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qaws [65]
4 years ago
11

1. Find the current ix in the circuits in Fig. E2.6.

Physics
1 answer:
artcher [175]4 years ago
6 0

It's hard to exaggerate how much easier that task would be,
if only I could see Fig. E2.6.

I can't imagine what reason you fantasized for why that figure
is printed alongside the problem, or whether the question ever
occurred to you.

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A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is
Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

           P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

         P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

            P₁ V₁ = P₂ V₂

            P_atm V₁ = (P_atm + ρ g H) V₂

             V₂ = V₁    P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

6 0
3 years ago
An archer shoots an arrow at a 74.0 m distant target, the bull's-eye of which is at same height as the release height of the arr
GuDViN [60]

A. The angle at which the arrow must be released to hit the bull's-eye is 20.7 °

B. The arrow will go over the branch.

<h3>A. How to determine the angle</h3>
  • Range (R) = 74 m
  • Initial velocity (u) = 33 m/s
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Angle (θ) = ?

R = u²Sine(2θ) / g

74 = 33² × Sine (2θ) / 9.8

Cross multiply

74 × 9.8 = 33² × Sine (2θ)

725.2 = 1098 × Sine (2θ)

Divide both sides by 1098

Sine (2θ) = 725.2 / 1098

Sine (2θ) = 0.6605

Take the inverse of sine

2θ = Sine⁻¹ 0.6605

2θ = 41.3

Divide both sides by 2

θ = 41.3 / 2

θ = 20.7 °

<h3>B. How to determine if the arrow will go over or under the branch</h3>

To determine if the arrow will go over or under the branch situated mid way, we shall determine the maximum height attained by the arrow. This can be obtained as follow:

  • Initial velocity (u) = 33 m/s
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Angle (θ) = 20.7 °
  • Maximum height (H) = ?

H = u²Sine²θ / 2g

H = [33² × (Sine 20.7)²] / (2 ×9.8)

H = 6.94 m

Thus, the maximum height attained by the arrow is 6.94 m which is greater than the height of the branch (i.e 3.50 m).

Therefore, we can conclude that the arrow will go over the branch

Learn more about projectile motion:

brainly.com/question/20326485

#SPJ1

3 0
2 years ago
If you push an 5 N object 2 m and then push a 10 N object 2 m. Which is TRUE?
Nadya [2.5K]

Answer:

Explanation:

You don't do the same amount of work. The work formula is F*d = W

W = work

F = force

d = the distance moved.

So you do more work when you move the 10N object because the Force (F) has doubled.

4 0
3 years ago
Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal po
Montano1993 [528]

Answer:

A) I = Io 0.578,   B) he light that leaves the polarized is completely polarized, being perpendicular to the axis of the second filter

Explanation:

A) Light passing through a polarizer must comply with the / bad law

          I = Io cos2 tea

Where is at the angle of the polarizer and incident light

          I = Io cos2 45

         I = Io 0.578

Therefore the beam intensity is 0.578 of the incident intensity

.B) the light that leaves the polarized is completely polarized, being perpendicular to the axis of the second filter

3 0
4 years ago
A brass statue with a mass of 0.40 kg and a density of 8.00×103kg/m3 is suspended from a string. When the statue is completely s
Mars2501 [29]

Answer:

ρ = 1469  kg/m³

Explanation:

given,

mass of statue = 0.4 Kg

density of statue = 8 x 10³ kg/m³

tension in the string = 3.2 N

density of the fluid = ?

Volume of the statue

V = \dfrac{0.4}{8\times 10^3}

V = 5 x 10⁻⁵ m³

W = ρ g V

W = ρ x 9.8 x 5 x 10⁻⁵

now, tension on the string will be equal to

T = mg - W

3.2 = 0.4 x 9.8 - ρ x 9.8 x 5 x 10⁻⁵

ρ x 9.8 x 5 x 10⁻⁵ = 0.72

ρ = 1469  kg/m³

8 0
3 years ago
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