Ask question

Ask question

All categories

3 years ago

15

Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves f

rom point X and travels counter-clockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?

1 answer:

stealth61 [152]3 years ago

4 0

Answer:

20.96 h

Explanation:

The perimeter of the track is 2*pi*r = 20pi miles

In 10 hours, car B would have moved 20miles. So, when Car A leaves from point X, car B is 20pi - 20 miles from point X counter-clockwise and car A.

From here, we can express the distance of A from X like this:

xa = 3t

And the distance of B would be:

xb = 20pi - 20 - 2t

The time t where they would passed each other and put 12 miles between them would be the one where xa - xb is equal to 12:

xa - xb = 12

3t - (20pi - 20 - 2t) = 12

5t = 20 pi - 8

t = (20pi - 8)/5 = 10.96 h

Remember to add this value to the 10 hours car B had already been racing:

t = 20.96h

You might be interested in

FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0

1 year ago

Can someone help me?

Gwar [14]

24- series
25- parallel
26- no, because they’re connected in series
27- yes, because they’re connected in parallel

7 0

3 years ago

The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto was subsequently discover

givi [52]

Answer:

Acceleration due to gravity, $a=4.61\times 10^{-14}\ m/s^2$

Explanation:

It is given that,

Mass of Pluto, $m=1.4\times 10^{22}\ kg$

Distance between Neptune and Pluto, $r=4.5\times 10^{12}\ m$

The force of gravity is balanced by the gravitational force between Neptune and Pluto. It is given by :

$a=\dfrac{Gm}{r^2}$

$a=\dfrac{6.67\times 10^{-11}\times 1.4\times 10^{22}}{(4.5\times 10^{12})^2}$

$a=4.61\times 10^{-14}\ m/s^2$

So, the acceleration due to gravity at Neptune due to Pluto is $4.61\times 10^{-14}\ m/s^2$ . Hence, this is the required solution.

4 0

4 years ago

When we come out in the sun, we feel warm. How does the heat from the sun reach us?

dsp73

Answer:

<h2>Radiation</h2>

Explanation:

<h3>Greetings !</h3>

The Sun reaches us by propagating through the vacuum of space. Sunlight reaches the Earth in about 8 minutes and 20 seconds. When this energy reaches the Earth's atmosphere, both conduction and convection play key roles to scatter it throughout the planet.

3 0

2 years ago

Read 2 more answers

If you drop an object from a height of 1.9 m, it will hit the ground in 0.62 s. if you throw a baseball horizontally with an ini

faltersainse [42]

That's the cool thing about free fall. The amount of time it takes to fall remains the same.

In this case, a ball that is simply dropped from rest will fall at the same rate as a ball that had some umph in the horizontal direction.

6 0

3 years ago