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2 years ago

Thirty-five percent of all school-age pedestrians killed in school transportation-related crashes were between the ages of

Physics

1 answer:

BartSMP [9]2 years ago

6 0

My drivers ed book says 15-19 years of age, hope it's right still.

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Grade 11 Physics

xeze [42]

1. Displacement = 50 m North. Distance = 150m
Displacement is a vector. Distance is a scalar.

2. Displacement is 100 miles East.

6 0

3 years ago

A dog can hear sounds in the range from 15

sergij07 [2.7K]

<h3>The wavelength corresponds to the lower cut-off point of the sounds is 22.93 meter</h3>

<em><u>Solution:</u></em>

Given that,

A dog can hear sounds in the range from 15 to 50,000 Hz

The speed of sound at a temperature of $20^{\circ}C$ is: 344 m/s

To find: wavelength

$\lambda= \frac{v}{f}$

Where,

v is speed of sound

f is the frequency

f = lower cut off point = 15 Hz

$\lambda = \frac{344}{15}\\\\\lambda = 22.93$

Thus the wavelength corresponds to the lower cut-off point of the sounds is 22.93 meter

8 0

2 years ago

What is the SI unit for acceleration?

qwelly [4]

The base unit for acceleration is m s ^-2

6 0

2 years ago

What type of climate do most islands within Oceania have? A. tundra B. tropical humid C. humid continental D. desert

Tomtit [17]

B
Most of the islands have a tropical humid climate

4 0

3 years ago

Read 2 more answers

A converging lens (f1 = 12.5 cm) is located 28.8 cm to the left of a diverging lens (f2 = -6.18 cm). A postage stamp is placed 3

k0ka [10]

Answer:

The final image relative to the diverging lens at 9.0 cm.

Explanation:

Given that,

Focal length of diverging lens = -6.18 cm

Focal length of converging lens = 12.5 cm

Distance of object = 36.9 cm

We need to calculate the image distance of converging lens

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{12.5}-\dfrac{1}{-36.9}$

$\dfrac{1}{v}=\dfrac{988}{9225}$

$v=9.33\ cm$

We need to calculate the image distance of diverging lens

Now, object distance is

$u=28.8-9.33=19.47 cm$

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{-6.18}-\dfrac{1}{-19.47}$

$\dfrac{1}{v}=-\dfrac{22150}{200541}$

$v=-9.0\ cm$

The image is formed 9.0 cm to the left of the diverging lens.

Hence, The final image relative to the diverging lens at 9.0 cm.

4 0

2 years ago