All categories

1 year ago

A bike with tires of radius 0.330 m speeds up from rest to 5.33 m/s in 6.27 s. What is the angular acceleration of the tires?

1 answer:

8 0

The angular acceleration of a bike with tires of radius 0.330 m speeds up from rest to 5.33 m/s in 6.27 s will be 2. 57 rad/s²

<h3>What is angular acceleration?</h3>

Angular acceleration can be defined as the time it takes for a change in angular velocity. It is denoted as 'α' with a unit of rads/s²

It is expressed thus;

α= Δω ÷ Δt

Where α = angular acceleration

Δw = change in angular velocity = velocity ÷ radius

Δ t = change in time

<h3>How to calculate the angular acceleration</h3>

Using the formula:

α= Δω ÷ Δt

v = 5. 33m/s, r = 0. 33m and t = 6.27s

Substitute the values to get the angular velocity

Δw = v÷ r = 5. 33 ÷ 0.330 = 16. 15 m/s

Substitute the value of Δw into the equation

α= Δω ÷ Δt = 16. 15 ÷ 6. 27 = 2. 57 rad/s²

Therefore, the angular acceleration of a bike with tires radius of 0. 330m, speed of 5. 33mls in 6. 27s is 2. 57 rad/s²

Learn more about angular acceleration here:

brainly.com/question/21278452

#SPJ1

You might be interested in

block with of mass m is at rest on horizontal frictionless surface at time t=0. A force given by F=Bt+C is applied horizontally

Alexeev081 [22]

Answer:

$v_{2} =\frac{1}{2}$

Explanation:

From the second law of Newton movement laws, we have:

$F=m*a$ , and we know that a is the acceleration, which definition is:

$a=\frac{dv}{dt}$ , so:

$F=m*\frac{dv}{dt}\\\frac{dv}{dt}=\frac{F}{m}=\frac{\frac{1}{2}(t+1)}{4}=\frac{t+1}{8}$

The next step is separate variables and integrate (the limits are at this way because at t=0 the block was at rest (v=0):

$dv=\frac{1}{8}(t+1)dt\\\int\limits^{v_{2}}_0 \, dv=\int\limits^{2}_{0} {\frac{1}{8}(t+1)} \, dt$

$v_{2}=\frac{1}{8}*(\frac{t^{2}}{2}+t)$ (This is the indefinite integral), the definite one is:

$v_{2}=\frac{1}{8}*(2+2)=\frac{1}{2}$

3 0

3 years ago

An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance

marshall27 [118]

Answer:

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Explanation:

Let the linear charge density of the charged wire is given as

$\frac{q}{L} = \lambda$

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

$\int E. dA = \frac{q}{\epsilon_0}$

here we have

$E \int dA = \frac{\lambda L}{\epsilon_0}$

$E. 2\pi r L = \frac{\lambda L}{\epsilon_0}$

so we have

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

4 0

2 years ago

Derived quantities depend on...........physical quantity

Vesnalui [34]

Derived quantities depend on.( fundamental)..........physical quantity

Are you from Nepal?

7 0

2 years ago

Read 2 more answers

Why would an insurance company want people to avoid risk, like Nick Cannon having to avoid skydiving?

Lina20 [59]

An insurance company wants people to avoid risks because if the person meets with an accsident the expense of the medicines and treatment should be covered by the insurance company.

5 0

2 years ago

Read 2 more answers

Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag

nadya68 [22]

Answer:

F = 147,78*10⁻⁹ [N]

Explanation:

By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .

The angle β ( angle between the line running through one of the charges in y axis and the charge in x axis) is

tan β = 0,5/0,7

tan β = 0,7142 then β = arctan 0,7142 ⇒ β = 35 ⁰

cos β = 0,81

d = √ (0,5)² + (0,7)² d1stance between charges

d = √0,25 + 0,49

d = √0,74 m

d = 0,86 m

Now Foce between two charges is:

F = K* q₁*q₂/ d² (1)

Where K = 9*10⁹ N*m²/C²

q₁ = 2,5* 10⁻⁹C

q₂ = 3,0*10⁻⁹C

d² = 0,74 m²

Plugging these values in (1)

F = 9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74 [N*m²/C²]*C*C/m²

F = 91,21 * 10⁻⁹ [N]

And Fx = F*cos β

Fx = 91,21 * 10⁻⁹ *0,81

Fx =73,89*10⁻⁹ [N]

Then total force acting on charge located at x = 0,7 m is:

F = 2* Fx

F = 2*73,89*10⁻⁹ [N]

F = 147,78*10⁻⁹ [N]

8 0

3 years ago