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velikii [3]
3 years ago
10

The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other ob

ject will be dropped with no initial speed. The objects are initially 179 cm above the ground and separated by 142 cm. What will be the difference in the landing locations of the two objects?
Physics
1 answer:
Crazy boy [7]3 years ago
4 0

Answer:

104.3 cm  or 179.7

Explanation:

First find time that it takes for the object to hit the ground

\sqrt{(2H)/g}  ->   \sqrt{(2 x 179)/ 9.8} = 6.04s\\*

Then find xf of projectile xf= 5.9(6.04) = 37.7\\\\

not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7   or    142+37.7  

hope that helps

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When developing an experiment design, which could a scientist take to improve the quality of the results?
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Answer:

B. Identify any biases about the answer to the scientific question.

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3 0
3 years ago
Calculate the power of the eye in D when viewing an object 5.70 m away. (Assume the lens-to-retina distance is 2.00 cm. Enter yo
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Answer:

Power=50.17dioptre

Power=50.17D

Explanation:

P=1/f = 1/d₀ + 1/d₁

Where d₀ = the eye's lens and the object distance= 5.70m=

d₁= the eye's lens and the image distance= 0.02m

f= focal length of the lense of the eye

We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m

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4 years ago
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