In a closed system that has 45 J of mechanical energy, the gravitational

A Turtle and a Snail are 360 meters apart, and they start to move towards each other at 3 p.m. If the Turtle is 11 times as fast

Neko [114]

Answer:

Snail's speed = $\frac{30m}{2400s}$ = 0.0125m/s

Turtle's speed = $\frac{330m}{2400s}$ = 0.1375m/s

Explanation:

Let the snail's speed be x m/s

The turtle's speed then is 11x m/s

Speed = Distance ÷ Time

Since speed and distance are directly proportional;

The ratio of the distances snail and turtle cover before they meet is x:11x respectively.

Simplified, the ratio of snail distance : turtle distance = 1:11

So snail covers a distance of $\frac{1}{12}$ × 360 = 30m

And turtle covers a distance of $\frac{11}{12}$ × 360 = 330m

The time each took before they met is 40 × 60 = 2400 seconds

Snail's speed = $\frac{30m}{2400s}$ = 0.0125m/s

Turtle's speed = $\frac{330m}{2400s}$ = 0.1375m/s

8 0

2 years ago

Read 2 more answers

A carousel - a horizontal rotating platform - of radius r is initially at rest, and then begins to accelerate constantly until i

OLga [1]

Answer:

α = (ω²)/8π

Explanation:

The angular acceleration(α) of the carousel can be determined by using rotational kinematics:

ω² =ωo² + 2αθ

Let's make α the subject of this equation ;

ω² - ωo² = 2αθ

α = (ω² −ωo²)/2θ

Now, from the question, since initially at rest, thus, ωo = 0

Also,since 2 revolutions, thus, θ = 2 x 2π = 4π since one revolution is 2π

Plugging in the relevant values to get ;

α = (ω²)/2(4π)

α = (ω²)/8π

7 0

2 years ago

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person

Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

$C=2\pi r$ (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

$r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m$

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

$L=Lo[1+\alpha \Delta T]$ (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

$L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m$

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

$r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m$

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

4 0

2 years ago

A car travels forward with constant velocity. It goes over a small stone, which gets stuck in the groove of a tire. The initial

puteri [66]

Answer:

A) vertically upward

Explanation:

Since the tyre is rotating with uniform angular speed and moving with constant linear speed

So as soon as a small stone is stuck into the groove of the tyre the speed of the stone is same as that of the tyre

so now we can say that stone will start revolving with the tyre of the car at constant angular speed and moving with uniform speed also

so here just after that the tangential acceleration of the stone must be zero while radial acceleration must be towards the center of the tyre given as

$a_c = \omega^2 R$

so we will have direction of net acceleration is towards its center so correct answer will be

A) vertically upward

7 0

2 years ago

a graph compares the weight and height of several different breeds of cat. What type of model is the graph?

pickupchik [31]

Answer: Mathematical Model

Explanation:

Took the test

3 0

2 years ago