All categories

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

You might be interested in

A car speedometer has a 4% uncertainty. What is the range of possible speeds (in km/h) when it reads 110 km/h?

Kruka [31]

4% of 110 is 4.4. So the possible range of speeds is the interval from 110-4.4 till 110+4.4.
105.6 till 114.4

4 0

3 years ago

Two cars are traveling along perpendicular roads, car A at 40 mi/hr, car B at 60 mi/hr. At noon, when car A reaches the intersec

Serggg [28]

Answer:

$\frac{dD}{dt} = -4 miles/hour$

negative sign indicates that the distance is decreasing with time

Explanation:

Let at any time t after noon that is 12 p.m.

distance traveled by car A = 40t

distance traveled by car B = 90-60t

then distance between the two cars at time t

$D^2= (40t)^2+(90-60t)^2$ ............1

also, at time 1 p.m.

distance $D^2= (40\times1)^2+(90-60\times1)^2$

D=50 Km

differentiating equation 1 w.r.t. t we get

$2D\frac{dD}{dt}= 2\times40t\times40+2(90-60t)(-60)$

put t= 1 and D= 50 we get

$2\times50\frac{dD}{dt}= 3200\times1-3600\times1$

$\frac{dD}{dt} = -4 miles/hour$

3 0

3 years ago

What wa can friction do to things like brake pads

andrey2020 [161]

Ware them down, its like rubbing two pieces of chalk together.

4 0

3 years ago

Read 2 more answers

Table C: Average Speeds for Lower Racetrack

Blababa [14]

Answer:

Centripetal Acceleration = v^2/r

= (circumference/time)^2/r

= (2*pi*r/t)²)/r

= ((2³.14*50/14.3)²)/50

= 9.64 m/s²

brainlist?

Explanation:

5 0

3 years ago

An automobile with a tangential speed of 54.1 km/h follows a circular road that has a radius of 41.6 m. The automobile has a mas

Viefleur [7K]

1) Available force of friction: 6174 N

2) No

Explanation:

The magnitude of the frictional force between the car's tires and the pavement of the road is given by

$F_f=\mu mg$

where

$\mu$ is the coefficient of friction

m is the mass of the car

g is the acceleration of gravity

For the car in this problem, we have:

$\mu=0.500$ (coefficient of friction)

m = 1260 kg (mass of the car)

$g=9.8 m/s^2$

Therefore, the force of friction is

$F_f=(0.500)(1260)(9.8)=6174 N$

In order to mantain the car in circular motion, the force of friction must be at least equal to the centripetal force.

The centripetal force is given by

$F=m\frac{v^2}{r}$

where

m is the mass of the car

v is the tangential speed

r is the radius of the curve

In this problem, we have

m = 1260 kg

$v=54.1 km/h =15.0 m/s$ is the tangential speed

r = 41.6 m is the radius of the curve

Therefore, the centripetal force is

$F=(1260)\frac{15.0^2}{41.6}=6814 N$

Therefore, the force of friction is not enough to keep the car in the curve, since $F_f$

4 0

3 years ago