Ask question

Ask question

All categories

3 years ago

5

If a storm is 7.5 kilometers away, how much time is expected between observations of lightning and thunder? Round your answer to

one decimal place. seconds

1 answer:

Genrish500 [490]3 years ago

3 0

Answer:

22.7 seconds

Explanation:

I just took the test...

You might be interested in

Type of energy transformed into chemical energy by plants is called type of energy transformed into chemical energy by plants is

yuradex [85]

I think the correct answer is light energy. It is light energy that is transformed into chemical energy by plants by the process called photosynthesis. In this process, plants<span> take in water, carbon dioxide, and sunlight and </span>turn<span> them </span>into<span> glucose and oxygen.</span>

6 0

3 years ago

What is a distraction that a pedestrian may engage in while crossing the street?

goldenfox [79]

Answer:

a dog walking or their phone rings or heard a neighbor talking to them

6 0

2 years ago

Teens realize that things may be “shades of gray” after they enter Piaget’s

expeople1 [14]

Teens realize that things may be “shades of gray” after they enter Piaget’s formal operational stage. The Piaget's approach for Adolescent, also referred to asFormal operational stage is the stage in which adolescents develop their ability to think about abstract concepts, they become more argumentative and <span>indecisive.</span>

5 0

3 years ago

Read 2 more answers

I really don’t get this one

worty [1.4K]

Compounds are molecules with 2 or more elements

So the answer would be the third one

CO2;H2O

4 0

2 years ago

At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

$\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

$\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

so

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

5 0

2 years ago