This table shows data collected by three devices: a thermometer on a weather balloon, a surface thermometer, and a barometer.

1 ) Starting from rest, a toy rocket accelerates at 12 m/sec/sec for exactly 4.0 seconds. It reaches 48 m/sec. Find the distance

dusya [7]

Displacement equals (Velocity times Time) plus half times the (acceleration times time squared). =. (48 * 4) + 1/2 * (12 *12^2) = 288meters

3 0

2 years ago

Una tractomula se desplaza con rapidez de 69 km/h. Cuando el conductor ve una vaca atravesada enmedio de la carretera, acciona l

Anestetic [448]

Answer:

Los datos que tenemos:

Rapidez: 69km/h

Tiempo que tarda en frenar = 4s.

Distancia inicial entre la tracto-mula y la vaca = 25m

Ok, la ecuación de desaceleración es:

D = (sf - si)/t

sf = velocidad final = 0m/s

si = velocidad inicial = 69km/h

t = tiempo = 4s

D = -69km/h/4s

ok, 1h = 3600s

D = (-69km/s)*1/(4*3600s) = -0.0048 km/s^2

Entonces la ecuación de aceleración es:

a(t) = -0.0048 km/s^2

Para la velocidad, integramos sobre el tiempo

v(t) = (-0.0048 km/s^2)*t + v0

donde v0 es la velocidad inicial, en este caso v0 = 69km/3600s = 0.0191km/s

v(t) = (-0.0048 km/s^2)*t + 0.0191km/s

Para la posición volvemos a integrar sobre el tiempo, esta vez suponemos la posición inicial igual a cero.

p(t) = (1/2)*(-0.0048 km/s^2)*t^2 + 0.0191m/s*t

Ahora, si p(t=4s) < 25m, esto implica que la tracto-mula no impacto con la vaca.

p(4s) = (1/2)*(-0.0048 km/s^2)*(4s)^2 + 0.0191km/s*4s = 0.038km

y 1km = 1000m

0.038km = 0.038*1000m = 38m

Entonces si, atropello a la vaca.

4 0

3 years ago

You are an engineer helping to design a roller coaster that carries passengers down a steep track and around a vertical loop. Th

vova2212 [387]

Answer:

h >5/2r

Explanation:

This problem involves the application of the concepts of force and the work-energy theorem.

The roller coaster undergoes circular motion when going round the loop. For the rider to stay in contact with the cart at all times, the roller coaster must be moving with a minimum velocity v such that at the top the rider is in a uniform circular motion and does not fall out of the cart. The rider moves around the circle with an acceleration a = v²/r. Where r = radius of the circle.

Vertically two forces are acting on the rider, the weight and normal force of the cart on the rider. The normal force and weight are acting downwards at the top. For the rider not to fall out of the cart at the top, the normal force on the rider must be zero. This brings in a design requirement for the roller coaster to move at a minimum speed such that the cart exerts no force on the rider. This speed occurs when the normal force acting on the rider is zero (only the weight of the rider is acting on the rider)

So from newton's second law of motion,

W – N = mv²/r

N = normal force = 0

W = mg

mg = ma = mv²/r

mg = mv²/r

v²= rg

v = √(rg)

The roller coaster starts from height h. Its potential energy changes as it travels on its course. The potential energy decreases from a value mgh at the height h to mg×2r at the top of the loop. No other force is acting on the roller coaster except the force of gravity which is a conservative force so, energy is conserved. Because energy is conserved the total change in the potential energy of the rider must be at least equal to or greater than the kinetic energy of the rider at the top of the loop

So

ΔPE = ΔKE = 1/2mv²

The height at the roller coaster starts is usually higher than the top of the loop by design. So

ΔPE =mgh - mg×2r = mg(h – 2r)

2r is the vertical distance from the base of the loop to the top of the loop, basically the diameter of the loop.

In order for the roller coaster to move smoothly and not come to a halt at the top of the loop, the ΔPE must be greater than the ΔKE at the top.

So ΔPE > ΔKE at the top. The extra energy moves the rider the loop from the top.

ΔPE > ΔKE

mg(h–2r) > 1/2mv²

g(h–2r) > 1/2(√(rg))²

g(h–2r) > 1/2×rg

h–2r > 1/2×r

h > 2r + 1/2r

h > 5/2r

5 0

3 years ago

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B. Projectile on cliff (range)

dimulka [17.4K]

Answer:

x = 41.28 m

Explanation:

This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.

Let's start by using trigonometry to find the initial velocity

cos 25 = v₀ₓ / v₀

sin 25 = Iv_{oy} / v₀

v₀ₓ = v₀ cos 25

v_{oy} = v₀ sin 25

v₀ₓ = 22 cos 25 = 19.94 m / s

v_{oy} = 22 sin 25 = 0.0192 m / s

let's use movement on the vertical axis

y = y₀ + v_{oy} t - ½ g t²

when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m

0 = 21 + 0.0192 t - ½ 9.81 t²

4.905 t² - 0.0192 t - 21 = 0

t² - 0.003914 t - 4.2813 =0

we solve the quadratic equation

t = $\frac{ 0.003914\ \pm \sqrt{0.003914^2 + 4 \ 4.2813 } }{2}$

t = $\frac{0.003914 \ \pm 4.13828}{2}$

t₁ = 2.07 s

t₂ = -2.067 s

since time must be a positive scalar quantity, the correct result is

t = 2.07 s

now we can look up the distance traveled

x = v₀ₓ t

x = 19.94 2.07

x = 41.28 m

5 0

3 years ago

What color is reflected off of most plant leaves?

serg [7]

Green is reflected off of most plant leaves.

3 0

3 years ago

Read 2 more answers