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katovenus [111]
3 years ago
13

Water that soaks into Earth may become _____ under the surface.

Physics
1 answer:
larisa [96]3 years ago
5 0
If it soaks into the earths surface it becomes ground water

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HELP ASAP PLEASEEEEEEEEEEEE
ira [324]
It's the second graph!
it's the only one with a negative gradient.
so the temperature of the ball will fall in water as it looses its heat.

activate windows,:-P
8 0
4 years ago
A ball is dropped from a building taking 3sec to fall to the ground. Calculate:
GenaCL600 [577]

Answer:

Vf = 29.4 m/s

h = 44.1 m

Explanation:

Data:

  • Initial Velocity (Vo) = 0 m/s
  • Gravity (g) = 9.8 m/s²
  • Time (t) = 3 s
  • Final Velocity (Vf) = ?
  • Height (h) = ?

==================================================================

Final Velocity

Use formula:

  • Vf = g * t

Replace:

  • Vf = 9.8 m/s² * 3s

Multiply:

  • Vf = 29.4 m/s

==================================================================

Height

Use formula:

  • \boxed{h=\frac{g*(t)^{2}}{2}}

Replace:

  • \boxed{h=\frac{9.8\frac{m}{s^{2}}*(3s)^{2}}{2}}

Multiply time squared:

  • \boxed{h=\frac{9.8\frac{m}{s^{2}}*9s^{2}}{2}}

Simplify the s², and multiply in the numerator:

  • \boxed{h=\frac{88.2m}{2}}

It divides:

  • \boxed{h=44.1\ m}

What is the velocity when falling to the ground?

The final velocity is <u>29.4 meters per seconds.</u>

How high is the building?

The height of the building is <u>44.1 meters.</u>

3 0
3 years ago
II.
Alekssandra [29.7K]

Answer:

I just need to get points soorry

Explanation:

3 0
3 years ago
A scooter is traveling at a constant speed v when it encounters a circular hill of radius r = 480 m. The driver and scooter toge
Grace [21]

Answer:

68.585m/sec , 779.1 N

Explanation:

To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.

From centripetal motion.

F =( mv^2)/2

But since we are dealing with weightlessness

r = 480m

g = 9.8m/s^2

M also cancels, so forget M.

V^2 = Fr

V = √ Fr

V =√ (9.8 x 480) = 4704

= 68.585m/sec.

b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960

= 4.9m/sec^2.

Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)

159kg × ( 9.8-4.9)

159kg × 4.9

= 779.1N

6 0
3 years ago
Evaluate the gravitational potential energy between two 5.00-kg spherical steel balls separated by a center-tocenter distance of
navik [9.2K]

Answer:

U = 8.30×10-⁹J

Explanation:

m1 = m2 = 5.00kg masses of the spheres

d = 15.0cm = 15×10-²m

r = 5.10cm = 5.10×10-²m

R = d + r = 15×10-² + 5.10×10-²

R = 20.10 ×10-²m = 0.201m

G = 6.67×10-¹¹Nm²/kg²

U = Gm1×m2/R = potential energybetween the spheres

U = 6.67×10-¹¹×5.00×5.00/0.201

U = 8.30×10-⁹J

7 0
3 years ago
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