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vampirchik [111]

3 years ago

14

4. With regard to the pH scale, a solution with a pH

1 answer:

pentagon [3]3 years ago

5 0

Answer:

A. Close to 14 is considered a strong base

Explanation:

A substance with a pH close to 14 is considered a strong base

<em>PLEASE</em><em> </em><em>DO MARK</em><em> </em><em>ME AS</em><em> </em><em>BRAINLIEST</em><em> </em><em>IF</em><em> </em><em>MY</em><em> </em><em>ANSWER</em><em> </em><em>IS</em><em> </em><em>HELPFUL</em><em> </em><em>;</em><em>)</em><em> </em>

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A series circuit contains a 9-volt battery, a 3-ohm resistor and a 2-ohm resistor. What is the voltage drop across the 2-ohm res

BARSIC [14]

Since everything in the circuit is in series .. .

-- The total resistance is (3 + 2) = 5 ohms.

-- The voltage across the 3-ohm resistor is 3/5 of the total voltage.

-- The voltage across the 2-ohm resistor is 2/5 of the total voltage.

(2/5) of (9 volts) = 18/5 = 3.6 volts .

7 0

3 years ago

Read 2 more answers

A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.

Anestetic [448]

Answer:

$\omega=0.37 [rad/s]$

Explanation:

We can use the conservation of the angular momentum.

$L=mvR$

$I\omega=mvR$

Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.

So we will have:

$(I_{proffesor - stool}+mR^{2})\omega=mvR$

Now, we just need to solve it for ω.

$\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}$

$\omega=\frac{1.5*2.7*0.4}{4.1+1.5*0.4^{2}}$

$\omega=0.37 [rad/s]$

I hope it helps you!

5 0

3 years ago

An MRI scanner is based on a solenoid magnet that produces a large magnetic field. The magnetic field doesn't stop at the soleno

GaryK [48]

Answer:

The maximum change in flux is $\Delta \o = 0.1404 \ Wb$

The average induced emf $\epsilon =0.11232 V$

Explanation:

From the question we are told that

The speed of the technician is $v = 0.80 m/s$

The distance from the scanner is $d = 1.0m$

The initial magnetic field is $B_i = 0T$

The final magnetic field is $B_f = 6.0T$

The diameter of the loop is $D = 19cm = \frac{19}{100} = 0.19 m$

The area of the loop is mathematically represented as

$A = \pi [\frac{D}{2} ]^2$

$= 3.142 \frac{0.19}{2}$

$= 0.02834 m^2$

At maximum the change in magnetic field is mathematically represented as

$\Delta \o = (B_f - B_i)A$

=> $\Delta \o = (6 -0)(0.02834)$

$\Delta \o = 0.1404 \ Wb$

The average induced emf is mathematically represented as

$\epsilon = \Delta \o v$

$= 0.1404 * 0.80$

$\epsilon =0.11232 V$

7 0

3 years ago

When you ride a bicycle at constant speed, nearly all the energy you expend goes into the work you do against the drag force of

ratelena [41]

Answer:

Power output = 96.506 watts

Explanation:

Drag coefficient (Cd) = 0.9

V = 7.3 m/s

Air density (ρ) = 1.225 kg/m^(3)

Area (A) = 0.45 m^2

Let's find the drag force ;

Fd=(1/2)(Cd)(ρ)(A)(v^(2))

So Fd = (1/2)(0.9)(1.225)(0.45)(7.3^(2)) = 13.22N

Drag power = Drag Force x Drag velocity.

Thus drag power, = 13.22 x 7.3 = 96.506 watts

8 0

3 years ago

A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me

loris [4]

To solve this exercise we will use the concept related to heat loss which is mathematically given as

$Q = mC_p \Delta T$

Where,

m = mass

$C_p$ = Specific Heat

$\Delta T =$ Change in temperature

Replacing with our values we have that

$m = 0.1g$

$C_p = 139J/Kg\cdot K \rightarrow$ Specific heat of mercury

$\Delta T = 8.5\°C-15\°C = -6.5\°C \Rightarrow -6.5K$

Replacing

$Q = (0.1*10^3)(138)(-6.5)\\Q = -0.09J$

Therefore the heat lost by mercury is 0.09J

5 0

3 years ago