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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.

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What force acts on all objects, all the time on Earth?

TiliK225 [7]

The force that acts on all objects, all the time on Earth is gravitational force.

The force that surface exert on an object perpendicularly is normal reaction.

<h3>What force acts on all objects, all the time on Earth?</h3>

Force due to gravity is gravitational pull on objects due to its position on earth's surface.

The force due to gravity on object's is calculated by applying Newton's second law of motion as follows;

F = mg

where;

m is the mass of the object
g is acceleration due to gravity

The force that surface exert on an object perpendicularly is normal reaction.

Thus, the force that acts on all objects, all the time on Earth is gravitational force.

Learn more about force of gravity here: brainly.com/question/2537310

7 0

3 years ago

Help me with Economics please and thank you the question is going to be down in a attached file while the answers on here

Tresset [83]

I think is A or B it depends on like what the trying to answer

6 0

3 years ago

Read 2 more answers

A gas has an initial volume of 2.5 L at a temperature of 275 K and a pressure of 2.1 atm. The pressure of the gas increases to 2

olchik [2.2K]

Answer:

2.10L

Explanation:

Given data

V1= 2.5L

T1= 275K

P1= 2.1atm

P2= 2.7 atm

T2= 298K

V2= ???

Let us apply the gas equation

P1V1/T1= P2V2/T2

substitute into the expression we have

2.1*2.5/275= 2.7*V2/298

5.25/275= 2.7*V2/298

Cross multiply

275*2.7V2= 298*5.25

742.5V2= 1564.5

V2= 1564.5/742.5

V2= 2.10L

Hence the final volume is 2.10L

7 0

3 years ago

In classical physics, consider a 2 kg block hanging on a spring with a spring constant of 50 N/m. Ignore air resistance. The blo

RUDIKE [14]

Answer:

v = 0

Explanation:

This problem can be solved by taking into account:

- The equation for the calculation of the period in a spring-masss system

$T = \sqrt{\frac{m}{k} }$ ( 1 )

- The equation for the velocity of a simple harmonic motion

$x = \frac{2\pi }{T}Asin(\frac{2\pi }{T}t)$ ( 2 )

where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block

Hence

$T = \sqrt{\frac{2 kg}{50 N/m}} = 0.2 s$

and by reeplacing it in ( 2 ):

$v = \frac{2\pi }{0.2s}(14cm)sin(\frac{2\pi }{0.2s}(0.9s)) = 140\pi sin(9\pi ) = 0$

In this case for 0.9 s the velocity is zero, that is, the block is in a position with the max displacement from the equilibrium.

5 0

3 years ago

Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and

lawyer [7]

1 kilometre=1000 metre

1 hour = 3600 second

$1\ km/hr=\frac{1000}{3600} m/s$

$1\ km/hr=\frac{5}{18} m/s$

The initial velocity of car A is 35.0 km/hr i.e

$35.0\ km/hr=35*\frac{5}{18} m/s$

= 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as $25\ km/hr^2$

$=\ 25*\frac{1000}{3600*3600} m/s^2$

$=0.00192901234 m/s^2$

The time taken by car A = 15 min.

From equation of kinematics we know that-

v= u+at [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A, v = 9.72 m/s +[0.00192901234×15×60]m/s

=11.456111106 m/s

The acceleration of B is given as $15\ km/hr^2$

$=0.00115740740740 m/s^2$

The time taken by car B =20 min

The final velocity of B is -

v= u+at

= u-at [Here a is negative due to deceleration]

=12.5 m/s +[0.0011574074074×20×60]

=13.8888888.....

=13.9

The acceleration of C is given as $40\ km/hr^2$

$=\ 0.003086419753 m/s^2$

The time taken by car C =30 min

The final velocity of C is-

v = u+at

=8.89 m/s+[0.003086419753×30×60] m/s

=14.4455555555..m/s

=14.45 m/s

The car C is decelerating.The deceleration is given as- $60\ km/hr^2$

$=0.0046296296296m/s^2$

The time taken by car D= 45 min.

The final velocity of the car D is-

v =u+at

=30.56 -[0.00462962962962×45×60]m/s

=18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

3 0

3 years ago