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3 years ago

What do you understand by upthrust Force

Physics

1 answer:

Sedaia [141]3 years ago

7 0

Answer:

hope you like it

Explanation:

An object that is partly, or completely, submerged experiences a greater pressure on its bottom surface than on its top surface. This causes a resultant force upwards. This force is called upthrust . The upthrust force is equal in size to the weight of the fluid displaced by the object.

Buoyancy or upthrust, is an upward force exerted by a fluid that opposes the weight of an immersed object.It is the force that pushes an object up. The upthrust, or buoyancy, keeps ships afloat. The upthrust, or buoyancy, keeps swimmers on top of the water.

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How does a sound wave transfer energy to your ears?

inessss [21]

Answer:

I think it's C!

Explanation:

Sound waves travel at 343 m/s through the air and faster through liquids and solids. The waves transfer energy from the source of the sound, e.g. a drum, to its surroundings. Your ear detects sound waves when vibrating air particles cause your eardrum to vibrate. The bigger the vibrations the louder the sound.

Hope this helps!

4 0

3 years ago

The compressed-air tank ab has a 250-mm outside diameter and an 8-mm wall thickness. it is fitted with a collar by which a 40-kn

valentinak56 [21]

Assume: neglect of the collar dimensions. Ď_h=(P*r)/t=(5*125)/8=78.125 MPa ,Ď_a=Ď_h/2=39 MPa Ď„=(S*Q)/(I*b)=(40*ă€–10ă€—^3*Ď€(ă€–0.125ă€—^2-ă€–0.117ă€—^2 )*121*ă€–10ă€—^(-3))/(Ď€/2 (ă€–0.125ă€—^4-ă€–0.117ă€—^4 )*8*ă€–10ă€—^(-3) )=41.277 MPa @ Point K: Ď_z=(+M*c)/I=(40*0.6*121*ă€–10ă€—^(-3))/(8.914*ă€–10ă€—^(-5) )=32.6 MPa Using Mohr Circle: Ď_max=(Ď_h+Ď_a)/2+âš(Ď„^2+((Ď_h-Ď_a)/2)^2 ) Ď_max=104.2 MPa, Ď„_max=45.62 MPa

3 0

3 years ago

An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo

Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =16321.0769 N/C

The electric field strength at point (x,y) = (0cm, 20 mm) is =35321.58999 N/C

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=16321.0769 N/C

Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷ 0.000000152881

=35321.58999 N/C

8 0

3 years ago

True or false Scientific endeavor is driven by both simple curiosity as well as societal demands

ICE Princess25 [194]

Answer:

its true that Scientific endeavor is driven by both simple curiosity as well as societal demands.

Explanation:

When a scientist has a curiosity about something he carried out a research. and when their is a demand of something in society that time scientific research is carried out. Therefore its true that a scientific endeavor is driven by simple curiosity or societal demand.

For example

in society, there is demand of a medicine which can completely kill the cancer and a scientist has curiosity to know how to kill cancer cell. In this way a scientific endeavor for cancer medicine can be carried out by both simple curiosity as well as societal demands.

6 0

3 years ago

A 66-kg diver jumps off a 9.7-m tower. (a) Find the diver's velocity when he hits the water. (b) The diver comes to a stop 2.0 m

Mariana [72]

Answer:

(a) 13.795 m/s.

(b) -3140.28 N.

Explanation:

(a) Using newton's equation of motion,

v² = u² + 2gs.......................... Equation 1

Where v = final velocity, u = initial velocity, s = height of the tower, g = acceleration due to gravity.

Given: s = 9.7 m, u = 0 m/s ( jump from a height), g = 9.81 m/s².

Substitute into equation 1

v² = 0² + 2×9.81×9.7

v² = 190.314

v = √(190.314)

v = 13.795 m/s.

Hence the velocity of the driver when he hits the water = 13.795 m/s.

(b)

F = ma.................... Equation 2

Where F = force exerted on the diver, m = mass of the diver, a = acceleration of the diver below the water surface.

Also using

v² = u² + 2as ............ Equation 3

Note: At the point when the diver enters the water, u = 13.795 m/s, and at the point when the diver comes to a complete stop, v = 0 m/s

Given: s = 2.0 m, u = 13.795 m/s, v = 0 m/s

Substitute into equation 3

0² = 13.795²+2(2a)

0 = 190.30203 + 4a

-4a = 190.30203

a = 190.30203/-4

a = -47.58 m/s²

Also given: m = 66 kg,

Substitute into equation 3

F = (-47.58)(66)

F = -3140.28

Note: The Force is negative because it act against the motion of the diver.

Hence the net force exerted on the diver while in the water = -3140.28 N.

7 0

3 years ago

What do you understand by upthrust Force​

What do you understand by upthrust Force