To develop this problem we will apply the concepts related to angular kinematic movement, related to linear kinematic movement. Linear velocity can be described in terms of angular velocity as shown below,

$v = r\omega \rightarrow \omega = \frac{v}{r}$

Here,

v = Lineal velocity

$\omega$ = Angular velocity

r = Radius

Our values are

$v = 6/ms$

$r = \frac{d}{2} = \frac{120*10^{-3}}{2} = 0.06m$

Replacing to find the angular velocity we have,

$\omega = \frac{6m/s}{0.06m}$

$\omega = 100rad/s$

Convert the units to RPM we have that

$\omega = 100rad/s (\frac{1rev}{2\pi rad})(\frac{60s}{1m})$

$\omega = 955.41rpm$

Therefore the angular speed of the wheels when the scooter is moving forward at 6.00 m/s is 955.41rpm

4 0

3 years ago

How do lone pairs of electrons affect the bond angle differently than electrons shared in a bond?

lubasha [3.4K]

Answer:

Lone pairs cause bond angles to deviate away from the ideal bond angles

Explanation:

Bonded electrons are stabilized and clustered between the bonding electrons meaning they are much closer together. Non-bonding electrons however are not being shared between any atoms which allows them to roam a little further spreading the charge density over a larger space and therefore interfering with what would be an expected bond angle

3 0

2 years ago

A rigid tank contains 1 kg of air (ideal gas) at 15 °C and 210 kPa. A paddle wheel supplies work input to the air such that fina

lisov135 [29]

Answer:

-58.876 kJ

Explanation:

m = mass of air = 1 kg

T₁ = Initial temperature = 15°C

T₂ = Final temperature = 97°C

Cp = Specific heat at constant pressure = 1.005 kJ/kgk

Cv = Specific heat at constant volume = 0.718 kJ/kgk

W = Work done

Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)

ΔU = Change in internal energy

Q = W+ΔU

⇒Q = W+mCvΔT

⇒0 = W+mCvΔT

⇒W = -mCvΔT

⇒Q = -1×0.718×(97-15)

⇒Q = -58.716 kJ

5 0

2 years ago