It is 25 miles from your house to a friend's house. Which unit conversion

Reference frame definitely changes when what also changes?

kompoz [17]

Explanation is in the file

tinyurl.com/wpazsebu

8 0

2 years ago

Read 2 more answers

Laws that implemented the consumers' right to be informed forbid __________.

Reptile [31]

Laws that implemented the consumers' right to be informed forbid misleading advertising.

Answer is C.

4 0

3 years ago

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Between 1911 and 1990, the top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of 1.2 mm/y.T

Anit [1.1K]

Answer:

Angular speed, $\omega=6.90\times 10^{-13}\ rad/s$

Explanation:

It is given that,

The top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of, v = 1.2 mm/yr

$1\ mm/yr=3.171\times 10^{-11}\ m/s$

Velocity, $v=3.80\times 10^{-11}\ m/s$

Height of the tower, h = 55 m

The height of the tower is equivalent to the radius. Let $\omega$ is the angular speed of the tower’s top about its base. The relation between the angular speed and the angular speed is given by :

$v=r\omega$

$\omega=\dfrac{v}{r}$

$\omega=\dfrac{3.80\times 10^{-11}\ m/s}{55\ m}$

$\omega=6.90\times 10^{-13}\ rad/s$

So, the average angular speed of the tower’s top about its base is $6.90\times 10^{-13}\ rad/s$ . Hence, this is the required solution.

6 0

3 years ago

A bike with 15cm diameter wheels accelerates uniformly from rest to a speed of 7.1m/s over a distance of 35.4m. Determine the an

Finger [1]

Answer:

9.47 rad/s^2

Explanation:

Diameter = 15 cm, radius, r = diameter / 2 = 7.5 cm = 0.075 m, u = 0, v = 7.1 m/s,

s = 35.4 m

let a be the linear acceleration.

Use III equation of motion.

v^2 = u^2 + 2 a s

7.1 x 7.1 = 0 + 2 x a x 35.4

a = 0.71 m/s^2

Now the relation between linear acceleration and angular acceleration is

a = r x α

where, α is angular acceleration

α = 0.71 / 0.075 = 9.47 rad/s^2

8 0

3 years ago

A 0.5 m diameter wagon wheel consists of a thin rim having a mass of 7 kg and six spokes, each with a mass of 1.2 kg. 1.2 kg 7 k

Arte-miy333 [17]

Explanation:

It is given that,

Mass of the rim of wheel, m₁ = 7 kg

Mass of one spoke, m₂ = 1.2 kg

Diameter of the wagon, d = 0.5 m

Radius of the wagon, r = 0.25 m

Let I is the the moment of inertia of the wagon wheel for rotation about its axis.

We know that the moment of inertia of the ring is given by :

$I_1=m_1r^2$

$I_1=7\times (0.25)^2=0.437\ kgm^2$

The moment of inertia of the rod about one end is given by :

$I_2=\dfrac{m_2l^2}{3}$

l = r

$I_2=\dfrac{m_2r^2}{3}$

$I_2=\dfrac{1.2\times (0.25)^2}{3}=0.025\ kgm^2$

For 6 spokes, $I_2=0.025\times 6=0.15\ kgm^2$

So, the net moment of inertia of the wagon is :

$I=I_1+I_2$

$I=0.437+0.15=0.587\ kgm^2$

So, the moment of inertia of the wagon wheel for rotation about its axis is $0.587\ kgm^2$ . Hence, this is the required solution.

4 0

3 years ago