All categories

2 years ago

An object is in free fall. After 10 seconds calculate the distance fallen? *

Physics

2 answers:

ch4aika [34]2 years ago

5 0

*For free fall,

Initial velocity (u) = 0

g = 9.8m/s²

time(t) = 10s

Using the second equation of free fall;

s = ut + 1/2gt²

where 's' is the distance fallen.

s = 0(10) + 1/2(9.8)(10)²

s = 1/2(9.8)(100)

s = 490m.

*Initial horizontal velocity (Ux) = 3m/s

Initial vertical velocity (Uy) = 0

since it did not start with any vertical velocity.

Using the second equation of vertical downward motion:

h = Uyt + 1/2gt²

where h is the height of the cliff = 60m.

g = 9.8m/s²

60 = 0(t) = 1/2 (9.8)(t²)

60 = 4.9t²

t = √(60/4.9) = 3.5s.

Citrus2011 [14]2 years ago

3 0

Answer:

9.81 × 10 = 98.1 meters

vertical displacement is s=1/2 at^2 + vt

initial vertical velocity is 0 so s=1/2 at^2

a in this instance is gravitational acceleration so 60m= 1/2 (9.81)t^2

solve for t, t = 3.497s. //I corrected this answer as just now I misread horizontal as vertical.

You might be interested in

_____________ circular motion occurs when an object is traveling with constant speed in a circle.

emmasim [6.3K]

Your answer is ''Uniform''.

Hope this helps :)

7 0

2 years ago

What is specific heat capacity

finlep [7]

It is the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).

3 0

2 years ago

Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m

Natali [406]

<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

7 0

2 years ago

Why did boy (the observer) hear the pitch of the sound change? What does this tell us about the frequency of the sound? Support

timofeeve [1]

Answer:

<u>because of the doppler effect</u>

Explanation:

<em>Remember</em>, the doppler effect refers to the changes in sound (frequency of sound) observed by a person who is in a position relative to the wave source.

In this example, we notice as the train comes closer to the boy, the sound becomes louder also increasing the pitch slightly, the doppler effect sets in when the train passes the boy because the boy notices a decrease in the pitch of the moving train.

We learn from the change in the observed sound of the train that the frequency of the sound is determined by the distance of the observer from the wave source.

In other words, the closer the source of the sound to the observer; the faster it travels to the observer, however, the farther it is; the lesser it is; the greater the sound heard.

5 0

2 years ago

Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan

Eddi Din [679]

To increase the centripetal acceleration to $2.00 m/s^2$ , you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

$a=0.50 m/s^2$

We want to increase it by a factor of 4, i.e. to

$a'=2.00 m/s^2$

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

This way, the new acceleration is

$a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a$