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3 years ago

Which scientist first stated the relationship between buoyant force and weight of displaced fluid

Physics

2 answers:

spayn [35]3 years ago

6 0

He was Archimedes of Syracuse.

ExtremeBDS [4]3 years ago

5 0

Answer:

Archimedes

Explanation:

Archimedes suggested in his work, called On Floating Bodies that "Any object, totally or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object." This was later known as the Archimedes principle.

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A 6 kg weight is lifted off the ground to a height that gives it 70.56 j of gravitational potential energy. what is its height?

Marina CMI [18]

GPE= 70.56 J -------------------> GPE= mgh-------------> X= height
70.56 = 6(kg) * 9.8(m/s/s) * X
70.56 = 58.8X
70.56/58.8= 58.8X/58.8
X= 1.2
The height is 1.2 feet or meters (whatever unit you are using in this problem)

5 0

3 years ago

Read 2 more answers

A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of 9.6 V/m and points eastwar

lara31 [8.8K]

Answer:

The values is $B = 3.2 *10^{-8} \ T$

The direction is out of the plane

Explanation:

From the question we are told that

The magnitude of the electric field is $E = 9.6 \ V/m$

The magnitude of the magnetic field is mathematically represented as

$B = \frac{E}{c}$

where c is the speed of light with value

$B = \frac{ 9.6}{3.0 *10^{8}}$

$B = 3.2 *10^{-8} \ T$

Given that the direction off the electromagnetic wave( c ) is northward(y-plane ) and the electric field(E) is eastward(x-plane ) then the magnetic field will be acting in the out of the page (z-plane )

3 0

3 years ago

0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a

klio [65]

Answer:

The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

$(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}$

$(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }$

where $h_{fg$ is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu

T is the temperature ( 15 + 460 )R

m is the mass of water ( 0.5 Ibm )

$V_{fg$ is specific volume ( 1.5 ft³ )

we substitute

$(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5})$ / $( (15+460)\frac{1.5}{0.5})$

$(\frac{dP}{dT} )_{sat } =$ 272.98 Ibf-ft²/R

Now,

$(\frac{dP}{dT} )_{sat } =$ $(\frac{P_2 - P_1}{T_2 - T_1})_{sat$

where P₁ is the initial pressure ( 50 psia )

P₂ is the final pressure ( 60 psia )

T₁ is the initial temperature ( 15 + 460 )R

T₂ is the final temperature = ?

we substitute;

$T_2$ $= ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}$

$T_2 = 475 + 5.2751\\$

$T_2 =$ 480.275 R

Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

3 0

3 years ago

In a football game, the running back takes a handoff and begins running toward midfield at 3.21 yards/s . As he moves through hi

Inessa [10]

Given Information:

Initial speed = u = 3.21 yards/s

Acceleration = α = 1.71 yards/s²

Final speed = v = 7.54 yards/s

Required Information:

Distance = s = ?

Answer:

Distance = s = 13.61

Explanation:

We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.

We know from the equations of motion,

v² = u² + 2αs

Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.

Re-arranging the above equation for distance yields,

2αs = v² - u²

s = (v² - u²)/2α

s = (7.54² - 3.21²)/2×1.71

s = 46.55/3.42

s = 13.61 yards

Therefore, the runner traveled a distance of 13.61 yards before being tackled.

8 0

3 years ago

A 5.0 kg object moving at 10 m/s on a frictionless surfaces collides with but does not stick to a 2.0 kg object that is initiall

lilavasa [31]

Answer:

I= 20 i {N.s}

Explanation:

In order to obtain the impulse on the 2 kg ball, you have to apply the equation of Impulse:

I=FΔt

Where I is the impulse vector, F is the net force and Δt is the interval of time when the force is applied.

In this case:

Δt=0.01 s

F= 2000 i N

where i is the unit vector in the x direction.

Replacing the values in the formula:

I=(2000)(0.01)i

Therefore:

I= 20 i {N.s}

3 0

4 years ago