4.

When a force of 20.0 N is applied to a spring, it elongates 0.20 m. Determine the period of oscillation of a 4.0-kg object suspe

mash [69]

Answer:

1.26 secs.

Explanation:

The following data were obtained from the question:

Force (F) = 20 N

Extention (e) = 0.2 m

Mass (m) = 4 Kg

Period (T) =.?

Next, we shall determine the spring constant, K for spring.

The spring constant, K can be obtained as follow:

Force (F) = 20 N

Extention (e) = 0.2 m

Spring constant (K) =..?

F = Ke

20 = K x 0.2

Divide both side by 0.2

K = 20/0.2

K = 100 N/m

Finally, we shall determine the period of oscillation of the 4 kg object suspended on the spring. This can be achieved as follow:

Mass (m) = 4 Kg

Spring constant (K) = 100 N/m

Period (T) =..?

T = 2π√(m/K)

T = 2π√(4/100)

T = 2π x √(0.04)

T = 2π x 0.2

T = 1.26 secs.

Therefore, the period of oscillation of the 4 kg object suspended on the spring is 1.26 secs.

6 0

3 years ago

A ball is traveling uphill with an initial velocity of 5.0 m/s and an acceleration of -2.0 m/s^2. A) How fast is the ball travel

Rzqust [24]

Answer:

A) The ball is traveling at 5.0 m/s (magnitude) when the ball returns to its release point.

B) The maximum uphill position is at 6.25 m from the release point.

C) On the way up, the velocity of the ball at x = 6.0 m is 1 m/s and on the way down it is - 1m/s.

Explanation:

Hi there!

The position and velocity of the ball can be calculated using the following equations:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of th ball at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

v = velocity at time t.

A) Let´s place the origin of the frame of reference at the point at which the ball has a velocity of 5.0 m/s. Then, x0 = 0.

When the ball returns to the initial point, its position will be 0. Then using the equation of position we can calculate at which time the ball is at x = 0:

x = x0 + v0 · t + 1/2 · a · t²

0 m = 5.0 m/s · t - 1/2 · 2.0 m/s² · t²

0 m = 5.0 m/s · t - 1.0 m/s² · t²

0 m = t (5.0 m/s - 1.0 m/s² · t)

t = 0 (this is logic becuase the ball starts at x = 0)

and

5.0 m/s - 1.0 m/s² · t = 0

t = -5.0 m/s / -1.0 m/s²

t = 5.0 s

With this time, we can calculate the velocity of the ball:

v = v0 + a · t

v = 5.0 m/s - 2.0 m/s² · 5.0 s

v = -5.0 m/s

The ball is traveling at 5.0 m/s when the ball returns to its release point.

B) Let´s use the equation of velocity to obtain the time at which the ball is at its maximum uphill position:

v = v0 + a · t

0 = 5.0 m/s - 2.0 m/s² · t

-5.0 m/s/ -2.0 m/s² = t

t = 2.5 s

Now, using the equation of position, let´s find the position of the ball at t = 2.5 s. This position will be the maximum uphill position because at that time the velocity is 0:

x = x0 + v0 · t + 1/2 · a · t²

x = 5.0 m/s · 2.5 s - 1/2 · 2.0 m/s² · (2.5 s)²

x = 6.25 m

The maximum uphill position is at 6.25 m from the release point.

C) First, let´s find the time at which the ball is 6.0 meters uphill from the releasing point:

x = x0 + v0 · t + 1/2 · a · t²

6.0 m = 5.0 m/s · t - 1/2 · 2 m/s² · t²

0 = -1 m/s² · t² + 5.0 m/s · t - 6.0 m

Solving the quadratic equation using the quadratic formula:

a = -1

b = 5

c = -6

t = [-b ± √(b² - 4ac)]/2a

t₁ = 2 s (on its way up)

t₂ = 3 s (on its way down)

Now, let´s calculate the velocity of the ball at those times:

v = v0 + a · t

v = 5.0 m/s - 2 m/s² · 2 s = 1 m/s

v = 5.0 m/s - 2 m/s² · 3 s = -1 m/s

On the way up, the velocity of the ball at x = 6.0 m is 1 m/s and on the way down it is - 1m/s.

7 0

3 years ago

A ball rolls 50 meters in

-Dominant- [34]

The speed of the ball is B. 5 meters per second

Explanation:

For an object in uniform motion (= at constant velocity), the speed can be calculated using the equation

$v=\frac{d}{t}$

where

v is the speed

d is the distance covered

t is the time taken

For the ball in this problem:

d = 50 m is the distance covered

t = 10 s is the time taken

Substituting, we find the speed:

$v=\frac{50}{10}=5 m/s$

Learn more about speed:

brainly.com/question/8893949

#LearnwithBrainly

3 0

3 years ago

How to calculate sound of an echo

bonufazy [111]

by an echo meter

please flw me and thank my answers

#Genius kudi

5 0

3 years ago

Gravity is a force that every mass exerts on every other mass. When you jump up in the air, not only does the Earth exert a grav

sesenic [268]

Answer:

C. Your mass is very small compared to Earth's mass.

Explanation:

Newton's third law of motion states that:

"When an object A exerts a force (action force) on an object B, object B exerts an equal and opposite force (reaction force) on object A".

If we apply this law to the situation described in the problem, we see that:

- The action force is the gravitational force exerted by the Earth on you

- The reaction force is the gravitational force exerted by you on the Earth

And according to the 3rd law, the magnitude of the two forces is equal:

$F_1 = F_2$ (1)

We also know that, according to Newton's second law of motion, the force on an object is equal to the product between its mass (m) and its acceleration (a):

$F=ma$

So we can rewrite (1) as

$ma = MA$

where

m is the your mass

a is your acceleration

M is the Earth's mass

A is the Earth's acceleration

For the term on the left, we see that m is small, so a is larger (therefore, your acceleration is visible). However, for the term on the right, we see that the mass of the Earth is very large (M is very large), therefore, A is very small, which means that the acceleration of the Earth is almost negligible because the Earth's mass is very large.

4 0

4 years ago