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Paha777 [63]
3 years ago
8

On Planet Y, which has no air, a dropped object falls 9 m in 3 seconds. What is g, the acceleration due to gravity, on that plan

et?
9.8 m/s2
3 m/s2
2 m/s2
Physics
1 answer:
slavikrds [6]3 years ago
7 0

Answer:

a = 2 m/s^2

which agrees with the third answer option provided.

Explanation:

Recall the kinematic formula for displacement under the action of a constant acceleration "a":

yf - yi = 1/2 a  t^2

using the information provided this equation becomes:

9 = 1/2 a (3)^2

solve for a:

9 * 2 / 9 = a

then a = 2 m/s^2

which agrees with the third answer option provided.

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an athlete sprints from 150 m south of the finish line to 65 m south of the finish line in 5.0s what is his average velocity
MatroZZZ [7]
  • Total displacement=150-65m=85m
  • Time=5s

\\ \rm\longrightarrow Avg\:Velocity=\dfrac{Total\:Displacement}{Total\:Time}

\\ \rm\longrightarrow Avg\:Velocity=\dfrac{85}{5}

\\ \rm\longrightarrow Avg\:Velocity= 17m/s

6 0
3 years ago
Read 2 more answers
A subway train is traveling at 22.2 m/s when it approaches a slower train 50m ahead traveling in the same direction at 6.94 m/s.
Amiraneli [1.4K]

Answer:

Time that they collide = 4.99s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Explanation:

We will use the equations of motion to obtain the solution required

At time t = 0

speed of first train = 22.2 m/s

Initial space between the two trains = 50 m

Speed of second train = 6.94 m/s

For the first car, distance covered by the first train = y

y = distance covered between the beginning of the deceleration and the point where the the two trains hit one another.

u = initial velocity = 22.2 m/s

t = time taken for all this to happen

a = deceleration = - 2.1 m/s²

y = ut + (1/2)at²

y = 22.2t - 1.05t² (eqn 1)

For the second train,

At t = 0, y = 50 m

Let the new distance moved by the second train before collision = (y - 50)

u = initial velocity = 6.94 m/s

t = time taken = t

a = acceleration of the second train = 0 m/s² (constant velocity)

(y - 50) = ut + (1/2)at²

y - 50 = 6.94t

y = 6.94t + 50 (eqn 2)

substituting for y in eqn 2 using the expression obtained in eqn 1

y = 22.2t - 1.05t²

y = 6.94t + 50

22.2t - 1.05t² = 6.94t + 50

1.05t² - 15.26t + 50 = 0

Solving this quadratic equation

t = 4.99 s or 9.54 s

The position of the two trains are the same at those two times, but the first time is when they hit each other.

t = 4.99 s

At 4.99 s, the the velocity of the first train

v = u + at

v = 22.2 + (-2.1×4.99) = 11.721 m/s in the same direction as the second train.

Relative velocity at this point will be

= 11.721 - 6.94 = 4.781 m/s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Hope this Helps!!!

4 0
3 years ago
In what subject could we see cross cutting concepts
meriva
A Framework for K–12 Science Education: Practices, Crosscutting Concepts, and Core Ideas (Framework) recommends science education in grades K–12 be built around three major dimensions: science and engineering practices, crosscutting concepts that unify the study of science and engineering through their common application across fields, and core ideas in the major disciplines of natural science.
4 0
3 years ago
A 1.3-kg model airplane flies in a circular path on the end of a 23-m line. The plane makes
storchak [24]

(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in

(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s

(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:

(2<em>π</em> (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s

(c) The plane accelerates toward the center of the path with magnitude

<em>a</em> = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²

(d) By Newton's second law, the tension in the line is

<em>F</em> = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N

4 0
3 years ago
When using a liquid as a solvent, what could be the state(s) of matter of the solute?
sattari [20]

Answer:

Explanation:

it can be a gas like carbon dioxide in soda

it can be liquid like alcohol in water

it can be solid like sugar and salt

8 0
3 years ago
Read 2 more answers
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