Why wheat flour is usually passed near a magnet before being packed?

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in the linear speed

$\Delta t$ is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

$v=8322 m/s$

Therefore, its linear speed is not changing, so the change in linear speed is zero:

$\Delta v=0$

And therefore, the tangential acceleration is zero as well:

$a_t=\frac{0}{\Delta t}=0 m/s^2$

5 0

2 years ago

Hii please help i’ll give brainliest if you give a correct answer please please hurry

fiasKO [112]

Answer:

B most likely

Explanation:

tell me if this is right or wrong

8 0

2 years ago

1. A boat initially moving at 10 m/s accelerates at 2 m/s2 for 10 s. What is the velocity of the boat after 10 seconds?

DIA [1.3K]

Answer:

5 m/s is the answer so yeah

5 0

2 years ago

Consider a box with two gases separated by an impermeable membrane. The membrane can move back and forth, but the gases cannot p

padilas [110]

Answer:

The answer is " $0.3336\ m^3$ "

Explanation:

Using the Promideal gas law:

$P_A=P_B\\\\P_A(V_A-\eta_A b)= \eta_A RT......(1)\\\\P_B V_B=\eta_B \bar{R}T........(2)\\\\From (1) \zeta (2)\\\\$

$\frac{\eta_A}{V_A-\eta_A b}=\frac{\eta B}{V B}\\\\ \frac{V A- \eta_A b}{V B}=\frac{\eta A}{\eta B }\\\\ \frac{V A-b}{V B}=\frac{1}{2}\\\\V A+V B=1\\\\V B =1- V A\\\\\frac{V A-b}{1-V A}=\frac{1}{2}\\\\2V A-2b=1-V A\\\\3 V A=1+2b\\\\V A=\frac{1+2b}{3}\\\\$

$=\frac{1+2(4\times 10^{-4})}{3}\\\\=0.3336\ m^3$

8 0

2 years ago

A spring with a spring constant of 1200 N/m has a 55-g ball at its end. The energy of the system is 5.5J. What is the amplitude

Pepsi [2]

The energy of the system is E=5.5 J. This energy is the same at every moment of the oscillation. When the stretch x of the spring is maximum (so, when the stretch is equal to the amplitude: x=A) the velocity of the spring is zero, so all this energy is just elastic potential energy of the spring:
$E= \frac{1}{2}kA^2$
From which we find
$A= \sqrt{ \frac{2E}{k} }= \sqrt{ \frac{2 \cdot 5.5 J}{1200 N/m} }=0.096 m$

4 0

2 years ago