Ask question

Ask question

All categories

Vladimir79 [104]

3 years ago

11

A vertical spring with stiffness k originally is at rest with no mass attached. Then, a mass M is attached, and the spring rocks

back and forth until it is at rest at a new equilibrium position. What is this position in terms of g, M, and k?

1 answer:

Alina [70]3 years ago

8 0

Answer:

The position of the spring in terms of g, m & k is $x = \frac{m g}{k}$

Explanation:

Stiffness of the spring = k

Mass = m

When a mass m is attached with the spring then spring stretched. in that case the force exerted on the spring is equal to weight of the mass attached.

⇒ Force exerted on the spring F = k x

⇒ m g = k x

⇒ $x = \frac{m g}{k}$

This is the position of the spring in terms of g, m & k.

You might be interested in

After the driver first notices the obstacle, the car moves uniformly for a time interval t1−t0=t before the brakes are applied.

loris [4]

Answer:

V(t1-t0)

Explanation:

Moving 'uniformly' means constant velocity (speed). the formula for constant speed motion is $V = \frac{distance}{time}$ =( change in position/ change in time)

where,

V is speed

given in the statement :

change in time = t = t1-t0

let the constant speed be ' V '

disance = X = X1-X0

applying the above mentioned formula: V = $\frac{X}{t}$

V = X/t

X = Vt

the distance X1-X0 = Vt =V(t1-t0)

3 0

3 years ago

A gang of robbers is escaping across city roofs at night. They come to the edge of one building and need to drop down to their g

REY [17]

Answer:

a) They will hit the ground with a speed of 19.6 m/s.

b) They are at a height of 20 m.

c) It is not a safe jump.

Explanation:

Hi there!

a) The equations of height and velocity in function of time of a free falling body are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering downward as negative direction).

v = velocity of the object at time t.

Using the equation of velocity, let's find the velocity at which they will hit the ground. The pebble is dropped (initial velocity = 0) and it takes 2 s to reach the ground:

v = v0 + g · t (v0 = 0)

v = g · t

v = -9.8 m/s² · 2.0 s

v = -19.6 m/s

They will hit the ground with a speed of 19.6 m/s.

b)Now, we have to use the equation of height:

h = h0 + v0 · t + 1/2 · g · t²

If we place the origin of the frame of reference on the ground, we have to find the initial height (h0) knowing that at t = 2.0 s, h = 0 m

0 m = h0 - 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 20 m

They are at a height of 20 m.

c)According to a NASA paper (Issues on Human Acceleration Tolerance After Long-Duration Space Flights, figure 10), if you fall with a vertical velocity greater than 17 m/s it is unlikely that you will survive. So, it is not a safe jump.

3 0

3 years ago

What and where is the asteroid belt?<br><br> Please ANSWER THIS

Rudiy27

Answer:

The asteroid belt is a region of our solar system, between the orbits of Mars and Jupiter, in which many small bodies orbit our sun.

Explanation:

Hope this helps!

3 0

3 years ago

Read 2 more answers

A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it

KatRina [158]

Answer:

$t = 1.82$

Explanation:

Given

$u = 7.70m/s$ -- initial velocity

$s = 30.2m$ --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

$s = ut + \frac{1}{2}gt^2$

Where

$g = 9,8m/s^2$

So, we have:

$30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2$

$30.2 = 7.70t + 4.9 * t^2$

Subtract 30.2 from both sides

$30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9t^2 - 30.2$

$7.70t + 4.9t^2 - 30.2 = 0$

$4.9t^2 + 7.70t - 30.2 = 0$

Solve using quadratic formula:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = 4.9;\ b = 7.70;\ c = -30.2$

$t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}$

$t = \frac{-7.70\±\sqrt{651.21}}{9.8}$

$t = \frac{-7.70\±25.52}{9.8}$

Split the expression

$t = \frac{-7.70+25.52}{9.8}$ or $t = \frac{-7.70-25.52}{9.8}$

$t = \frac{17.82}{9.8}$ or $t = -\frac{33.22}{9.8}$

Time can't be negative; So, we have:

$t = \frac{17.82}{9.8}$

$t = 1.82$

Hence, the time to hit the ground is 1.82 seconds

7 0

2 years ago

Describe how the coriolis effect acts differently based on location on earth

sammy [17]

Explanation:

The Coriolis effect happens when an entity is perceived from a moving reference frame going in a straight path. The changing reference frame makes the object appear as if it were moving along a curved road.

Circulation is counter-clockwise in the northern hemisphere. Circulation is clockwise in the southern hemisphere, and it is the equator, it is straight down without circulation.

6 0

3 years ago