Ask question

Ask question

All categories

2 years ago

13

You drop a rock down a well and hear a splash 3 s later. As Charlie Brown would say, the well is 3 seconds deep. How fast is the

rock moving when it hits the water?

1 answer:

vampirchik [111]2 years ago

5 0

Since it was dropped, it should be the speed of gravity which is 9.8 meters/second

You might be interested in

The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 54/(7 + x2 + y2), where T is measured in °C and x,

ANTONII [103]

Answer:

dt/dx = -0.373702

dt/dy = -1.121107

Explanation:

Given data

T(x, y) = 54/(7 + x² + y²)

to find out

rate of change of temperature with respect to distance

solution

we know function

T(x, y) = 54 /( 7 + x² + y²)

so derivative it x and y direction i.e

dt/dx = -54× 2x / (7 +x² + y²)² .........................1

dt/dy = -54× 2y / (7 + x² + y²)² .........................2

now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2

dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²

dt/dx = -0.373702

and

dt/dy = -54× 2(3) / (7 + (1)² + (3)²)²

dt/dy = -1.121107

7 0

2 years ago

He throws a second ball (B2) upward with the same initial velocity at the instant that the first ball is at the ceiling. c. How

Gwar [14]

Answer:

hello your question has some missing parts

A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.

answer : c) 0.39 sec

d) 2.25 m

e) 1.92 m/sec

Explanation:

The initial velocity of the first ball = 7.67 m/sec ( calculated )

Time required for first ball to reach ceiling = 0.78 secs ( calculated )

Determine how long after the second ball is thrown do the two balls pass each other

Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 = 9.8t^2 / 2

hence d = 4.9t^2 ----- ( 1 )

Initial speed of second ball = first ball initial speed = 7.67 m/sec

3 - d = 7.67t - 4.9t ---- ( 2 )

equating equation 1 and 2

3 = 7.67t therefore t = 0.39 sec

Determine how far the balls are above the Juggler's hands ( when the balls pass each other )

form equation 1 ;

d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m

therefore the height the balls are above the Juggler's hands is

3 - d = 3 - 0.75 = 2.25 m

determine their velocities when the pass each other

velocity = displacement / time

velocity = d / t = 0.75 / 0.39 sec = 1.92 m/sec

7 0

3 years ago

(b) The distance of mass from mass A if there is no gravitational force acted on C

shepuryov [24]

Answer:

(a) The force, acting on object 'C' is approximately 2.66972 × 10⁻¹⁰ Newtons

(b) The distance of 'C' from 'A', in the direction particle 'B' if there is no meters gravitational force acting on 'C' is appromimately 0.829 meters or 1.877 meters

Explanation:

The given parameters are;

The mass of particle, A, m₁ = 2 kg

The mass of particle, B, m₂ = 0.3 kg

The mass of particle, C, m₃ = 0.05 kg

The distance between particle 'A' and particle 'B', r₁ = 0.15 m

The distance between particle 'B' and particle 'C', r₂ = 0.05 m

(a) The gravitational force, 'F', is given as follows;

$F =G \times \dfrac{m_{1} \times m_{2}}{r^{2}}$

Where;

F = The force between the two masses

G = The gravitation constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m₁ = The mass of object 1

m₂ = The mass of object 2

If 'C' is placed at 0.05 m from 'B', we have;

F₂₃ = 6.67430 × 10⁻¹¹ × 0.05 × 0.3/(0.05²) ≈ 4.00458 × 10⁻¹⁰

The gravitational force between force between particle 'B' and particle 'C', F₂₃ = 4.00458 × 10⁻¹⁰ N (towards the right)

F₁₃ = 6.67430 × 10⁻¹¹ × 0.05 × 2/(0.1²) ≈ × 10⁻¹⁰

The gravitational force between force between particle 'A' and particle 'B', F₁₃ = 6.6743 × 10⁻¹⁰ N (towards the left)

The force, 'F', acting on object 'C' = F₁₃ - F₂₃

F = (6.6743 - 4.00458) × 10⁻¹⁰ = 2.66972 × 10⁻¹⁰ N

The force, acting on object 'C' ≈ 2.66972 × 10⁻¹⁰ N

(b), When there is no gravitational force acting on 'C', let the distance of 'C' from 'A' = x

We have;

F₂₃ = F₁₂

$F_{23} =G \times \dfrac{m_{1} \times m_{2}}{r_1^{2}} = F_{13} =G \times \dfrac{m_{1} \times m_{3}}{r_2^{2}}$

By plugging in the values and removing like terms, we get;

$\dfrac{0.3 \times 0.05}{(1.15 - x)^{2}} = \dfrac{2 \times 0.05}{x^2}$

(1.15 - x)² × 2 × 0.05 = 0.3 × 0.05 × x²

0.1·x² - 0.23·x + 1.3225 = 0.015·x²

0.1·x² - 0.23·x + 1.3225 - 0.015·x² = 0

0.085·x² - 0.23·x + 0.13225= 0

x = (0.23± √((-0.23)² - 4 × 0.085 × ( 0.13225)))/(2 × 0.085))

x ≈ 0.829, or x ≈ 1.877

Therefore, the distance of 'C' from 'A', if there is no gravitational force acting on 'C', x ≈ 0.829 m, or x = 1.877 m, in the direction of 'B'

7 0

2 years ago

Vision is blurred if the head is vibrated at 29 Hz because the vibrations are resonant with the natural frequency of the eyeball

Likurg_2 [28]

Answer:

k = 4422.35 KN/m

Explanation:

Given that

Frequency ,f= 29 Hz

m = 7.5 g

Natural frequency ω

ω = 2 π f

We also know that for spring mass system

ω ² m =k

k=Spring constant

So we can say that

( 2 π f)² = m k

By putting the values

(2 x π x 29)² = 7.5 x 10⁻³ k

33167.69 = 7.5 x 10⁻³ k

k=4422.35 x 10³ N/m

k = 4422.35 KN/m

Therefore spring constant will be 4422.35 KN/m

4 0

3 years ago

One difference between a solar flare and a CME is that a solar flare is composed of ___________, while a CME is composed of ____

JulsSmile [24]

Answer:

magnetic energy (proton) and magnetic plasma.

Explanation:

The solar fare consists of bright light that occurs in various wavelengths and is observed at the surface.

They are not as strong as compared to the coronal mass ejection or CME. The solar fares consist of 10²² joules, while the plasma is ejected from the solar corona and can be clearly seen from a distance.

The Solar flares represent an atmospheric disturbance and plasms are the medium for the growth and development of solar flare and lead to solar activity.

6 0

3 years ago