Elements in the periodic table are arranged in order of increasing...

A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the

ser-zykov [4K]

Answer:

The maximum length during the motion is $L_{max} = 1.45m$

Explanation:

From the question we are told that

The mass is $m =0.5 kg$

The vertical spring length is $L = 1.10m$

The unstretched length is $L_{un} = 1.30m$

The initial speed is $v_i = 1.3m/s$

The new length of the spring $L_{new} = 1.30 m$

The spring constant k is mathematically represented as

$k = -\frac{F}{y}$

Where F is the force applied $= m * g = 0.5 * 9.8=4.9N$

y is the difference in weight which is $=1.10-0.50=0.6m$

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

Now substituting values accordingly

$k = \frac{4.9}{0.6}$

$= 8.17 N/m$

The elastic potential energy is given as $E_{PE} = \frac{1}{2} k D^2$

where D is this the is the displacement

Since Energy is conserved the total elastic potential energy would be

$E_T = initial \ elastic\ potential \ energy + kinetic \ energy$

$E_T = \frac{1}{2} k D_{max}^2 = \frac{1}{2} k D^2 + \frac{1}{2} mv^2$

Substituting value accordingly

$\frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2$

$4.085 * D_{max}^2 = 3.69$

$D^2_{max} = 0.9033$

$D_{max} = 0.950m$

So to obtain total length we would add the unstretched length

So we have

$L_{max} = 0.950 + 0.5 = 1.45m$

5 0

3 years ago

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Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram

Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

Given the following data:

Force 1 = 500 Newton

Distance 1 = 1.2 meter

Force 2 = 500 Newton

Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

$Work\;done = Force \times distance$

For student 1:

$Work\;done = 500 \times 1.2$

Work done = 600 Joules

For student 2:

$Work\;done = 500 \times 5$

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

7 0

3 years ago

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By counting the number of waves that pass a certain point each second, we will know the _______ of the wave.

Travka [436]

The answer is A. Frequency

6 0

3 years ago

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Check all that apply. The magnetic force on the current-carrying wire is strongest when the current is parallel to the magnetic

dedylja [7]

Answer:

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.

The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.

Explanation:

The magnitude of the magnetic force exerted on a current-carrying wire due to a magnetic field is given by

$F=ILB sin \theta$ (1)

where I is the current, L the length of the wire, B the strength of the magnetic field, $\theta$ the angle between the direction of the field and the direction of the current.

Also, B, I and F in the formula are all perpendicular to each other. (2)

According to eq.(1), we see that the statement:

"The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines."

is correct, because when the current is perpendicular to the magnetic field, $\theta=90^{\circ}, sin \theta = 1$ and the force is maximum.

Moreover, according to (2), we also see that the statements

"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field. "

"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current. "

because F (the force) is perpendicular to both the magnetic field and the current.

5 0

3 years ago

n the figure, a uniform ladder 12 meters long rests against a vertical frictionless wall. The ladder weighs 400 N and makes an a

Leto [7]

Since the ladder is standing, we know that the coefficient of friction is at least something. This [gotta be at least this] friction coefficient can be calculated. As the man begins to climb the ladder, the friction can even be less than the free-standing friction coefficient. However, as the man climbs the ladder, more and more friction is required. Since he eventually slips, we know that friction is less than what's required at the top of the ladder.

The only "answer" to this problem is putting lower and upper bounds on the coefficient. For the lower one, find how much friction the ladder needs to stand by itself. For the most that friction could be, find what friction is when the man reaches the top of the ladder.

Ff = uN1

Fx = 0 = Ff + N2

Fy = 0 = N1 – 400 – 864

N1 = 1264 N

Torque balance

T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)

N2 = 439 N

Ff = 439= u N1

U = 440 / 1264 = 0.3481

3 0

3 years ago

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