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lisov135 [29]
2 years ago
10

“Active listening” means to _______________. a. be quiet when someone is speaking b. check your understanding of what someone sa

id c. ask questions about what someone said and meant d. all of the above Please select the best answer from the choices provided A B C D
Physics
1 answer:
DIA [1.3K]2 years ago
4 0

Answer:

All of the above is the correct answer!!!

Explanation:

mark me brainliest

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Suppose to a scientist was able to construct a barometer with a liquid being twice denser than mercury, then how would the liqui
exis [7]

Answer:

           h = h₀/2,  the liquid column rises half the height

Explanation:

Pressure is defined as force per area,  

         P = F / A

in a barometer the force is the weights of the liquid column

         F = W = mg

         P = m g/A

if we use the definition of density

         ρ = \frac{m}{A h}

         \frac{m}{A} = ρ h

we substitute

         P = ρ g h

Let's use the index o for the mercury

         P₀ = ρ₀ g h₀

when we change the liquid for another with

          ρ = 2ρ₀

the pressure expression is

         P = 2ρ₀ g h

in the problem they indicate that the pressure is equal to the initial P = P₀

we substitute

           ρ₀ g h₀ = 2ρ₀ g h

           h₀ = 2h

           h = h₀/2

the liquid column rises half the height

3 0
3 years ago
Is a glacier nothing more than a huge mineral flowing downhill?
NNADVOKAT [17]
Yes. A glacier is nothing more than a huge mineral flowing downhill.
4 0
4 years ago
Two parallel wires run in a north-south direction. The eastern wire carries 15.0 A northward while the western wire carries 6.0
djyliett [7]

Answer:

The magnitude  and direction of the  magnetic field is 2.7 x 10⁻⁵ T upwards

Explanation:

Given;

current in the eastern wire, I_e = 15 A

current in the western wire, I_w = 6 A

distance between the wires, d = 30 cm = 0.3 m

The magnetic field at a distance R from a line current I, is given as;

B = \frac{\mu_o I }{2 \pi R}

The magnetic field between the wires, are in opposite directions, and since the currents are also in opposite directions, the magnetic fields of the wires will be added.

The total field = magnetic field (east) + magnetic field (west);

B = \frac{\mu_o I_e}{2 \pi R_e} + \frac{\mu_0 I_w}{2 \pi R_w} \\\\B = \frac{\mu_o}{2\pi} (\frac{I_e}{R_e} + \frac{I_w}{R_w})

where;

R_w is the distance of the field from west = 10cm = 0.1 m

R_e is the distance of the field on east from west = d - 10cm = 30cm - 10cm = 20cm = 0.2 m

The total magnetic field is;

B = \frac{\mu_o}{2\pi} (\frac{I_e}{R_e} + \frac{I_w}{R_w})\\\\B = \frac{4\pi *10^{-7}}{2\pi} (\frac{15}{0.2} + \frac{6}{0.1})\\\\B = 2*10^{-7}(75 + 60)\\\\B =  2*10^{-7}(135)\\\\B = 2.7*10^{-5} \ T

Since total magnetic field is positive, the direction of the field is upwards (positive y direction)

Therefore, the magnitude  and direction of the  magnetic field is 2.7 x 10⁻⁵ T upwards

3 0
3 years ago
. Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 1.00 × 104 kg. The thrust of its engin
Viktor [21]

Answer:

a) The module's acceleration in a vertical takeoff from the Moon will be 1.377 \frac{m}{s^2}

b) Then we can say that a thrust of 3*10^{4} N won't be able to lift off the module from the Earth because it's smaller than the module's weight (9.8 *10^{4} N).

Explanation:

a) During a vertical takeoff, the sum of the forces in the vertical axis will be equal to mass times the module's acceleration. In this this case, the thrust of the module's engines and the total module's weight are the only vertical forces. (In the Moon, the module's weight will be equal to its mass times the Moon's gravity acceleration)

T-(m*g)=m*a

Where:

T= thrust =3 *10^{4} N

m= module's mass =1 *10^{4} N

g= moon's gravity acceleration =1.623 \frac{m}{s^2}

a= module's acceleration during takeoff

Then, we can find the acceleration like this:

a=\frac{T}{m} -g=\frac{3*{10}^4 N}{1*{10}^4 kg}-1.623\frac{m}{s^2}

a=1.377 \frac{m}{s^2}

The module's acceleration in a vertical takeoff from the Moon will be 1.377 \frac{m}{s^2}

b) To takeoff, the module's engines must generate a thrust bigger than the module's weight, which will be its mass times the Earth's gravity acceleration.

weight=m*g=(1*{10}^4 kg)*(9.8 \frac{m}{s^2})=9.8 *10^{4} N

Then we can say that a thrust of 3*10^{4} N won't be able to lift off the module from the Earth because it's smaller than the module's weight (9.8 *10^{4} N).

8 0
4 years ago
A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational force
Alexeev081 [22]

Answer:

This distance is measured from the center of the earth   r = 3.4 10⁸ m

Explanation:

The equation for gravitational attraction force is

      F = G m1 m2 / r²

Where g is the universal gravitation constant, m are the masses of the body and r is the distance between them, remember that this force is always attractive

Let's write the sum of force on the ship and place the condition that is balanced

    F1 -F2 = 0

    F1 = F2

Let's write this equation for our case

   G m Me / r² = G m Mm / (r'.)²

   

The distance r is measured from the center of the earth and the distance r' is measured from the center of the moon,

   

      r' = 3.85 10⁸ m

Let's simplify and calculate the distance

     Me / r² = Mm / / (3.85 108- r)²

     Me / Mm (3.85 108- r)² = r²

     √ 81.4 (3.85 108 -r) = r

     √ 81.4  3.85 108 = r (1 + √ 81.4)

     34.74 108 = r (10.02)

     r = 34.74 10⁸ / 10.2

     r = 3.4 10⁸ m

This distance is measured from the center of the earth

6 0
4 years ago
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