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2 years ago

Select the correct answer from each drop-down menu.

Physics

2 answers:

Zina [86]2 years ago

6 0

Answer:

I agree with the other guy, I think its lower to higher.

Explanation:

fgiga [73]2 years ago

5 0

Low place to a high place I think

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Water enters a student's house 10.0 m above the ground through a pipe with a cross section area of 1.00 x 10-4m2 at ground. Insi

dezoksy [38]

Answer:

(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)

(b). P₂ = 236500 Pa

Explanation:

This is quite straight-forward so let us begin by defining the terms given.

Given that;

The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.

Let us make the velocity of water inside the house be V₁

such that the Volume of water entering the per second is = A₁V₁

Therefore, in 90sec:

45 L = 90 A₁V₁

V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴

V₁ = 10m/s (velocity of water inside the house)

From the continuity equation we have that;

A₁V₁ = A₂V₂

0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂

V₂ = 5m/s (velocity at ground level)

(b). We are told to calculate the water pressure in the pipeline at the ground level.

Using Bernoulli's equation;

P₁ + pgh₁ + 1/2PV₁² (inside) = P₂ + pgh₂ + 1/2PV₂² (ground level)

1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²

P₂ (pressure) = 1.01*10⁵Pa

Therefore we have;

101000 + 98000 + 50000 = P₂ + 12500

P₂ = 236500 Pa

cheers I hope this helped !!

3 0

2 years ago

Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s

ozzi

Answer:

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe ..... (1)

T Cosθ = mg .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

$tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}$

$tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}$

$tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}$

$tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}$

As θ is very small, so tanθ and Sinθ is equal to θ.

$\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}$

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

7 0

2 years ago

Batteries involve the flow of electricity from one metal to another. What happens to these metals to cause the batteries to, eve

Nataly [62]

The metals will start to rust lol. i think. because this messes up how the metals conduct the flow of the electricity.

3 0

2 years ago

Using the law of conservation of angular momentum, estimate how fast a collapsed stellar core would spin if its initial spin rat

Nataly_w [17]

Answer:

$\omega_{f} = 1000000\,\frac{rev}{day}$

Explanation:

The law of conservation of angular momentum states that angular momentum remains constant when there is no external moment or forces applied to the system. Let assume that star can be modelled as an sphere, then:

$\frac{2}{5}\cdot M\cdot R_{o}^{2} \cdot \omega_{o} = \frac{2}{5}\cdot M\cdot R_{f}^{2} \cdot \omega_{f}$

The final angular speed is:

$\omega_{f} = \omega_{o}\cdot (\frac{R_{o}}{R_{f}})^{2}$

$\omega_{f} = (1\,\frac{rev}{day} )\cdot (\frac{10000\,km}{10\,km} )^{2}$

$\omega_{f} = 1000000\,\frac{rev}{day}$

3 0

2 years ago

A green truck is moving to the right. A red truck is moving to the left with a speed of 6 m/s. The mass of the red truck is 1,00

Contact [7]

Answer:

2 m/s

Explanation:

$m_1$ = Mass of red truck = 1000 kg

$m_2$ = Mass of green truck= 3000 kg

$u_1$ = Initial Velocity of red truck = 6 m/s

$u_2$ = Initial Velocity of green truck

$v$ = Velocity with which they move together = 0

For elastic collision

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow m_1u_1 + m_2u_2 =0\\\Rightarrow m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow m_1u_1 + m_2u_2 =0\\\Rightarrow u_2=-\frac{m_1u_1}{m_2}\\\Rightarrow u_2=-\frac{1000\times 6}{3000}\\\Rightarrow u_2=-2\ m/s$

Velocity of the green truck is 2 m/s

8 0

2 years ago