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Elena L [17]
2 years ago
9

Aquifers are an example of which two are spheres interacting?

Physics
1 answer:
mash [69]2 years ago
3 0
<h2>Answer::</h2>

Humans (biosphere) built a dam out of rock materials (geosphere). Water in the lake (hydrosphere) seeps into the cliff walls behind the dam, becoming groundwater (geosphere), or evaporating into the air (atmosphere).','.

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Assume an 8-kg bowling ball moving at 2m/s bounces off a spring at the same speed that it had before bouncing.What is the moment
aliya0001 [1]

Answer:-16kgm/s

Explanation:

-mv=-2×8=-16kgm/s

5 0
3 years ago
What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long water tank that can hold 50.0 kg of water by the weight
svetoff [14.1K]

Answer:

1088.9N/m2

Explanation:

Calculation for What pressure is exerted

First step is to find the area of bottom of the tank using formula

Area=Width*breadth

Let plug in the formula

Area=0.5*0.9

Area=0.45m2

Now let calculate what pressure is exerted using this formula

Pressure=Force/Area

Where,

Force=Mass *Gasoline

Area=Width of the tank* Length of the tank

Let plug in the formula

Pressure=50*9.8/0.5*0.9

Pressure=490/0.45

Pressure=1088.9N/m2

Therefore What pressure is exerted is 1088.9N/m2

3 0
2 years ago
you apply the same amount of heat to five grams of water and five grams of aluminum the temperature of the aluminum increases mo
Triss [41]
I can conclude that the aluminum heats up faster then the water, since how they are made up is more different, each has a different compound to it, plus water is more denser<span />
6 0
3 years ago
Read 2 more answers
A proton enters a uniform magnetic field of strength 1 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’
scZoUnD [109]

A charged particle moving in a magnetic field experiences a force equal to:

\vec{F}=q\vec{v}\times \vec{B}

Thus, the magnitude of the force that the proton experiences is given by:

F=qvBsin\theta

The magnetic field is perpendicular to the proton's velocity, therefore, we have \theta=90^\circ. Replacing the given values, we obtain:

F=1.6*10^{-19}C(300\frac{m}{s})(1T)sin(90^\circ)\\F=4.8*10^{-17}N

3 0
3 years ago
A block of mass 12.2 kg is sliding at an initial velocity of 3.9 m/s in the positive x-direction. The surface has a coefficient
Studentka2010 [4]

Answer:

Explanation:

a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force

R = mg where m is mass of the block

Force of friction F = μ x mg

= .173 x 12.2 x 9.8

= 20.68 N

b ) Only force of friction is acting on the body so

deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²

acceleration = - 1.7 m /s²

c )

v² = u² - 2 a s

v = 0 , u = 3.9 m /s

a = 1.7 m /s

0 = 3.9² - 2 x 1.7 x s

s = 4.47  m

5 0
2 years ago
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