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Black_prince [1.1K]

3 years ago

12

In downtown Chicago, the east-west blocks are 400 ft long while the north-south blocks are 280 ft long. Because of the many one-

way streets, it can be challenging to get around. Veronica starts at the corner of Dearborn and Ohio Streets. She drives four blocks north to Superior, two blocks east to Wabash, then a block south to get to her destination at Wabash and Huron.
Required:
What is the straight-line distance from her starting point?

1 answer:

pickupchik [31]3 years ago

8 0

Answer:

The answer is "1160 ft".

Explanation:

Using the Pythagoras theorem:

$\to a= 280 \times 3= 840\\\\\to b= 400 \times 2= 800\\\\$

$\bold{\to y^2= a^2+b^2}\\\\$

$=840^2 +800^2\\\\= 705600+640000\\\\=1345600\\\\=1160 \ ft\\$

$\to y=1160 \ ft$

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3 years ago

A piston absorbs 25 J of heat from its surroundings while expanding from

Ivahew [28]

The work done is -30 J
The heat is 25 J

<h3>What is the heat and the work?</h3>

We know that the work done by a gas could be positive or negative same as the heat. If the work done is positive then work is done on the system.

The work done is obtained from;

W = PΔV

W = 1.0 x 105 Pa(0.0006 m³ - 0.0003 m³)

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6 0

1 year ago

The earth travels around the sun once a year in an approximately circular orbit whose radius is 1.50x10^11 m. From this data det

seraphim [82]

(a) Determine the circumference of the Earth through the equation,
C = 2πr
Substituting the known values,
C = 2π(1.50 x 10¹¹ m)
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Then, divide the answer by time which is given to a year which is equal to 31536000 s.
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Hence, the orbital speed of the Earth is ~29883.307 m/s.

(b) The mass of the sun is ~1.9891 x 10³⁰ kg.

8 0

3 years ago

A 1.20-m cylindrical rod of diameter 0.570 cm is connected to a power supply that maintains a constant potential difference of 1

nasty-shy [4]

(a) $1.72\cdot 10^{-5} \Omega m$

The resistance of the rod is given by:

$R=\rho \frac{L}{A}$ (1)

where

$\rho$ is the material resistivity

L = 1.20 m is the length of the rod

A is the cross-sectional area

The radius of the rod is half the diameter: $r=0.570 cm/2=0.285 cm=2.85\cdot 10^{-3} m$ , so the cross-sectional area is

$A=\pi r^2=\pi (2.85\cdot 10^{-3} m)^2=2.55\cdot 10^{-5} m^2$

The resistance at 20°C can be found by using Ohm's law. In fact, we know:

- The voltage at this temperature is V = 15.0 V

- The current at this temperature is I = 18.6 A

So, the resistance is

$R=\frac{V}{I}=\frac{15.0 V}{18.6 A}=0.81 \Omega$

And now we can re-arrange the eq.(1) to solve for the resistivity:

$\rho=\frac{RA}{L}=\frac{(0.81 \Omega)(2.55\cdot 10^{-5} m^2)}{1.20 m}=1.72\cdot 10^{-5} \Omega m$

(b) $8.57\cdot 10^{-4} /{\circ}C$

First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is

I = 17.5 A

So the resistance is

$R=\frac{V}{I}=\frac{15.0 V}{17.5 A}=0.86 \Omega$

The equation that gives the change in resistance as a function of the temperature is

$R(T)=R_0 (1+\alpha(T-T_0))$

where

$R(T)=0.86 \Omega$ is the resistance at the new temperature (92.0°C)

$R_0=0.81 \Omega$ is the resistance at the original temperature (20.0°C)

$\alpha$ is the temperature coefficient of resistivity

$T=92^{\circ}C$

$T_0 = 20^{\circ}$

Solving the formula for $\alpha$ , we find

$\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C$

5 0

3 years ago

Water ﬂows through a horizontal nozzle in steady ﬂow at the rate of 10m3/s. The inlet and outlet diameters are d1 = 0.5m and d2

Dvinal [7]

Answer:

P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,

Explanation:

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Q = A v

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v₁ = 50.93 m / s

v₂ = Q / A₂

v₂ = 10 4 /π 0.25²

v₂ = 203.72 m / s

Now we can use Bernoulli's equation in the colon

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear

P₁ = P2 + ½ rho (v₂² - v₁²)

P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)

P₁ = 1.013 10⁵ + 2.205 10⁷

P₁ = 2.215 10⁷ Pa

la definicion de presion es

P₁ = F₁/A₁

F₁ = P₁ A₁

F₁ = 2.215 10⁷ pi d₁²/4

F₁ = 2.215 10⁷ pi 0.5²/4

F₁ = 4.3 106 N

6 0

3 years ago