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lys-0071 [83]
3 years ago
14

I NEED THIS ASAP!!! A ball is thrown straight up with an initial velocity of 4.40 m/s. Assuming there is no air friction, what i

s the maximum height the ball will reach? 40pts pls
Physics
1 answer:
Lilit [14]3 years ago
5 0

Answer:

I think it is 80m/s

Explanation:

d = ½ g t2

  = ½ (10 m/s2) (4 s)2

  = (5 m/s2) (16 s2)

  = 80 ms

75% sure

Hope this helps!!!

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A golfer hits a golf ball upwards at an angle. We can ignore air resistance on the ball.
Natalka [10]

Answer:

Check the diagram from the photo

Explanation:

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3 years ago
Winds that blow from the north and south poles
Gennadij [26K]
Winds that blow from the north and south poles would be called k<span>atabatic winds. I'm not sure if I spelled that right, but that's the answer I hope.</span>
8 0
3 years ago
The half-life of cobalt-60 is 5.26 years. If 50 g are left after 15.8 years, how many
PtichkaEL [24]

Answer:

400 g

Explanation:

The computation of the number of grams in the original sample is shown below:

Given that

half-life = 5.26 years

total time of decay = 15.8 years

final amount = 50.0 g

Now based on the above information  

number of half-lives past is

=  15.8 ÷ 5.26

= 3 half-lives

Now

3 half-lives = 1 ÷ 8 remains = 50.0 g

So, the number of grams would be

= 50.0 g × 8

= 400 g

4 0
2 years ago
A plane starting from rest accelerates at 3m/s2 for 25s. Calculate the increase in velocity after:
horsena [70]
  • Initial velocity=u=0m/s
  • Acceleration=a=3m/s^2

Time not needed now

Case-1:-

  • Time=1s
  • Final velocity=v

\\ \rm\hookrightarrow v=u+at

\\ \rm\hookrightarrow v=0+3(1)

\\ \rm\hookrightarrow v=3m/s

Case-2:-

  • Time=3s

\\ \rm\hookrightarrow v=0+3(3)

\\ \rm\hookrightarrow v=9m/s

Case-3:-

  • Time=25s

\\ \rm\hookrightarrow v=0+3(25)

\\ \rm\hookrightarrow v=75m/s

5 0
2 years ago
Two objects are dropped from different heights at the same instant. If the first object takes 10.7 seconds to hit the ground and
Lubov Fominskaja [6]

Answer:

The answer to your question is : 521.8 m

Explanation:

Data:

Different heights

Time first object (tfo) = 10.7 s

Time second object (tso)= 14.8 s

Initial speed of both objects(vo) = 0 m/s

a = 9.81 m/s²

Formula:

h = vot + 1/2 (a)(t)² but vo = 0    so, h = 1/2 (a)(t)²

Then, height fo      h = 1/2 (9.81)(10.7)² = 561.6 m

          height so     h = 1/2(9,81)(14.8)² = 1074.4 m

Difference in their heights =  1074.4 m - 561.6 m = 521.8 m

5 0
3 years ago
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