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zavuch27 [327]
3 years ago
11

Explain how buoyancy can defeat the force of gravity​

Physics
1 answer:
Olenka [21]3 years ago
7 0

Answer:

Buoyancy causes things to float in water or liquid which appears to defeat the force of gravity. Hope this helps.

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You can find electric power lines under the ground by looking for magnetic fields at ground level. This is best explained by whi
luda_lava [24]
What was the answer?

8 0
3 years ago
A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the contai
Ugo [173]

Answer:

A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

a)  Let us assume the depth be h

As we know that

Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h

After solving this,  

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

b)

Given that  

height of the extra fluid is

h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}

h' = 0.355 m

Now let us assume the pressure at the bottom is Pb'

so, the equation would be

Pb' - Pat = d \times g \times  (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000

Pb' = 122 KPa

3 0
3 years ago
Both the lens and the cornea of the eye have a primary function of
3241004551 [841]

Answer: B. bending light

Explanation:

The phenomenom of vision in human eye is thanks to refraction (when light changes its direction as it passes through one medium to another), and this is what the cornea and the lens do.

When the ray of light encounters the eye, the first thing it finds is the <u>cornea</u>, which<u> bends this ray and begins to form an image</u>, then light passes through the <u>pupil</u>, which is in charge of regulating the amount of light that enters in the eye.  

After light travels through pupil it passes through the <u>lens</u>, where <u>the rays of light change the direction again in order to focus the formed image on the retina. </u>

At this point it is important to note the formed image is downward, then the retina transforms light into electrical impulses that are sent to the brain through the optic nerve and finally the brain interprets these messages, and forms a right upward image.

In the image attached these parts can be seen.

6 0
3 years ago
An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1
goblinko [34]

Answer:

v"_{1} = v_{1} tanΘ

v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}Θ

Θ = tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )

Explanation:

Applying the law of conservation of momentum, we have:

Δp_{x = 0}

p_{x} = p"_{x}

m_{1}v_{1} = m_{2}v"_{2} cosΘ (Equation 1)

Δp_{y} = 0

p_{y} = p"_{y}

0 = m_{1} v"_{1} - m_{2} v"_{2} sinΘ (Equation 2)

From Equation 1:

v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}Θ

From Equation 2:

m_{2} v"_{2}sinΘ = m_{1} v_{1}

v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }

Replacing Equation 3 in Equation 4:

v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}

v"_{1}=v_{1}\frac{sinΘ}{cosΘ}

v"_{1}=v_{1}tanΘ (Equation 5)

And we found Θ from the Equation 5:

tanΘ=\frac{v"_{1}}{v_{1}}

Θ=tan^{-1}(\frac{v"_{1}}{v_{1}})

7 0
3 years ago
At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is
n200080 [17]

a) 8.93\cdot 10^{-30} N, downward

b) 1.76\cdot 10^{-17} N, upward

c) 1.20\cdot 10^{-17} N, downward

Explanation:

a)

The gravitational force of an object (also known as weight of the object) is the attractive force with which the object is pulled towards the Earth's centre.

For an object near the Earth's surface, the magnitude of the gravitational force is given by the equation

F=mg

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

m=9.11\cdot 10^{-31} kg is the mass of the electron

g=9.8 m/s^2 is the acceleration due to gravity

Therefore, the gravitational force on the electron is:

F=(9.11\cdot 10^{-31})(9.8)=8.93\cdot 10^{-30} N

And the direction is downward, towards the Earth's centre.

b)

The electric force on a charged particle is the force produced by the presence of an electric field.

In particular, this force is:

- Repulsive (away from the source of the field) if the charge has the same sign of the charge source of the field

- Attractive (towards the source of the field) if the charge has opposite sign to the charge source of the field

The magnitude of the electric force is given by:

F=qE

where

q is the charge of the particle

E is the strength of the electric field

In this problem:

q=1.6\cdot 10^{-19} C is the charge of the electron

E=110 N/C is the strength of the electric field

Substituting,

F=(1.6\cdot 10^{-19})(110)=1.76\cdot 10^{-17} N

And since the electron has negative charge, the  direction of the force is opposite to that of the electric field, so upward.

c)

When a charged particle is moving in a magnetic field, it experiences a force which is perpendicular to both the direction of motion of the charge and to the direction of the magnetic field.

The magnitude of this force is given by (if the charge moves perpendicular to the magnetic field)

F=qvB

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

In this problem:

q=1.6\cdot 10^{-19} C is the charge of the electron

v=1.50\cdot 10^6 m/s is the velocity

B=50.0\mu T = 50.0 \cdot 10^{-6} T is the strength of the magnetic field

Substituting,

F=(1.6\cdot 10^{-19})(1.50\cdot 10^6)(50.0\cdot 10^{-6})=1.20\cdot 10^{-17} N

And the direction can be determined using the right-hand rule:

- Index finger: direction of velocity (eastward)

- Middle finger: direction of magnetic field (northward)

- Thumb: direction of force (upward) --> however the electron has negative charge, so the direction of the force is reversed --> downward

5 0
3 years ago
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