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3 years ago

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A wheel rotates with a constant angular acceleration of 3.45 rad/s^2. Assume the angular speed of the wheel is 1.85 rad/s at ti

= 0.
(a) Through what angle does the wheel rotate between t = 0 and t = 2.00 s

1 answer:

lesantik [10]3 years ago

8 0

Answer:

θ =607.33°

Explanation:

Given that

Angular acceleration α = 3.45 rad/s²

Initial angular speed ,ω = 1.85 rad/s

The angle rotates by wheel in time t

$\theta=\omega t +\dfrac{1}{2}\alpha t^2$

Now by putting the values

$\theta=\omega t +\dfrac{1}{2}\alpha t^2$

$\theta=1.85\times 2 +\dfrac{1}{2}\times 3.45\times 2^2$

θ = 10.6 rad

$\theta=\dfrac{180}{\pi}\times 10.6\ degree$

θ =607.33°

Therefore angle turn by wheel in 2 s is θ =607.33°

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Sasha sits on a horse on a carousel 3.5 m from the center of the circle. She makes a revolution once every 8.2 seconds. What is

Leokris [45]

Answer: 2.7 m/s

Explanation:

Given the following :

Period (T) = 8.2 seconds

Radius = 3.5 m

The tangential speed is given as:

V = Radius × ω

ω = angular speed = (2 × pi) / T

ω = (2 × 22/7) / 8.2

ω = 6.2857142 / 8.2

ω = 0.7665505

Therefore, tangential speed (V) equals;

r × ω

3.5 × 0.7665505 = 2.6829268 m/s

2.7 m/s

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3 years ago

The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T

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a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

f) The two trains avoid collision if the acceleration of the freight train is at least $0.35 m/s^2$

Explanation:

a)

We can describe the position of the passenger train at time t with the equation

$x_p(t)=u_p t + \frac{1}{2}at^2$

where

$u_p = 25.0 m/s$ is the initial velocity of the passenger train

$a=-0.100 m/s^2$ is the deceleration of the train

On the other hand, the position of the freight train is given by

$x_f(t)=x_0 + v_f t$

where

$x_0=200 m$ is the initial position of the freight train

$v_f = 15.0 m/s$ is the constant velocity of the train

The collision occurs if the two trains meet, so

$x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0$

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

b)

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

$x_f(t)=x_0 +v_ft$

And substituting t = 22.5 s, we find:

$x_f(22.5)=200+(15)(22.5)=537.5 m$

We can verify that the passenger train is at the same position at the time of the collision:

$x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m$

So, the two trains collide at x = 537.5 m.

c)

In the graph in attachment, the position-time graph of each train is represented. We have:

The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

d)

In order to avoid the collision, the freight train must have a initial position of

$x_0'$

such that the two trains never meet.

We said that the two trains meet if:

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t$

Re-arranging,

$\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0$

Substituting the values for the acceleration and the velocity,

$0.05t^2-10t+x_0'=0$

The solution of this equation is given by the formula

$t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}$

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

$10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m$

Therefore, the freight train must have a head start of 500 m.

e)

In this case, we want to find the acceleration $a'$ of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t$

Re-arranging

$\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0$

Substituting,

$-0.5at^2-10t+200=0$

The solution to this equation is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}$

Again, the two trains never meet if the discriminant is negative, so

$10^2-4\cdot (-0.5a') \cdot (200)$

So, the deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

f)

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

$x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2$

where $a_f$ is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2$

Re-arranging,

$\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0$

And the solution is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}$

Again, the two trains avoid collision if the discriminant is negative, so

$10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2$

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3 years ago

A 5.0-nC point charge is embedded at the center of a nonconducting sphere (radius = 2.0 cm) which has a charge of -8.0 nC distri

Igoryamba

Answer:

3.6 × 10⁵ N/C = 360 kN/C

Explanation:

Let R = 2.0 cm be the radius of the sphere and q = -8.0 nC be the charge in it. Let q₁ be the charge at radius r = 1.0 cm. Since the charge is uniformly distributed, the volume charge density is constant. So, q/4πR³ = q₁/4πr³

q₁ = q(r/R)³. The electric field due to q₁ at r is E₁ = kq₁/r² = kq(r/R)³/r² = kqr/R³

The electric field due to the point charge q₂ = 5.0 nC is E₂ = kq₂/r².

So, the magnitude of the total electric field at r = 1.0 cm is

E = E₁ + E₂ = kqr/R³ + kq₂/r² = k(qr/R³ + q₂/r²)

E = 9 × 10⁹(-8 × 10⁻⁹ C × 1 × 10⁻² m/(2 × 10⁻² m)³ + 5 × 10⁻⁹ C/(1 × 10⁻² m)²)

E = 9 × 10⁹(-1 × 10⁻⁵ + 5 × 10⁻⁵)

E = 9 × 10⁹(4 × 10⁻⁵)

E = 36 × 10⁴ N/C = 3.6 × 10⁵ N/C = 360 kN/C

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