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Svetradugi [14.3K]
3 years ago
9

A truck with 0.420-m-radius tires travels at 32.0 m/s. what is the angular velocity of the rotating tires in radians per second?

what is this in rev/min?
Physics
1 answer:
andrey2020 [161]3 years ago
4 0

Angular velocity of the rotating tires can be calculated using the formula,

v=ω×r

Here, v is the velocity of the tires = 32 m/s

r is the radius of the tires= 0.42 m

ω is the angular velocity

Substituting the values we get,

32=ω×0.42

ω= 32/0.42 = 76.19 rad/s

= 76.19×\frac{1}{2\pi} *60 revolution per min

=728 rpm

Angular velocity of the rotating tires is 76.19 rad/s or 728 rpm.

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A jogger accelerates from rest to 4.86 m/s in 2.43 s. A car accelerates from 20.6 to 32.7 m/s also in 2.43 s. (a) Find the magni
Aleonysh [2.5K]

Explanation:

It is given that,

Initially, the jogger is at rest u₁ = 0

He accelerates from rest to 4.86 m, v₁ = 4.86 m

Time, t₁ = 2.43 s

A car accelerates from u₂ = 20.6 to v₂ = 32.7 m/s in t₂ = 2.43 s

(a) Acceleration of the jogger :

a=\dfrac{v-u}{t}

a=\dfrac{4.86\ m/s-0}{2.43\ s}

a₁ = 2 m/s²

(b) Acceleration of the car,

a=\dfrac{v-u}{t}

a=\dfrac{32.7\ m/s-20.6\ m/s}{2.43\ s}

a₂ = 4.97 m/s²

(c) Distance covered by the car,

d_1=u_1t_1+\dfrac{1}{2}a_1t_1^2

d_1=0+\dfrac{1}{2}\times 2\times (2.43)^2

d₁ = 5.904 m

Distance covered by the jogger,

d_2=u_2t_2+\dfrac{1}{2}a_2t_2^2

d_2=20.6\times 2.43+\dfrac{1}{2}\times 4.97\times (2.43)^2

d₂ = 64.73 m

The car further travel a distance of, d = 64.73 m - 5.904 m = 58.826 m

Hence, this is the required solution.

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2 years ago
In a chemical equation, where do the products appear
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