2) Um gás ideal sofre uma determinada transformação, conforme mostra o gráfico abaixo. Considere

during a journey, a car travels at 40 km in 2.5 hours, next 62 km in 3 hours, then took a break for 30 minutes, again travelled

wlad13 [49]

45mph is the answer if you do the math right

6 0

3 years ago

A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that

VMariaS [17]

Answer:

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

Explanation:

As we know that the string is horizontal here

so the tension force in the string is due to electrostatic force on it

now we will have

$F = qE$

so here the force is tension force on it

$F = 6.57 \times 10^{-2} N$

$Q = 6.80 \times 10^3 C$

now we have

$6.57 \times 10^{-2} = (6.80 \times 10^3)E$

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

4 0

4 years ago

Drag the tiles to the correct boxes to complete the pairs.

Brut [27]

Answer:

5 percent = normal matter

68 percent = dark energy

27 percent = dark matter

Explanation:

3 0

2 years ago

What water pressure must a pump that is located on the first floor supply to have water on the thirteenth of a building with a p

irga5000 [103]

The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

The given parameters;

<em>Pressure on the 13 th floor, P₁ = 35 PSI</em>
<em>Distance between each floor, d = 10 ft</em>

The vertical pressure of the water is calculated as follows;

$P = \rho gh\\\\\frac{P}{h} = \rho g\\\\\frac{P}{h} = k\\\\\frac{P_1}{h_1} = \frac{P_2}{h_2} \\\\$

The vertical height of the first floor from the 13th floor = 130 ft

The vertical height of the 13 ft floor = 10 ft

$P_1 = \frac{P_2 h_1}{h_2} \\\\P_1 = \frac{35 \times 130}{10} \\\\P_1 = 455 \ PSI$

Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

Learn more about vertical height and pressure here: brainly.com/question/15691554

3 0

2 years ago

A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p

Kipish [7]

Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

1/u = 5.0 m⁻¹

u = 1/5.0 m⁻¹

u = 0.2 m

u = 20 cm

So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.

8 0

3 years ago