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2 years ago

1.2 m long spring with a spring force of 200.0 N/m, 12.0 kg mass is attached to it, find the length

Physics

2 answers:

lesantik [10]2 years ago

6 0

Answer: 1.232

Explanation:I remember having this as a hw and this was the answer correct me if I was wrong

Crank2 years ago

3 0

From Hooke's law, the length of the spring after extension and after the mass is attached is 1.788 meters. Option B is the answer.

<h3>HOOKE'S LAW</h3>

Hooke's law state that in an elastic material, the force applied is directly proportional to the extension provided that the elastic limit is not exceeded.

Given that a 12.0 kg mass is attached to 1.2 m long spring with a spring constant of 200.0 N/m.

The given parameters are:

mass m = 12kg

Initial length $L_{1}$ = 1.2 m

Spring constant K = 200 N/m

Extension e = ?

According to Hooke's law

F = Ke

But F = mg

mg = Ke

Substitute all the parameters into the formula to get extension e

12 x 9.8 = 200e

e = 117.6 / 200

e = 0.588 m

The length of the spring after extension and after the mass is attached will be

$L_{2}$ = $L_{1}$ + e

$L_{2}$ = 1.2 + 0.588

$L_{2}$ = 1.788 m

Therefore, the correct answer is option B because the length of the spring after extension and after the mass is attached is 1.788 meters.

Learn more about Hooke's law here: brainly.com/question/12253978

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Find the equilibrant of two 10.0-N forces acting upon a body when the angle between the forces is 90° Solve graphically using a

bazaltina [42]

The equilibrant force of the two given forces is 14.14 N.

<h3 /><h3 /><h3>What is equilibrant force?</h3>

This is a single force that balances other given forces.

The given parameters:

First force, F₁ = 10 N
Second force, F₂ = 10 N
Angle between the forces, θ = 90⁰

The equilibrant force of the two given forces is calculated as follows;

$F = \sqrt{F_1 ^2 + F_2 ^2} \\\\F = \sqrt{(10)^2 + (10)^2} \\\\F = 14.14 \ N$

Thus, the equilibrant force of the two given forces is 14.14 N.

Learn more about equilibrant force here: brainly.com/question/8045102

5 0

2 years ago

How does vegetation in the soil affect the rate of erosion

avanturin [10]

Answer:

Plants slow down water as it flows over the land and this allows much of the rain to soak into the ground. Plant roots hold the soil in position and prevent it from being blown or washed away. Plants break the impact of a raindrop before it hits the soil, reducing the soil's ability to erode.

Explanation:

4 0

3 years ago

Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. At the start of a play, Matthews runs down

MA_775_DIABLO [31]

Answer:

a) D_ total = 18.54 m, b) v = 6.55 m / s

Explanation:

In this exercise we must find the displacement of the player.

a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components

sin 45 = y₁ / d

cos 45 = x₁ / d

y₁ = d sin 45

x₁ = d sin 45

y₁ = 8 sin 45 = 5,657 m

x₁ = 8 cos 45 = 5,657 m

The second offset is d₂ = 12m at 90 of the 50 yard

y₂ = 12 m

x₂ = 0

total displacement

y_total = y₁ + y₂

y_total = 5,657 + 12

y_total = 17,657 m

x_total = x₁ + x₂

x_total = 5,657 + 0

x_total = 5,657 m

D_total = 17.657 i^+ 5.657 j^ m

D_total = Ra (17.657 2 + 5.657 2)

D_ total = 18.54 m

b) the average speed is requested, which is the offset carried out in the time used

v = Δx /Δt

the distance traveled using the pythagorean theorem is

r = √ (d1² + d2²)

r = √ (8² + 12²)

r = 14.42 m

The time used for this shredding is

t = t1 + t2

t = 1 + 1.2

t = 2.2 s

let's calculate the average speed

v = 14.42 / 2.2

v = 6.55 m / s

8 0

3 years ago

If a simple truss member is known to carry a tensile force, then the internal force drawn on a free-body at a section cut when u

mote1985 [20]

Answer: b) pointed toward and parallel to the member.

Explanation:

It is shown in the picture attached

6 0

2 years ago

7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee

wlad13 [49]

Answer:

$N_{top}=9.8N\\N_{bottom}=33.4N$

b) $v_{min}=4.4m/s$

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is $a_{cp}=\frac{v^2}{R}$ . At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

$-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg$

Which means:

$N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N$

The limit for falling off would be $N_{top}=0$ , so the minimum speed would be:

$0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s$

3 0

2 years ago