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melamori03 [73]
3 years ago
8

If an object is thrown in an upward direction from the top of a building 1.6 x 10^2 ft. high at an initial velocity of 21.82 mi/

h what is its final velocity when it hits the ground? ( disregard wind resistance. Round answer to nearest whole number and do not reflect negative direction in your answer.)
A. 101 ft/s
B. 96 ft/s
C 110 ft/s
Physics
2 answers:
Goshia [24]3 years ago
7 0

If an object is thrown in an upward direction from the top of a building 1.60 x 102 ft. high at an initial velocity of 21.82 mi/h, what is its final velocity when it hits the ground? (Disregard wind resistance. Round answer to nearest whole number and do not reflect negative direction in your answer.)


this question is troubling me i guessed 96 ft/s

can someone help me out and explain it thanks so much!!!!!!



svp [43]3 years ago
4 0

Question: if an object is thrown in an upward direction from the top of a building 1.6 x 10^2 ft. high at an initial velocity of 21.82 mi/h what is its final velocity when it hits the ground? ( disregard wind resistance. Round answer to nearest whole number and do not reflect negative direction in your answer.)

Answer: 110 ft/s

Explanation: Disregard wind resistance round answer to nearest whole number and do not reflect negative direction in your answer is 110

question answered by

(jacemooris04)

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What is the acceleration of a 10kg mass pushed by a 5N force?​
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Answer:

<h2>0.5 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

From the question we have

a =  \frac{5}{10} =  \frac{1}{2}   \\  = 0.5

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<h3>0.5 m/s²</h3>

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8. Which explains how a plane mirror works?
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What do scientitsts do to easily share mesurment data they can understand
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There are a few ways to do this- unfortunately different fields are better at it than others! Medical research is generally pretty good, some other fields likewise very good, some not as much. Basically, though, what they do is use standadisation- they agree on the terminology, units of data, statistical measures, and so forth, that will be used in that scientific field. As much as possible, every scientist in the field uses those standards so everyone working in the field should recognise it. For instance, in clinical trials, there is very good agreement worldwide on what the different metrics we use are- e.g. in cancer research, we usually want to know the 5-year survival rate (meaning the percentage of patients still alive 5 years after diagnosis). So anyone with the right training should be able to pick up a clinical trial report and understand what the results are and what the report is saying.
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A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the ho
Lady_Fox [76]

Answer:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>  

The ball rotates 6.78 revolutions.

     

Explanation:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>        

At the bottom the ball has the following angular speed:

\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m

To find the revolutions we need the time, which can be found using the following equation:                

v_{f} = v_{0} + at  

t = \frac{v_{f} - v_{0}}{a} (1)

So first, we need to find the acceleration:

v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}    (2)  

By entering equation (2) into (1) we have:

t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}

Since it starts from rest (v₀ = 0):  

t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s

Finally, we can find the revolutions:  

\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!                                                                                                                                                                                          

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