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3 years ago

Based on Archimedes' principle, the greatest buoyant force an object can experience in water is determined by which quantity?

Physics

1 answer:

ValentinkaMS [17]3 years ago

6 0

Answer:

B. The object's volume

Explanation:

When an object is immersed in a fluid, it experiences an upward force which is called buoyant force. The magnitude of the buoyant force is given by:

$B=\rho_f V_{disp} g$

where

$\rho_f$ is the density of the fluid in which the object is immersed

$V_{disp}$ is the volume of the fluid displaced by the object

$g$ is the acceleration due to gravity

When the object is totally immersed in the fluid, $V_{disp}$ corresponds to the volume of the object; when the object is only partially immersed, $V_{disp}$ corresponds only to the volume of the part of the object immersed.

From the formula, we see that the greatest buoyant force is experienced by the object when it is fully immersed. Moreover, we see that the buoyant force depends only on one property of the object: its volume. Therefore, the correct choice is

B. The object's volume

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If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli

Aleks04 [339]

Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N= $=\frac{m(V)^{2}}{R}=m*(w)^2*R$

$Friction force\\Ff = k*N\\Ff= k*m*w^2*R$

$Gravitational force \\Fg = m*g$

These must be balanced (the net force on the people will be 0) so set them equal to each other.

$Fg = Ff$

$m*g = k*m*w^2*R$

$g=k*w^{2}*R$

$w^2 =\frac{g}{k*R}$

$w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}$

There are 2*pi radians in 1 revolution so:

$RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88$

So you need about 30 RPM to keep people from falling out the bottom

7 0

3 years ago

Si pudieras viajar a la Luna:

Nostrana [21]

Answer:

i) Distancia, ii) La cinta métrica es impracticable.

Explanation:

i) El concepto físico que se construye únicamente del punto de salida y el punto de llegada a la Luna es el concepto de desplazamiento, definido como la distancia en línea recta de un punto en el espacio con respecto a un punto de referencia (la Tierra en este caso).

La distancia puede involucrar trayectorias curvilíneas entre los puntos mencionados.

ii) Por último, el uso de una cinta métrica es impracticable debido a la cantidad de material a utilizar y los efectos gravitacionales, electromagnéticos y mecánicos que inducen a una deflexión o una ruptura de esa cinta debido a la magnitud de la distancia entre las superficies del planeta y el satélite, respectivamente.

En este caso, es mejor utilizar la medición con tecnología láser, basadas en el fenómeno del electromagnetismo.

4 0

3 years ago

I need help can someone help me pls

pychu [463]

Answer:

copper

Explanation:

4 0

3 years ago

Read 2 more answers

A runner moves 2.88 m/s north. She accelerates at 0.350 m/s^2 at a -52.0 angle. At the point in the motion where she is running

MrRissso [65]

The runner has initial velocity vector

$\vec v_0=\left(2.88\dfrac{\rm m}{\rm s}\right)\,\vec\jmath$

and acceleration vector

$\vec a=\left(0.350\dfrac{\rm m}{\mathrm s^2}\right)(\cos(-52.0^\circ)\,\vec\imath+\sin(-52.0^\circ)\,\vec\jmath)$

so that her velocity at time $t$ is

$\vec v=\vec v_0+\vec at$

She runs directly east when the vertical component of $\vec v$ is 0:

$2.88\dfrac{\rm m}{\rm s}+\left(0.350\,\dfrac{\rm m}{\mathrm s^2}\right)\sin(-52.0^\circ)\,t=0\implies t=10.4\,\rm s$

It's not clear what you're supposed to find at this particular time... possibly her position vector? In that case, assuming she starts at the origin, her position at time $t$ would be

$\vec x=\vec v_0t+\dfrac12\vec at^2$

so that after 10.4 s, her position would be

$\vec x=(10.1\,\mathrm m)\,\vec\imath+(17.2\,\mathrm m)\,\vec\jmath$

which is 19.9 m away from her starting position.

8 0

3 years ago

Read 2 more answers

Give an example of mass making a difference in the amount of gravitational energy. Tell how you know the gravitational energy is

Andreas93 [3]

Answer:

The Gravitational potential energy at large distances is directly proportional to the masses and inversely proportional to the distance between them. The gravitational potential energy increases as r increases.

Examples of Gravitational Energy

A raised weight.

Water that is behind a dam.

A car that is parked at the top of a hill.

A yoyo before it is released.

5 0

2 years ago