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2 years ago

5

Sarah drives at a constant speed of 15 m/s around a circular horizontal curve of diameter 60 m. What are the magnitude and direc

tion of her acceleration?

1 answer:

mojhsa [17]2 years ago

5 0

Answer:

7.5 m/s^2

Explanation:

I hope my answer helps u.

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Assuming two hypothetical maps that each cover a standard 8.5 by 11-inch sheet of paper, the larger-scale map would cover a larg

sukhopar [10]

Answer:

<em>b) false</em>

Explanation:

The scale of a map is the ratio of a distance on the map to the corresponding distance on the ground. Scaling allow us to capture a large geographical area on a reduced platform while still retaining the relative sizes and positioning of places on the map to their real life sizes and positioning. If both maps cover a standard 8.5 by 11-inch sheet of paper, then the map with the smaller ratio will have the bigger geographical area.

To understand better, let us assume two geographical areas A and B. A is bigger than B. If we were to put them both on the same area of map paper, then we'll have to scale up the smaller geographical area B so as to fit into the map paper. This means that the geographical area with the smaller area B will have the larger scale on the map.

8 0

3 years ago

An Olympic runner completes the 200-meter sprint in 23 seconds. What is the runnerâ€™s average speed? (Round your answer to the

Amanda [17]

The average speed of the runner is 8.7m/s.

Therefore, Option C) 8.7m/s is the correct answer.

Given the data in the question;

Distance covered; $d = 200m$

Time taken; $t = 23s$

Average speed; $s = \ ?$

Speed is the rate at which an object covers a certain distance. It is expressed as:

$s = \frac{d}{t}$

Where s is speed, d is distance and t is time taken.

We substitute our given distance into the equation

$s = \frac{200m}{23s} \\\\s = 8.6956m/s\\\\s = 8.7m/s$

The average speed of the runner is 8.7m/s

Therefore, Option C) 8.7m/s is the correct answer.

Learn more: brainly.com/question/21503615

3 0

2 years ago

In 1977 off the coast of Australia, the fastest speed by a vessel on the water

fenix001 [56]

Answer: 154.08 m/s

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$a_{ave}=1.80 m/s^{2}$

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0$ the initial velocity and $V_{f}$ the final velocity

$\Delta t=85.6 s$

Then:

$a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}$

Since $V_{o}=0$ :

$a_{ave}=\frac{V_{f}}{\Delta t}}$

Finding $V_{f}$ :

$V_{f}=a_{ave} \Delta t$

$V_{f}=(1.80 m/s^{2})(85.6 s)$

Finally:

$V_{f}=154.08 m/s$

8 0

3 years ago

1. Which pair of physical quantities consists of two vectors?

Tems11 [23]

Answer:

Velocity and force consists of two vectors as they both have magnitude and direction

8 0

3 years ago

Read 2 more answers

An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airst

sweet-ann [11.9K]

Answer:

see explanation below

Explanation:

Given that,

$T_1 =$ 500°C

$T_2$ = 25°C

d = 0.2m

L = 10mm = 0.01m

U₀ = 2m/s

Calculate average temperature

$\\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5$

262.5 + 273

= 535.5K

From properties of air table A-4 corresponding to $T_{avg}$ = 535.5K $\approx 550K$

k = 43.9 × 10⁻³W/m.k

v = 47.57 × 10⁻⁶ m²/s

$P_r = 0.63$

A)

Number for the first strips is equal to

$R_e_x = \frac{u_o.L}{v}$

$R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4$

Calculating heat transfer coefficient from the first strip

$h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3$

$h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2$

The rate of convection heat transfer from the first strip is

$q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W$

The rate of convection heat transfer from the fifth trip is equal to

$q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2$

Calculating $h_o_-_4$

$h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2$

The rate of convection heat transfer from the tenth strip is

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2$

Calculating

$h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W$

The rate of convection heat transfer from 25th strip is equal to

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)$

Calculating $h_o_-_2_5$

$h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2$

Calculating $h_o_-_2_4$

$h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W$

6 0

4 years ago