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Basile [38]
3 years ago
12

An object is placed in a fluid and then released. Assume that the object either floats to the surface (settling so that the obje

ct is partly above and partly below the fluid surface) or sinks to the bottom. (Note that, for Parts A through D, you should assume that the object has settled in equilibrium.)
A.Consider the following statement: The magnitude of the buoyancy force is equal to the weight of fluid displaced by the object.
Under what circumstances is this statement true?

A. always
B. only for an object that floats
C. only for an object that sinks
D. never
Physics
2 answers:
Irina18 [472]3 years ago
7 0

Answer:

A. Always true

Explanation:

This is because, the buoyancy force is always present whenever and object is placed in a fluid. The magnitude of this  buoyancy force is always equal to the weight of the fluid    displaced by the object according to Archimedes' principle. This principle is true irrespective of whether the object floats or not. When any object is inserted in a fluid, the buoyancy force is always present irrespective of whether it floats or not.  

sp2606 [1]3 years ago
3 0

Answer:

The correct option is;

A. always

Explanation:

Archimedes' principle states that when an object is immersed in a fluid, and it is either partially or wholly submerged, the upward force acting on the object, that is the buoyancy force, is equal to the weight of the fluid displaced by the object.

That is the object immersed in the fluid is observed to have an apparent reduction in weight that is equal to the weight of the volume of the fluid displaced by the  immersed  object.

Therefore, the above principle always occurs.

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If the 50-kg crate starts from rest and achieves a velocity of v = 4 m/s
Kay [80]

Answer:

P = 227 N

Explanation:

Assuming the crate is on horizontal ground and subject to a horizontal force.

F = ma

P - μmg = ma

P = m(a + μg)

P = m(v²/2s + μg)

P = 50(4²/(2(5))+ 0.3(9.8))

P = 227 N

6 0
3 years ago
A crate is sliding on the floor. If there is a total force acting on the crate in the same direction as it is sliding, the crate
Leto [7]

But we do not know whether the force is pushing or pulling (the same direction (both forces are parallel) but: .........[ ]<-F-- or .......[ ]--F-->). I suppose the correct answer is B

7 0
3 years ago
Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller dia
Crank

Answer:

Wn = 9.14 x 10¹⁷ N

Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

<u>Wn = 9.14 x 10¹⁷ N</u>

3 0
3 years ago
Suppose an object is launched from a point 320 feet above the earth with an initial velocity of 128 ft/sec upward, and the only
Ne4ueva [31]

Answer:

(a)Therefore the highest altitude attained by the object is =576 ft .

(b)Therefore the object takes 6 sec to fall to the ground.

Explanation:

Initial velocity: Initial velocity is a velocity from which an object starts to move.

u is usually used for notation of initial notation.

Final velocity: Final velocity is a velocity of an object after certain second from starting.

The final velocity is denoted by v.

Acceleration: The difference of final velocity and initial velocity per unit time

The S.I unit of acceleration is m/s².

(a)

Given that u= 128 ft\sec and g = 32 ft/sec².

At highest point the velocity of the object is 0 i.e v=0

Since the displacement is opposite to the gravity.

Therefore acceleration( a)= -g = -32 ft/sec².

To find the time this to happen we use the following formula

v=u+at

Here v=0

⇒0=128+(-32) t

⇒32t=128

⇒t = 4 sec

To determine the height we use the following formula

s=ut+\frac{1}{2} at^2

\Rightarrow s= (128\times4)+\frac{1}{2}\times (-32) \times4^2

⇒s= 256 ft

Therefore the highest altitude attained by the object is =(320+256)ft=576 ft .

(b)

At the highest point the velocity of the object is 0.

so u=0. a=g= 32 ft/sec²  [ since the direction of gravity and the displacement are same] s= 576 ft

To determine the time to fall we use the following formula

s=ut+\frac{1}{2} at^2

\Rightarrow 576 = (0\times t)+\frac{1}{2} \times 32 \times t^2

\Rightarrow 16\times t^2=576

\Rightarrow t^2=\frac{576}{16}

\Rightarrow t^2=36

⇒t=6 sec

Therefore the object takes 6 sec to fall to the ground.

8 0
3 years ago
What is my displacement if i walked 200 meters west then turned around and walked 150 meters east
Alinara [238K]
You're walking in one direction, and then the exact opposite of that direction, so you simply have to subtract the two distances. 
200-150=50
You're 50 meters west of where you originally started.
You're west because 200 meters west is greater than 150 meters east. If the distance walked east was greater than the distance walked west, you would've been east of your starting position.
6 0
3 years ago
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