PHYSICS COLLEGE

Water flows at speed of 4.4 m/s through ahorizontal pipe of diameter 3.3 cm . The gauge
pressure P1 of the water in the pipe is 2 atm .
A short segment of the pipe is constricted to
a smaller diameter of 2.4 cm
(IMAGE)
What is the gauge pressure of the water
flowing through the constricted segment? Atmospheric pressure is 1.013 × 10^5 Pa . The density of water is 1000 kg/m^3
. The viscosity
of water is negligible.
Answer in units of atm

Answers

Answer 1

Answer:

Answer:

1.75 atm

Explanation:

Mass is conserved, so the mass flow before the constriction equals the mass flow after the constriction.

m₁ = m₂

ρQ₁ = ρQ₂

Q₁ = Q₂

v₁A₁ = v₂A₂

v₁ πd₁²/4 = v₂ πd₂²/4

v₁ d₁² = v₂ d₂²

Now use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Since h₁ = h₂:

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

Writing v₂ in terms of v₁:

P₁ + ½ ρ v₁² = P₂ + ½ ρ (v₁ d₁²/d₂²)²

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₁² (d₁/d₂)⁴

P₁ + ½ ρ v₁² (1 − (d₁/d₂)⁴) = P₂

Plugging in values:

P₂ = 2 atm + ½ (1000 kg/m³) (4.4 m/s)² (1 − (3.3 cm / 2.4 cm)⁴) (1 atm / 1.013×10⁵ Pa)

P₂ = 1.75 atm

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A single-turn circular loop of wire of radius 5.0 cm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.02500.0250-\text{s}s time interval, the magnitude of the field increases uniformly from 200 to 300 mT. Determine the magnitude of the emf induced in the loop

Answers

Given Information:

time = Δt = 0.0250 seconds

Radius = r = 5 cm = 0.05 m

Change in Magnetic field = ΔB = (0.300 - 0.200) T

Number of turns = N = 1

Required Information:

Magnitude of induced emf = ξ = ?

Answer:

Magnitude of induced emf = ξ = 3.141x10⁻² V

Explanation:

The EMF induced in a circular loop of wire in a changing magnetic field is given by

ξ = -NΔΦ/Δt

Where change in flux ΔΦ is given by

ΔΦ = ΔBA

ΔΦ = ΔBπr²

ΔΦ = (0.300 - 0.200)*π*(0.05)²

ΔΦ = 7.854x10⁻⁴ T.m²

ξ = -NΔΦ/Δt

ξ = -(1*7.854x10⁻⁴)/0.0250

ξ = -3.141x10⁻² V

The negative sign is due to Lenz law.

Answer:

-0.0314 V

Explanation:

Parameters given:

Initial magnetic field, Bini = 200 mT = 0.2T

Final magnetic field, Bfin = 300mT = 0.3 T

Number of turns, N = 1

Radius, r = 5 cm = 0.05 m

Time, t = 0.025 secs

Induced EMF is given as:

EMF = [-(Bfin - Bini) * N * pi * r²] / t

EMF = [-(0.3 - 0.2) * 1 * 3.142 * 0.05²] / 0.025

EMF = (-0.1 * 3.142 * 0.0025) / 0.025

EMF = -0.0314 V

A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the ground. What one of the following statements is true if air resistance is neglected? The acceleration of the rock is zero when it is at the highest point. The speed of the rock is negative while it falls toward the ground. As the rock rises, its acceleration vector points upward. At the highest point the velocity is zero, the acceleration is directed downward. The velocity and acceleration of the rock always point in the same direction.

Answers

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g glass ball charged to 4.7 nC is shot straight up at 4.8 m/s from the floor level. How high does the ball go if the ceiling voltage is +3.0x10^6V?

Answers

The height at which the ball goes for the given parameters is; 0.827 m

What is the height of the ball?

We are given;

distance between the metal plates; d = 3.1 m

mass of glass; m = 1.1g = 0.0011 kg

charge on the glass; q = 4.7 nC = 4.7 × 10⁻⁹ C

speed of the glass ball; v = 4.8 m/s

voltage of the ceiling; V = +3.0 × 10⁶ V

The repulsive force experienced by the ball is gotten from the formula;

F = qV/d

|F| = (4.7 × 10⁻⁹ × 3 × 10⁶)/3.1

|F| = 4.548 × 10⁻³ N

F = -4.548 × 10⁻³ N (negative because it is repulsive force)

The net horizontal force experienced by the ball is;

F_net = F - mg

F_net = (-4.548 × 10⁻³) - (0.0011 × 9.8)

F_net = -15.328 × 10⁻³ N

To get the height of the ball, we will use the formula;

F_net * h = ¹/₂mv²

h = (¹/₂ * 0.0011 * 4.8²)/(15.328 × 10⁻³)

We took the absolute value of F_net, hence it is not negative

h = 0.827 m

Read more about height of ball at; brainly.com/question/12446886

Answer:

The ball traveled 0.827 m

Explanation:

Given;

distance between the metal plates of the room, d = 3.1 m

mass of the glass, m = 1.1g

charge on the glass, q = 4.7 nC

speed of the glass ball, v = 4.8 m/s

voltage of the ceiling, V = +3.0 x 10⁶ V

The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;

F = qV/d

|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)

|F| = 4.548 x 10⁻³ N

F = - 4.548 x 10⁻³ N

The net horizontal force experienced by this ball is;

$F_(net) = F_c - mg\n\nF_(net) = -4.548 *10^(-3) - (1.1*10^(-3) * 9.8)\n\nF_(net) = -15.328*10^(-3) \ N$

The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.

$W = F_(net) *h\n\nW = 15.328 *10^(-3) * h$

W = K.E

$15.328*10^(-3) *h = (1)/(2)mv^2\n\n 15.328*10^(-3) *h = (1)/(2)(1.1*10^(-3))(4.8)^2\n\n 15.328*10^(-3) *h =0.0127\n\nh = (0.0127)/(15.328*10^(-3))\n\n h = 0.827 \ m$

Therefore, the ball traveled 0.827 m

A javelin is thrown in the air. Its height is given by ( ) 1 2 8 6 20 h x x x = − + + , where x is the horizontal distance in feet from the point at which the javelin is thrown. a. How high is the javelin when it was thrown? b. What is the maximum height of the javelin? c. How far from the thrower does the javelin strike the ground?'

Answers

The maximum height, the location on the ground and the initial vertical height of the javelin is required.

The initial height of the javelin is 6 feet.

The maximum height of the javelin is 326 feet.

The javelin strikes the ground at 160.75 feet.

The given equation is

$h(x)=-(1)/(20)x^2+8x+6$

where $x$ is the horizontal distance

At $x=$ we will get the initial vertical height.

$h(0)=-(1)/(20)*0+8* 0+6\n\Rightarrow h(0)=6$

Vertex of a parabola is given by

$x=-(b)/(2a)=-(8)/(2* -(1)/(20))\n\Rightarrow x=80$

$h(80)=-(1)/(20)(80)^2+8* 80+6=326$

At $h(x)=0$ the javelin will hit the ground

$0=-(1)/(20)x^2+8x+6\n\Rightarrow -x^2+160x+120=0\n\Rightarrow x=(-160\pm √(160^2-4\left(-1\right)*120))/(2\left(-1\right))\n\Rightarrow x=-0.75,160.75$

Learn more about parabolas from:

brainly.com/question/14477557

This question is incomplete, the complete question is;

A javelin is thrown in the air. Its height is given by h(x) = -1/20x² + 8x + 6

where x is the horizontal distance in feet from the point at which the javelin is thrown.

a. How high is the javelin when it was thrown?

b. What is the maximum height of the javelin?

c. How far from the thrower does the javelin strike the ground?'

Answer:

a. height of the javelin when it was thrown is 6 ft

b. the maximum height of the javelin is 326 ft

c. distance from the thrower is 160.75 ft

Explanation:

Given h(x) = -1/20x² + 8x + 6

we determine the height when x = 0

h(0) = -1/20(0)² + 8(0) + 6 = 6 ft

therefore height of the javelin when it was thrown is 6 ft

to determine the maximum height of the javelin;

we find the vertex of the quadratic

h = - [ 8 / ( 2(-1/20) ) ] = 80

therefore

h(80) = -1/20(80)² + 8(80) + 6

= -320 + 640 + 6 = 326 ft

therefore the maximum height of the javelin is 326 ft

Now the thrower is at the point ( 0,0 ) and the javelin comes down at another point ( x,0 )

this is possible by calculating h(x) = 0

⇒ -1/20x² + 8x + 6 = 0

⇒ x² - 160x - 120 = 0

⇒ x = [ -(-160) ± √( (-160)² - 4(1)(-120) ) ] / [ 2(1) ]

x = [ 160 ± √(25600 + 480) ] / 2

[x = 160.75 ; x = -0.75 ]

distance cannot be Negative

therefore distance from the thrower is 160.75 ft

In an 8.00km race, one runner runs at a steady 11.8 km/hr and another runs at 15.0 km/hr. How far from the finish line is the slower runner when the faster runner finishes the race?