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4 years ago

5

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 28.0 m/s. With what maximum sp

eed can it go around a curve having a radius of 73.0 m?

1 answer:

Dima020 [189]4 years ago

5 0

Given Information:

Radius = r₁ = 150 m

Radius = r₂ = 73 m

Maximum speed = v₁ = 28.0 m/s

Required Information:

Maximum speed = v₂ = ?

Answer:

Maximum speed v₂ = 19.53 m/s

Explanation:

The maximum acceleration of the truck is

a = v²₁/r₁

a = 28²/150

a = 5.226 m/s²

Now if we change the radius to r₂ = 73 m

Then the maximum speed of the truck is

v₂ = √a*r₂

v₂ = √5.226*73

v₂ = 19.53 m/s

So the speed of the truck is decreased when the radius of curve is decreased.

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Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

so

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$

this means, for the angular momentum measure from the rod:

$\frac{d\vec{L}_{rod}}{dt} = 0$

that means :

$\vec{L}_{rod} = constant$

So, the magnitude of initial angular momentum is :

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)$

but the angle is 90°, so:

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|$

$| \vec{L}_{rod_i} | = r_i * m * v_i$

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

$| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}$

$| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}$

For our final angular momentum we have:

$| \vec{L}_{rod_f} | = r_f * m * v_f$

and the radius is 0.250 m and the mass is 2.00 kg

$| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f$

but, as the angular momentum is constant, this must be equal to the initial angular momentum

$7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f$

$v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}$

$v_f = 15 \frac{m}{s}$

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3 years ago

Pls someone I need it urgently and explain Solving and explanation so I can understand Thank you

Temka [501]

Answer:

f = 6.37 Hz, T = 0.157 s

Explanation:

The expression you have is

y = 5 sin (3x - 40t)

this is the equation of a traveling wave, the general form of the expression is

y = A sin (kx - wt)

where A is the amplitude of the motion, k the wave vector and w the angular velocity

Angle velocity and frequency are related

w = 2π f

f = w / 2π

from the equation w = 40 rad / s

f = 40 / 2π

f = 6.37 Hz

frequency and period are related

f = 1 / T

T = 1 / f

T = 1 / 6.37

T = 0.157 s

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3 years ago