WHAT WAS THE ACCELERATION OF THE CART WITH THE LOW FAN SPEED?

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

Sav [3153]

Respuesta:

11.4 m/s

Explicación:

La fórmula para la aceleración centrípeta es:

$a=\frac{v^2}{R}$

donde, a es la aceleración, v la velocidad alrededor de la circunferencia y R el radio del círculo.

En este problema,

a = g = aceleración debida a la gravedad en la cima = $9.81\ m/s^2$

v = ?

R = 13.2 m

Por lo tanto,

$9.81=\frac{v^2}{13.2}$

$v^2=9.81\times {13.2}$

v = 11.4 m/s

8 0

2 months ago

Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.

kicyunya [3294]

Result:

$v=\sqrt{k\frac{e^2}{m_e r}}$ , $2.18\cdot 10^6 m/s$

Explanation:

The electromagnetic attraction between the electron and the proton in the nucleus is equivalent to the centripetal force:

$k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}$

where

k represents the Coulomb constant

e denotes the charge of the electron

e denotes the charge of the proton in the nucleus

r signifies the distance from the electron to the nucleus

v indicates the velocity of the electron

is the mass of the electron

Rearranging for v, we determine

$v=\sqrt{k\frac{e^2}{m_e r}}$

Inside a hydrogen atom, the distance separating the electron from the nucleus is roughly

$r=5.3\cdot 10^{-11}m$

while the mass of the electron is

$m_e = 9.11\cdot 10^{-31}kg$

and the charge is

$e=1.6\cdot 10^{-19} C$

By plugging in the values into the formula, we achieve

$v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s$

7 0

1 month ago

The ballistic pendulum was invented in 1742 by English mathematician Benjamin Robins. It consists of an initially stationary pen

serg [3582]

Answer:

What is your question?

Please feel free to modify or inquire in the comments.

Thank you.

Explanation:

7 0

2 months ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Yuliya22 [3333]

Details that are not provided: the problem figure is included.

We can address the exercise by applying Poiseuille's law. This law indicates that for a fluid flowing in a laminar manner within a confined pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ represents the pressure difference across the two ends

$\mu$ denotes the viscosity of the fluid

L signifies the length of the pipe

$Q=Av$ indicates the volumetric flow rate, where $A=\pi r^2$ is the cross-sectional area of the tube and $v$ refers to the fluid's velocity

r stands for the pipe's radius.

This law can be utilized for the needle, allowing us to compute the pressure difference between point P and the needle's end. In this scenario, we have:

$\mu=0.001 Pa/s$ is the dynamic viscosity of water at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Substituting these values into the formula yields:

$\Delta P = 3200 Pa$

This pressure difference specifies the value between point P and the needle's termination. As the end of the needle is under atmospheric pressure, the gauge pressure at point P, relative to atmospheric pressure, is exactly 3200 Pa.

8 0

2 months ago

The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of

kicyunya [3294]

3.258 m/s Explanation: The spring constant is assumed to be 263 N/m and the displacement of the spring is also assumed to be 0.7 m; the coefficient of friction between blocks is 0.4. The energy stored in the spring is described by . Given the conservation of energy in the system, the speed of the 8 kg block just prior to collision is 3.258 m/s.

7 0

1 month ago