PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (40pts)

Physics

2 answers:

In-s [12.5K]3 years ago

6 0

Answer: A

Explanation:

Just speed I think.

Speed is scalar, while velocity is a vector so you would need a direction. For acceleration you would need to know the specific speed at different points in time.

tia_tia [17]3 years ago

3 0

A cheetah can run 140 miles in 2 hours.

The information we can calculate about the cheetah given this information is the "speed" (A).

You might be interested in

Why water in earthern pot remain cool in summer

hjlf

Answer:

Since in summer, the eastern side do not face the sunlight and hence the water in eastern pot remain cool in summer.

8 0

3 years ago

Read 2 more answers

Im doing the BIg Bang for my project anyone know what it is, make it a ¨question¨ and an experiment.

Daniel [21]

The big bang is a theory of our universe and many others along with making our solar system. Meanwhile, in other universes have are solar system as well that is why we see stars at night because the sun and it's rays are so bright during the day it blocks them out.

6 0

3 years ago

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 initial velocity 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:

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