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3 years ago

Which most likely represents the process of dissolving carbon dioxide in soda?

Physics

2 answers:

JulijaS [17]3 years ago

6 0

Answer:B

Explanation:

baherus [9]3 years ago

3 0

Answer:increased pressure to increase solubility

Explanation:B on ed

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H. If the mass of an object is 390 g and its volume is 50 cm?, then its density is ....... g/cm' or ....... kg/m”.

beks73 [17]

Answer: See explanation

Explanation:

The density would like be calculated by dividing the mass by volume. This will be:

Density = Mass / Volume

Density = 390 / 50

Density = 7.8g/cm

3 0

2 years ago

A car moved 60 km East and 90 km West. What is the displacement?

Reptile [31]

Answer:

We have been given that car is moving 60 km in east

And 90 km in west

That means we are given two different directions

So, if distance in East is taken to be positive 60 km

Then the distance when car is moving in west direction will be taken as -90 km

So, the displacement means how far the object is from to initial to final position.

So, the displacement will be (-90)+60 = -30

So, it will be in -30 km (minus is west direction).

7 0

3 years ago

A metal ring 5.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend

labwork [276]

Answer:

Ein: 2.75*10^-3 N/C

Explanation:

The induced electric field can be calculated by using the following path integral:

$\int E_{in} dl=-\frac{\Phi_B}{dt}$

Where:

dl: diferencial of circumference of the ring

circumference of the ring = 2πr = 2π(5.00/2)=15.70cm = 0.157 m

ФB: magnetic flux = AB (A: area of the loop = πr^2 = 1.96*10^-3 m^2)

The electric field is always parallel to the dl vector. Then you have:

$E_{in}\int dl=E_{in}(2\pi r)=E_{in}(0.157m)$

Next, you take into account that the area of the ring is constant and that dB/dt = - 0.220T/s. Thus, you obtain:

$E_{in}(0.157m)=-A\frac{dB}{dt}=-(1.96*10^{-3}m^2)(-0.220T/s)=4.31*10^{-4}m^2T/s\\\\E_{in}=\frac{4.31*10^{-4}m^2T/s}{0.157m}=2.75*10^{-3}\frac{N}{C}$

hence, the induced electric field is 2.75*10^-3 N/C

8 0

3 years ago

The belt of one ramp moves at a constant speed such that a person who stands still on it leaves the ramp 64 s after getting on.

Bad White [126]

Answer:

0.74 m/s

Explanation:

given,

time taken by the ramp to cover cliff , t = 64 s

acceleration = 0.37 m/s²

distance traveled by the belt

$x = v_{belt}t_{belt}$

Clifford is moving with constant acceleration

$x = v_ot + \dfrac{1}{2}at^2$ ...(1)

initial velocity is equal to zero

$x =\dfrac{1}{2}at^2$ ........(2)

equating equation (1) and (2)

$v_{belt}t_{belt}=\dfrac{1}{2}at^2$

$v_{belt}t_{belt}=\dfrac{1}{2}a(\dfrac{1}{4}\times t_{belt})^2$

$v_{belt}=\dfrac{1}{32}a t_{belt}$

$v_{belt}=\dfrac{1}{32}\times 0.37 \times 64$

$v_{belt}= 0.74\ m/s$

Speed of the belt is equal to 0.74 m/s

3 0

3 years ago

Give some example acostic using "ELEMENTS"

Jet001 [13]

E=ERBIUM
L=LITHIUM
E=EINSTEINIUM
M=MAGNESIUM
E=EUROPIUM
N=NITROGEN
T=TITANIUM
S=SULPHURDIOXIDE

5 0

3 years ago