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svetoff [14.1K]

2 years ago

7

Richard walks from his home to his school at a constant speed. it takes him 4 min to travel 100 m. which of the following lines

in the following distance-time graph could show Richard's motion on the way to school?

1 answer:

Sindrei [870]2 years ago

5 0

Since his speed is constant,

B.

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A sound source is located somewhere along the x-axis. Experiments show that the same wave front simultaneously reaches listeners

guapka [62]

Answer:

a) x₀ = - 2 m , b) y = 4.47 m

Explanation:

A wave travels in the middle with constant speed, let's use the equation of uniform motion

v = d / t

t = d / v

The distance to the first listeners, see attached

d₁ = x₀-x

t = (x₀ +7) / v

The distance to the second listener

d₂ = x - x₀

t = (+ 3- x₀) / v

As the wave arrives at the same time, we can equal the two equations

(x₀ +7) / v = (3 -x₀) / v

x₀ + 7 = 3 - x₀

2 x₀ = 3 - 7

x₀ = -4/2

x₀ = - 2 m

b) The time it takes for the wave to reach the listeners of the x-axis, where the speed of sound is 340 m / s

t = 5/340

t = 0.0147 s

Let's look for the distance the wave travels for the listener axis and

v = d₃ / t

d₃ = v.t

d₃ = 340 * 0.0147

d₃ = 5 m

For the distance component we use the Pythagorean triangle

d₃² = x₀² + y²

y² = d₃² - x₀²

y = √ (d₃² -4)

y = √ (5² -4)

y = 4.47 m

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If al snow and glaiers melted, would the sea level rise or fall

Kobotan [32]

The sea level would rise because the snow and glaciers are water

4 0

3 years ago

How and why does air pressure change with altitude in the atmosphere?

wlad13 [49]

Pressure with Height: pressure decreases with incrementing altitude. The pressure at any caliber in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a homogeneous surface at lower calibers.

5 0

3 years ago

2. Find the time taken by the bus to reach the stop. need only group B, 2 answer

Molodets [167]

Answer:

t = 2 seconds

Explanation:

In 2nd question, the question is given the attached figure.

Initial speed of the bus, u = 0

Acceleration of the bus, a = 8 m/s²

Final speed, v = 16 m/s

We need to find the time taken by the car to reach the stop. Acceleration of an object is given by :

$a=\dfrac{v-u}{t}$

t is time taken

$t=\dfrac{v-u}{a}\\\\t=\dfrac{16-0}{8}\\\\t=2\ s$

The bus will take 2 seconds to reach the stop.

3 0

3 years ago

Can someone please help me with this physics question? I'm desperate!

Lelu [443]

Answer:

a) 2·√10 seconds

b) Linda should be approximately 30.6 meters

c) Jenny's speed at the 100-m mark is approximately 6.325 m/s

Explanation:

The speed with which Linda is running = 8.6 m/s

The point Jenny starts = The 80-m mark

The acceleration of Jenny = 1.0 m/s²

a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows

Δs = u·t + (1/2)·a·t²

Δs = Distance = 100-m - 80-m = 20-m

u = The initial velocity of Jenny = 0

a = Jenny's acceleration = 1.0 m/s²

∴ 20 = 0×t + (1/2) × 1 × t² = t²/2

20 = t²/2

t = √(20 × 2) = 2·√10

The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds

b) The distance Linda runs in t = 2·√10 seconds, d = v × t

Given that Linda's velocity, v = 8.6 m/s, we have;

d = 8.0 × 2·√10 = 16·√10

The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters

Therefore, Linda should be approximately (50.6 - 20) meters = 30.6 meters behind Jenny when Jenny starts running

c) Jenny's speed at the 100 m mark is given as follows;

v = u + a·t

t = 2·√10 seconds, a = 1.0 m/s², u = 0

∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325

Jenny's speed at the 100-m mark ≈ 6.325 m/s

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2 years ago