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svetoff [14.1K]
2 years ago
7

What is the z coordinate for the above vector?

Physics
1 answer:
Hitman42 [59]2 years ago
4 0
120 m would be he answer
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Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.
scoundrel [369]

Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

   Pressure at sea level = 1 atm = 101300 Pa

   we know

   P = ρ g h

   h = \dfrac{P}{\rho\ g}

   h = \dfrac{101300}{1.3\times 9.8}

          h = 7951.33 m

height of the atmosphere will be equal to 7951.33 m

b) when air density decreased linearly to zero.

  at x = 0  air density = 0

  at x= h   ρ_l = ρ_sl

 assuming density is zero at x - distance

 \rho_x = \dfrac{\rho_{sl}}{h}\times x

now, Pressure at depth x

dP = \rho_x g dx

dP = \dfrac{\rho_{sl}}{h}\times x g dx

integrating both side

P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx

P =\dfrac{\rho_{sl}\times g h}{2}

 now,

h=\dfrac{2P}{\rho_{sl}\times g}

h=\dfrac{2\times 101300}{1.3\times 9.8}

  h = 15902.67 m

height of the atmosphere is equal to 15902.67 m.

6 0
2 years ago
An empty rubber balloon has a mass of 0.0120 kg. The balloon is filled with helium at 0°C, 1 atm pressure, and a density of 0.17
Anuta_ua [19.1K]

Answer:

a) F_b = 6.62 N

b) F_net = 5.583 N

Explanation:

Given:

- Conditions of He gas:  T = 0 C , P = 1 atm , ρ = 0.179 kg/m^3

- The mass of balloon m = 0.012 kg

- The radius of balloon r = 0.5 m

Find:

a)What is the magnitude of the buoyant force acting on the balloon?

b)What is the magnitude of the net force acting on the balloon?

Solution:

- The buoyant force F_b acting on the balloon is equal to the weight of the air it displaces.The mass of the displaced air ρ*V is the volume of the balloon times the density of the. Multiplying that by acceleration due to gravity gives its weight.

                                 F_b = ρ*V*g

                                 F_b = 4*ρ*g*pi*r^3 / 3

                                 F_b = 4*1.29*9.81*pi*.5^3 / 3

                                 F_b = 6.62 N

- The net force will be the difference between the balloon’s weight and the buoyant force. The weight of the balloon is the density of the helium times the volume of the balloon added to the mass of the empty balloon.

                                 F_g = ρ*V*g + m*g

                                 F_g = 4*ρ*g*pi*r^3 / 3 + 0.012*9.81

                                 F_g = 4*0.179*9.81*pi*.5^3 / 3 + 0.012*9.81

                                 F_g = 1.037 N

- The net force is the difference between weight and buoyant force

                                F_net = F_g - F_b

                                F_net = 6.62 - 1.037

                                F_net = 5.583 N

7 0
3 years ago
A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top
Anton [14]

Answer:

y_{max} = 0.829\,m

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}

v = 2.913\,\frac{m}{s}

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

(3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}

After some algebraic handling, a second-order polynomial is formed:

12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s

1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0

The roots of the polynomial are, respectively:

\Delta s_{1} \approx 0.128\,m

\Delta s_{2} \approx -0.099\,m

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

\Delta s \approx 0.128\,m

The maximum height that the block reaches after rebound is:

(3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}

y_{max} = 0.829\,m

4 0
3 years ago
Read 2 more answers
If a sound wave is produced by a speaker at room temperature and has a
Oksana_A [137]

Answer: 183.8Hz

Explanation:

Given that,

wavelength of sound (λ) = 1.85 m

frequency of the sound (F) = ?

Recall that the speed of sound (V) in air is a constant with a value of 340m/s

So, apply the formula

V = F λ

340 m/s = F x 1.85 m

F = 340m/s / 1.85m

F = 183.8Hz

Thus, the frequency of the sound that is

generated is 183.8 hertz

3 0
2 years ago
6.2 x 10^18 is equal to<br><br> A one ohm<br> B. one coulomb<br> C. one volt<br> D. one ampere
zhannawk [14.2K]
<h3>Answer:</h3><h3>B.one coulomb</h3>

<h3>Explanation:</h3><h3>It takes an enormous number of electrons to equal one coulomb.In fact,it takes about 6.2x10¹⁸.Looking at this from the opposite perspective,one electron has a negative charge equal to -1.6x10¹⁹ coulombs.</h3>

5 0
2 years ago
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