All categories

3 years ago

The buoyant force exerted by a liquid is equal to the _____ of the fluid displaced. mass weight volume density

Physics

2 answers:

castortr0y [4]3 years ago

6 0

The buoyant force exerted by a liquid is equal to the weight of the fluid <span>displaced.</span>

Lunna [17]3 years ago

5 0

Answer:

weight

Explanation:

the buoyant force acting on the body of equal to the weight of the liquid displaced by the body.

When a body is immersed (partly or wholly) in a liquid, it experience an upward force. This is called buoyant force or upthrust force.

the buoyant force acting on the body is equal to the weight of the liquid displaced by the body.

The buoyant force is equal to the product of volume of immersed part of the body, density of liquid and the acceleration due to gravity.

You might be interested in

A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900

Whitepunk [10]

It would be 17 m/s

If we use

V2 = V1 + a*t
Sub in 5 for v1
2m/s*2 for a
And
6 for t
That should give you the answer.

5 0

3 years ago

Read 2 more answers

In the diagram, which is demonstrated by arrow 1? A. liquid water evaporating from the surface to the atmosphere B. water vapor

Shkiper50 [21]

I think D. liquid water moving along the surfac

6 0

3 years ago

Question 8

viktelen [127]

Answer: D(t) = $8.e^{-0.4t}.cos(\frac{\pi }{6}.t )$

Explanation: A harmonic motion of a spring can be modeled by a sinusoidal function, which, in general, is of the form:

y = $a.sin(\omega.t)$ or y = $a.cos(\omega.t)$

where:

|a| is initil displacement

$\frac{2.\pi}{\omega}$ is period

For a Damped Harmonic Motion, i.e., when the spring doesn't bounce up and down forever, equations for displacement is:

$y=a.e^{-ct}.cos(\omega.t)$ or $y=a.e^{-ct}.sin(\omega.t)$

For this question in particular, initial displacement is maximum at 8cm, so it is used the cosine function:

$y=a.e^{-ct}.cos(\omega.t)$

period = $\frac{2.\pi}{\omega}$

12 = $\frac{2.\pi}{\omega}$

ω = $\frac{\pi}{6}$

Replacing values:

$D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)$

The equation of displacement, D(t), of a spring with damping factor is $D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)$ .

3 0

3 years ago

A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500

kenny6666 [7]

Answer:

6 m/s

Explanation:

Given that :

mass of the block m = 200.0 g = 200 × 10⁻³ kg

the horizontal spring constant k = 4500.0 N/m

position of the block (distance x) = 4.00 cm = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e $\frac{1}{2} kx^2 = \frac{1}{2} mv^2$

$kx^2 = mv^2$

$4500* 0.04^2 = 200*10^{-3} *v^2$

$7.2 =200*10^{-3}*v^{2}$

$v^{2} =\frac{7.2}{200*10^{-3}}$

$v =\sqrt{\frac{7.2}{200*10^{-3}}}$

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is 6 m/s

5 0

4 years ago

What NE describes the way energy is stored inside atoms

drek231 [11]

Answer:

I believe its Nuclear Energy

Explanation:

N:nuclear

E:energy

3 0

3 years ago