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Pani-rosa [81]
3 years ago
7

Which of the following surface offers minimum friction ? Sand Water Ice Grass

Physics
2 answers:
Alchen [17]3 years ago
8 0
Answer: ice... welcome
sergij07 [2.7K]3 years ago
8 0

Answer:

it is Ice, of course

Explanation:

some text to answer

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A heavy rock is shot upward from the edge of a vertical cliff. It leaves the edge of the cliff with an initial velocity of 12 m/
jenyasd209 [6]

Answer:

The height of the building is 88.63 m.

Explanation:

Given;

initial component of vertical velocity, v_i = 12 m/s sin 26° = 5.26 m/s

initial horizontal component of the velocity, u_i = 12 m/s cos 26° =10.786 m/s

horizontal distance traveled by the rock, x = 40.4 m

time of flight is calculated as;

x = u_i t

t = x / u_i

t = 40.4 / 10.786

t = 3.75 s

Determine the final vertical velocity of the ball;

v_f = v_i + gt\\\\v_f = 5.26 + (9.8 *3.75)\\\\v_f = 42.01 \ m/s

Determine the height of the rock;

v_f^2 = v_i^2 + 2gh\\\\h = \frac{v_f^2 - v_i^2}{2g}\\\\ h = \frac{(42.01)^2 - (5.26)^2}{2*9.8}\\\\h = 88.63 \ m

Therefore, the height of the building is 88.63 m.

5 0
3 years ago
Please I need with this
guajiro [1.7K]
I think its b im not sure im in 8th grade
8 0
3 years ago
HELP
BaLLatris [955]
The part of the electromagnetic spectrum has a shorter wavelength than ultraviolet light is x-rays.
3 0
3 years ago
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A car accelerates from rest to a velocity of 5 meters/second in 4 seconds. What is its average acceleration over this period of
wlad13 [49]
The car's average acceleration would be 1.25m/s^2 or 1.25meters/second/second. That looks to be the fourth one you've listed.
7 0
3 years ago
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You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a
Zolol [24]

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

1. To find the horizontal velocity of the rock we need to use the following equation:

d = v*t \rightarrow v = \frac{d}{t}    

<u>Where</u>:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

t = \sqrt{\frac{2d}{g}}

<u>Where:</u>

g: is gravity = 9.8 m/s²

t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s

Now, the horizontal velocity of the rock is:

v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s      

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.          

2. To calculate the distance at which the projectile will land, first, we need to find the time:

t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s

So, the distance is:

d = v*t = 6.67 m/s*1.43 s = 9.54 m    

Therefore, the projectile will land at 9.54 m of the second cliff.

I hope it helps you!        

7 0
4 years ago
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