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matrenka [14]
3 years ago
6

What environments does tornado not occur in?

Physics
2 answers:
castortr0y [4]3 years ago
7 0
I think winter is right where i live we never have tornadoes
Nadusha1986 [10]3 years ago
5 0
The winter I think would be the answer
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The tallest sequoia sempervirens tree in California’s redwood national parks is 111 m tall. Suppose an object is thrown downward
ch4aika [34]

The object's <u>initial velocity</u> is equal to -10.59\frac{m}{s}

Why?

From the statement we know the height of the tree and the time it takes to reach the ground, so, if we need to calculate its initial velocity, we can use the following formula:

y=y_o-v_{o}*t-\frac{1}{2}g*t^{2}

Where,

y, is the final height (0 meters in this case)

yo, is the initial height (111 meters in this case)

t, is the time elapsed (3.8 seconds in this case)

vo, is the initial speed.

g, is the acceleration due to gravity (-9.81 m/s2)

Now, let's set the origin at the top of the tree, so, rewriting the formula, we have:

y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}

So, isolating the initial velocity, we have:

y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}

y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}\\\\v_{o}*t=y-y_o-\frac{1}{2}g*t^{2}\\\\v_{o}=\frac{y-y_o-\frac{1}{2}g*t^{2}}{t}

Finally, substituting and calculating, we have:

v_{o}=\frac{-111m-0-\frac{1}{2}(-9.8\frac{m}{s^{2}}) *(3.8s)^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-9.8\frac{m}{s^{2}})*(14.44s^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-141.51m)}{3.8s}=\frac{-111m+70.75m}{3.8s}\\\\v_{o}=\frac{-40.25m}{3.8s}=-10.59\frac{m}{s}

Hence, we have that the <u>initial velocity</u> of the object is -10.59\frac{m}{s}

Have a nice day!

7 0
4 years ago
What are limits that would be age appropriate for teens vs toddlers? give me some examples of limits that you may have for a tee
Pavlova-9 [17]

The limits which would be age appropriate for toddlers include being flexible and realistic while that of a teen should be more rigid.

<h3>Who are Toddlers?</h3>

This refers to a child whose age ranges from 1 year to 4 years and limits set for them should be flexible due to it helping them feel more secure and decreases anxiety.

Example of limits that you may have for a teen is reducing their screen time while an example of limits for toddlers is correcting their junk eating habits. Teens should have a rigid limit so as to enable them do the right things at all times due to their strong emotions.

Read more about Limits here brainly.com/question/1419949

#SPJ1

6 0
2 years ago
A car is traveling at 108 km/h, stuck behind a slower car. Finally the road is clear and the car pulls over to make a pass. The
mezya [45]

Answer:

The average acceleration of the car is 2.143 meters per square second.

Explanation:

Let assume that car accelerates uniformly, in that case, we can obtain the value of acceleration by using the following equation of motion:

v = v_{o}+a\cdot t

Where:

v_{o} - Initial velocity, measured in meters per second.

v - Final velocity, measured in meters per second.

a - Acceleration, measured in meters per square second.

t - Time, measured in seconds.

Now, we clear acceleration within expression:

a = \frac{v-v_{o}}{t}

Initial and final velocities are now converted from kilometers per hour into meters per second:

v_{o} = \left(108\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)

v_{o} = 30\,\frac{m}{s}

v = \left(135\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)

v = 37.5\,\frac{m}{s}

If we know that t = 3.5\,s, then, the average acceleration of the car is:

a = \frac{37.5\,\frac{m}{s}-30\,\frac{m}{s} }{3.5\,s}

a = 2.143\,\frac{m}{s^{2}}

The average acceleration of the car is 2.143 meters per square second.

8 0
3 years ago
A 230 g air-track glider is attached to a spring. The glider is pushed in 8.2 cm and released. A student with a stopwatch finds
Natasha_Volkova [10]

Answer:

4.3 N/m

Explanation:

m = 230 g = 0.230 kg, x = 8.2 cm

in 13 oscillations, time taken = 19 s

In 1 oscillation, time taken = 19 / 13 = 1.46 s

By the use of formula of time period , Let k be the spring constant.

T = 2\pi \sqrt{\frac{m}{k}}

1.46 = 2\pi \sqrt{\frac{0.230}{k}}

0.054 = 0.230 / k

k = 4.26 N/m

k = 4.3 N/m

8 0
3 years ago
The freezing point of water is 0ÁC, and the boiling point is 100ÁC. This range of temperatures can be found naturally on the Ear
viva [34]
C.
The range of temperatures on Earth allows water to exist in all of its states.
8 0
3 years ago
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