2 in. The inner cylinder len

Lesechka [4]

Answer:

The dynamic viscosity and kinematic viscosity are $1.3374\times 10^{-6}$ lb-s/in2 and $1.4012\times 10^{-3}$ in2/s.

Explanation:

Step1

Given:

Inner diameter is 2.00 in.

Gap between cups is 0.2 in.

Length of the cylinder is 2.5 in.

Rotation of cylinder is 10 rev/min.

Torque is 0.00011 in-lbf.

Density of the fluid is 850 kg/m3 or 0.00095444 slog/in³.

Step2

Calculation:

Tangential force is calculated as follows:

T= Fr

$0.00011 = F\times(\frac{2}{2})$

F = 0.00011 lb.

Step3

Tangential velocity is calculated as follows:

$V=\omega r$

$V=(\frac{2\pi N}{60})r$

$V=(\frac{2\pi \times10}{60})\times1$

V=1.0472 in/s.

Step4

Apply Newton’s law of viscosity for dynamic viscosity as follows:

$F=\mu A\frac{V}{y}$

$F=\mu (\pi dl)\frac{V}{y}$

$0.00011=\mu (\pi\times2\times2.5)\frac{1.0472}{0.2}$

$\mu =1.3374\times 10^{-6}$ lb-s/in².

Step5

Kinematic viscosity is calculated as follows:

$\upsilon=\frac{\mu}{\rho}$

$\upsilon=\frac{1.3374\times 10^{-6}}{0.00095444}$

$\upsilon=1.4012\times 10^{-3}$ in2/s.

Thus, the dynamic viscosity and kinematic viscosity are $1.3374\times 10^{-6}$ lb-s/in2 and $1.4012\times 10^{-3}$ in2/s.

4 0

4 years ago

How do I put a picture on here for my graph equation??

Pavel [41]

Answer:

Take a screenshot of it and when you go to question you will see a paperclip looking thing

Explanation:

8 0

4 years ago

Read 2 more answers

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any frictio

Dmitrij [34]

Complete Question

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any friction and calculate their speed v when they reach the horizontal bottom position. At the top position, the radius of the circular path of their mass centers is 21 m, and all six cars have the same mass.V = -18 m What is v?X km/h

Answer:

$v=23.6m/s$

Explanation:

Velocity $v_c=18.0km/h$

Radius $r=21m$

initial velocity u $u=19=>5.27778$

Generally the equation for Angle is mathematically given by

$\theta=\frac{v_c}{2r}$

$\theta=\frac{18}{2*21}$

$\theta=0.45$

$\theta=25.7831 \textdegree$

Generally

Height of mass

$h=\frac{rsin\theta}{\theta}$

$h=\frac{21sin25.78}{0.45}$

$h=20.3m$

Generally the equation for Work Energy is mathematically given by

$0.5mv_0^2+mgh=0.5mv^2$

Therefore

$v=\sqrt{u^2+2gh}$

$v=\sqrt{=5.27778^2+2*9.81*20.3}$

$v=23.6m/s$

3 0

3 years ago

1. A gas pressure difference is applied to the legs of a U-tube manometer filled with a liquid with S = 1.5. The manometer readi

julia-pushkina [17]

Answer:

1) The pressure difference is 4.207 kilopascals.

2) 2.5 pounds per square inch equals 5.093 inches of mercury and 5.768 feet of water.

Explanation:

1) We can calculate the gas pressure difference from the U-tube manometer by using the following hydrostatic formula:

$\Delta P = \frac{S\cdot \rho_{w}\cdot g \cdot \Delta h}{1000}$ (Eq. 1)

Where:

$S$ - Relative density, dimensionless.

$\rho_{w}$ - Density of water, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta h$ - Height difference in the U-tube manometer, measured in meters.

$\Delta P$ - Gas pressure difference, measured in kilopascals.

If we know that $S = 1.5$ , $\rho_{w} = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta h = 0.286\,m$ , then the pressure difference is:

$\Delta P = \frac{1.5\cdot \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.286\,m)}{1000}$

$\Delta P = 4.207\,kPa$

The pressure difference is 4.207 kilopascals.

2) From Physics we remember that a pound per square unit equals 2.036 inches of mercury and 2.307 feet of water and we must multiply the given pressure by corresponding conversion unit: ( $p = 2.5\,psi$ )

$p = 2.5\,psi\times 2.037\,\frac{in\,Hg}{psi}$

$p = 5.093\,in\,Hg$

$p = 2.5\,psi\times 2.307\,\frac{ft\,H_{2}O}{psi}$

$p = 5.768\,ft\,H_{2}O$

2.5 pounds per square inch equals 5.093 inches of mercury and 5.768 feet of water.

4 0

4 years ago

Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.

vaieri [72.5K]

Answer:21.3%

Explanation:

Given

80 % reduction in tool life

According to Taylor's tool life

$VT^n$ =c

where V is cutting velocity

T=tool life of tool

80 % tool life reduction i.e. New tool Life is 0.2T

Thus

$VT^{0.12}=V'\left ( 0.2T\right )^{0.12}$

$V'=\frac{V}{0.2^{0.12}}$

$V'=\frac{V}{0.824}$ =1.213V

Thus a change of 21.3 %(increment) is required to reduce tool life by 80%

6 0

3 years ago