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andrezito [222]
2 years ago
8

WILL MARK BRAINLIEST!!!!!!!!! 20 PTS!!!!!!!!!!!

Physics
2 answers:
polet [3.4K]2 years ago
6 0

Answer:

Ocean surface:0

Explanation:

qwelly [4]2 years ago
5 0

Answer:

My answer is=

Ocean surface:0

Seaweed:(-20)

Clownfish:(-23)

Squid:(-44)

Ocean floor:(-50)

(I'm very sorry if I was wrong T~T)

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A helicopter is traveling with a velocity of 12 m/s directly upward.
Irina-Kira [14]

Answer:

H = 27.35 m

Explanation:

From projectile motion, we know that;

h = (v_o)²/2g

Where;

v_o = 12 m/s

g = 9.8 m/s²

Thus;

h = 12²/(2 × 9.8)

h = 144/19.6

h = 7.35 m

Now, we are told that when the man is 20 m above the pillow, he lets go of

the rope.

Thus, greatest height reached by the man above the ground is;

H = 20 + h

H = = 20 + 7.35

H = 27.35 m

5 0
2 years ago
Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia
Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km  west

velocity V1 = 8.9 km/h

distance B = 4.5 km  east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7     ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x )  / 8.9      .............2

now equating equation 1 and 2

time A = time B

x / 7  =   ( 10.2 - x )  / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

4 0
3 years ago
QUICK!! What class of leer is shown below?
wlad13 [49]

Answer:

3rd class lever

HOPE IT HELPS YOU OUT PLEASE MARK IT AS BRAINLIEST AND FOLLOW ME PROMISE YOU TO FOLLOW BACK ON BRAINLY.IN

5 0
2 years ago
Read 2 more answers
50 points!!! Kinetics
iris [78.8K]

Answer:

98 m √

Explanation:

How about s = Vo * t + ½at² ?

s = h = Vo * 2s - 4.9m/s² * (2s)² = 2Vo - 19.6

and

h = Vo * 10s - 4.9m/s² * (10s)² = 10Vo - 490

Subtract 2nd from first:

0 = -8Vo + 470.4

Vo = 58.8 m/s

h = 58.8m/s * 2s - 4.9m/s² * (2s)² = 98 m

3 0
2 years ago
Read 2 more answers
a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone​
Dafna1 [17]

Answer:

The centripetal acceleration of the stone is 5 m/s²

Explanation:

The length of the string to which the stone is attached, r = 1 m

The speed with which the string is rotated, v = 5 m/s

The centripetal acceleration, a_c, is given as follows;

a_c = \dfrac{v^2}{r}

Therefore, the centripetal acceleration of the stone found as follows;

a_c = \dfrac{(5 \ m/s)^2}{1 \ m} = 5 \ m/s^2

The centripetal acceleration of the stone, a_c = 5 m/s².

5 0
2 years ago
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