Ask question

Ask question

All categories

2 years ago

12

A truck with 0.410 m radius tires travels at 25.0 m/s. What is the angular velocity of the rotating tires in radians per second?

1 answer:

Eduardwww [97]2 years ago

6 0

relation between linear velocity and angular velocity is given as

$v = R\omega$

here

v = linear speed

R = radius

$\omega$ = angular speed

now plug in all data in the equation

$25.0 = 0.410 \omega$

$\omega = \frac{25}{0.410}$

$\omega = 60.9 rad/s$

so rotating speed is 60.9 rad/s

You might be interested in

Acid precipitation is caused by a mix of

saveliy_v [14]

C. sulfur dioxide, carbon dioxide, and nitrogen oxides

These three chemical compounds play a role in the chemistry of acid rain or acid precipitation. These compounds are then in the process as they evaporate are converted into acids, mainly, nitric acid and sulfuric acid. These toxic gases that composes acid precipitation is actually caused by industries, factories or plants that utilizes such chemicals and then uses a humongous amount which then is released in during the process of certain products or manufactured objects.

4 0

2 years ago

Two curling stones collide on an ice rink. Stone 1 has a mass of 16 kg

nika2105 [10]

Answer:

53.57 kg or 54 kg

-.81 m/s

Explanation:

m2= $\frac{v}{vf2} \\$ (2m1) -m1

v1f= $(\frac{m1-m2}{m1+m2})v$

4 0

2 years ago

The columns in the periodic table are called groups. What do the elements in group 18 have in common

dimaraw [331]

Answer: D

Explanation:

they are all odorless, colorless, monatomic gases with very low chemical reactivity.

Brainliest?

7 0

2 years ago

132 projects in science fair. if 8 projects fit in a row. how many full rows

Umnica [9.8K]

The real answer would be 16.5 but since you want to have a full number it would be 16

3 0

2 years ago

Read 2 more answers

The acceleration of a particle is defined by the relation a 5 28 m/s2. Knowing that x 5 20 m when t 5 4 s and that x 5 4 m when

mamaluj [8]

Explanation:

It is given that,

The acceleration of a particle, $a=-8\ m/s^2$ (negative as the particle is decelerating)

Initial distance, x₁ = 20 m

Initial time, t₁ = 4 s

New distance x₂ = 4 m

Velocity, v = 10 m/s

(A) Calculating initial distance using second equation of motion as :

$x_1=ut_1+\dfrac{1}{2}at^2$

$20=4u+\dfrac{1}{2}(-8)\times 4^2$

u = 21 m/s

When velocity of the particle is zero, time taken is t (say). Using first equation of motion as :

$v=u+at$

$0=21+(-8)t$

t = 2.62 seconds

So, the velocity of the particle is zero at t = 2.62 seconds.

(B) Velocity at t = 11 s

$v=21+(-8)\times 11$

v = 13 m/s

Total distance covered at t = 11 s. The overall path travelled by the particle during its entire journey is called total distance covered.

$d=ut+\dfrac{1}{2}at^2+|ut+\dfrac{1}{2}at^2|$

$d=21\times 2.62+\dfrac{1}{2}\times (-8)(2.62)^2+|21\times 8.38+\dfrac{1}{2}\times (-8)(8.38)^2|$

d = 132.48 m

So, the distance travelled by the particle at t = 11 seconds is 132.48 meters.

5 0

2 years ago