С какой скоростью велосипедист проходит закругление велотрека радиусом 25 м, если центростремительное ускорение при этом 4 м/с2?

Which measurement is equivalent to 350 milliliters?

350 milliliters equals to .35 liters
Therefore, 3.5 times 10^-1 would be the only correct answer since that is the only one that matches .35 liters

4 0

3 years ago

In order to walk barefoot on hot coals without hurting your feet

siniylev [52]

Before a person walks through burning coal, the person will make sure their feet are very wet. When they start walking on the coal, this moisture will evaporate and form a protective gas layer underneath the person's feet. You can see examples of this if you happen to drip some water on a hot stove or any very hot surface. The water will very easily glide around on top of a newly formed layer of air underneath it -- like air hockey pucks on an air hockey table. Note that when someone walks through burning coal, typically this is also done very quickly to prevent a great deal of exposure to possible harm. By walking quickly, thinking positively, and letting the water cushion you from immediate danger over a short distance, such a task is possible. You may have also heard of physics teachers demonstrating how this principle works by sticking their hand first in a bucket of water and then quickly in a bucket of boiling molten lead. In the lead, their hand is protected briefly by a layer of gas from the evaporated water (the water vapor). I'm fairly sure that there is a name for this particular layer of gas, but I'm afraid the name is beyond me at the moment. In other words, water vapor has a low heat capacity and poor thermal conduction. Very often, the coals or wood embers that are used in fire walking also have a low heat capacity. Sweat produced on the bottom of people's feet also helps form a protective water vapor. All of this together makes it possible, if moving quickly enough, to walk across hot coals without getting burned. WARNING: Do not attempt to perform any of the actions described above. You can seriously injure yourself. Answered by: Ted Pavlic, Electrical Engineering Undergrad Student, Ohio St. (citing my source)

5 0

3 years ago

Can you please solve this for me urgently want to make sure if my answers are correct?

MA_775_DIABLO [31]

Answer:

Resolving horizontally. :

$\sum d_{x} = - (17.0 \cos 20.0) - (11.0 \cos 35.0) + (30.0 \cos 50.0) + 0 \\ { \underline{d _{x} = - 5.702 \: m}} \\ \\ \sum d _{y} = (17.0 \sin 20.0) + 12.0 - (11.0 \sin 35.0) - (30.0 \sin 50.0) \\ { \underline{d _{y} = - 11.476 \: m}}$

therefore, for resultant:

$d = \sqrt{ {d _{y} }^{2} + d _{x} {}^{2} }$

substitute:

$d = \sqrt{ {( - 5.702)}^{2} + {( - 11.476)}^{2} } \\ \\ d = \sqrt{164.211} \\ \\ { \boxed{ \boxed{ \bf{d = 12.8 \: m}}}}$

6 0

3 years ago

A point charge q1 = 3.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 10 m.

UkoKoshka [18]

Answer:

a) 1.6 mN b) -1.6 mN c) -1.6 mN d) 1.6 mN

Explanation:

The electrostatic force between 2 point charges, obeys the Coulomb's Law, that can be expressed as follows:

F₁₂ = k*q₁*q₂/(r₁₂)² (in magnitude)

The direction of the force, is along the line that joins the charges (along the x axis) and as q₁ and q₂ are of the same sign, aims away from both charges.

a) So, for the force on q₂, we have:

F₁₂ = 9*18*10⁻⁵ N = 1.6 mN (positive as it is aiming in the positive x direction)

b) The force on q1, according to Newton's 3rd Law, is just equal and opposite to the one on q2:

F₂₁ = (-9*18*10⁻⁵) N = -1.6 mN (towards the negative x direction, away from q1)

c) If q₂ were -6.0 μC, the force will be the same in magnitude, but as now both charges have different signs, they wil attract each other, so the direction of the forces will be exactly the opposite to the first case:

F₁₂ = -1.6 mN (going towards the origin, where q₁ is located)

F₂₁ = 1.6 mN (going in the positive x direction, towards q₂)

6 0

4 years ago

Read 2 more answers

un avión aterriza en la superficie de un portaaviones a 50 m/s y se detiene por completo en 120 metros, ¿cuál es la aceleración

sergejj [24]

Answer:

La aceleración necesaria para detener el avión es - 10.42 m/s².

Explanation:

Un movimiento uniformemente acelerado (M.U.A) es aquél cuya aceleración es constante y la velocidad de un objeto cambia a medida que el movimiento evoluciona.

Siendo la aceleración "a" el cambio de velocidad al tiempo transcurrido en un punto A a B, la velocidad inicial la velocidad que tiene un cuerpo al iniciar su movimiento en un período de tiempo y la velocidad final la velocidad que tiene un cuerpo al finalizar su movimiento en un período de tiempo, entonces en M.U.A se cumple:

Vf² - Vo² = 2*a*d

donde:

Vf: Velocidad final
Vo: Velocidad inicial
a: Aceleración
d: Distancia recorrida

En este caso:

Vf: 0 m/s, porque el avión se detiene
Vo: 50 m/s
a: ?
d: 120 m

Reemplazando:

(0 m/s)² - (50 m/s)² = 2*a*120 m

Resolviendo:

$a=\frac{(0 m/s)^{2} -(50 m/s)^{2} }{2*120 m}$

a= - 10.42 m/s²

La aceleración necesaria para detener el avión es - 10.42 m/s².

5 0

3 years ago