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Zinaida [17]
3 years ago
14

2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de

termine the distance from the edge of the well to the water's surface.​
Physics
1 answer:
kow [346]3 years ago
3 0

Answer:

S = 11.025 m

Explanation:

Given,

The time taken by the pebble to hit the water surface is, t = 1.5 s

Acceleration due to gravity, g = 9.8 m/s²

Using the II equations of motion

                          S = ut + 1/2 gt²

Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity

                                u = 0

Therefore, the equation becomes

                            S = 1/2 gt²

Substituting the given values in the above equation

                              S = 0.5 x 9.8 x 1.5²

                                 = 11.025 m

Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m

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Given the equation p2 = a3, what is the orbital period, in days, for the planet venus? (venus is located 0.72 au from the sun?)
melisa1 [442]

The correct answer is 223 days.

The relationship between the duration of revolution and the separation between the sun is shown by Kepler's third law. Using the notions of circular motion and the gravitational and centripetal forces, we may obtain this equation.

According to Kepler's third rule, the semi-major axis of an orbit is linked to the orbital period of a planet around the sun as follows:

p² = a³

where an is the semi-major axis/distance to the star and p is the orbital period in years.

It is said that a = 0.72 AU for Venus.

P= √(0.72 AU)^3 = 0.61 years.

365 days in a year = 222.9 ≈ 223 days.

To learn more about Kepler's third rule refer the link:
brainly.com/question/1608361

#SPJ4

5 0
1 year ago
An inventor develops a stationary cycling device by which an individual, while pedaling, can convert all of the energy expended
mr Goodwill [35]

Answer:

Q=94185\ J

Explanation:

Given:

  • mass of water, m=0.3\ kg
  • initial temperature of water, T_i=20^{\circ}C
  • final temperature of water, T_f=95^{\circ}C
  • specific heat of water, c=4186\ J.kg^{-1}.K^{-1}

<u>Now the amount of heat energy required:</u>

Q=m.c.\Delta T

Q=0.3\times 4186\times (95-20)

Q=94185\ J

Since all of the mechanical energy is being converted into heat, therefore the same amount of mechanical energy is required.

4 0
3 years ago
A force of 80 N is exerted on an object on a frictionless surface for a distance of 4 meters. If the object has a mass of 10 kg,
solmaris [256]
There are two ways to find energy. Energy=F*d=mv^2. We can use this relationship to find v:
Fd=0.5mv^{2} \\ 80*4=0.5*10v^{2} \\ 64=v^{2} \\ v=8 m/s
6 0
3 years ago
A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e
mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

 ∑ M = 0

mass of the log will be concentrated at the center  

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)

200 \times 0.1 L = 53.5 \times (x - 0.6 L)

0.374 L =x - 0.6 L

       x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

7 0
3 years ago
A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.0 km, the je
jenyasd209 [6]
<h2>The answer got is reasonable.</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 300 m/s  

Acceleration, a = ?

Final velocity, v = 400 m/s  

Displacement,s = 4 km = 4000 m

Substituting  

v² = u² + 2as

400² = 300² + 2 x a x 4000

a = 8.75 m/s² = 8.8 m/s²

The acceleration is 8.8 m/s²

The answer got is reasonable.

7 0
3 years ago
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