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4 years ago

If the work done to stretch an ideal spring by 4.0 cm is 6.0 j, what is the spring constant (force constant) of this spring?

Physics

1 answer:

Alisiya [41]4 years ago

4 0

Answer:

The spring constant is 3750 N/m

Explanation:

Use the following two relationships:

(Work) = (Force) x (Displacement)

(Force) = (Spring constant) x (Displacement)

(Spring constant) = (Force) / (Displacement) = (Work) / (Displacement)^2

(Spring constant) = 6.0 kg.(m^2/s^2) / 0.0016 m^2 = 3750 N/m

The spring constant is 3750 N/m

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Why do boys like boys

Llana [10]

Answer:

maybe they're gay maybe they act gay so they can be around there girl crush cause a lot of girls like to hang out around gay guys

Explanation:

4 0

3 years ago

A commuter airplane starts from an airport and takes theroute. The plane first flies to city A, located 175 km away in a directi

bearhunter [10]

Answer:

245.45km in a direction 21.45° west of north from city A

Explanation:

Let's place the origin of a coordinate system at city A.

The final position of the airplane is given by:

rf = ra + rb + rc where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:

rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km

rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km

The module of this position is:

$rf = \sqrt{rfX^2+rfY^2} = 245.45km$

And the angle measure from the y-axis is:

$\alpha =atan(rfX/rfY) = 21.45\°$

So the answer is 245.45km in a direction 21.45° west of north from city A

6 0

4 years ago

What is a load force

vladimir2022 [97]

Answer:

A push or pull exerted on an object

8 0

4 years ago

A 4 kg bowling ball moving at 1.4 m/s east impacts a 400 g pin that is stationary. After the impact, the ball is moving at 0.5 m

nignag [31]

The speed of the pin after the elastic collision is 9 m/s east.

<h3>Final speed of the pin</h3>

The final speed of the pin is calculated by applying the principle of conservation of linear momentum as follows;

m1u1 + mu2 = m1v1 + m2v2

where;

m is the mass of the objects
u is the initial speed of the objects
v is the final speed of the objects

4(1.4) + 0.4(0) = 4(0.5) + 0.4v2

5.6 = 2 + 0.4v2

5.6 - 2 = 0.4v2

3.6 = 0.4v2

v2 = 3.6/0.4

v2 = 9 m/s

Thus, The speed of the pin after the elastic collision is 9 m/s east.

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

3 0

2 years ago

A solenoid with 385 turns per meter and a diameter of 17.0 cm has a magnetic flux through its core of magnitude 1.28 × 10-4 (T·m

shepuryov [24]

Answer:

(a)The current passes through the solenoid is 11.7 A.

(b) The new current will be one-fourth of the initial current.

Explanation:

Given that,

The number of turns per meter = 385

Diameter of solenoid = 17.0 cm = $17\times 10^{-2}$ m

Magnetic flux through core of solenoid $\phi$ = $1.28\times 10^{-4}$ Tm²

(a)

Magnetic field B= $\mu_0nI$

$\mu_0= 4\pi \times 10^{-7}$ T/amp m

Cross section area of the solenoid A= $\pi \frac{d^2}{4}$

$=\pi\frac{ (17\times 10^{-2})^2}{4}$ m²

The angle between magnetic field and cross section of the solenoid is $\theta =0^\circ$

The magnetic flux through a area A with magnetic fie;d B is

$\phi = BA cos\theta$

$\Rightarrow \phi =( \mu_0nI)(\pi \frac{d^2}4)cos \theta$

$\Rightarrow I =\frac{\phi}{(\mu_0\pi n \frac{d^2}4cos\theta)}$

$=\frac{4\phi}{(\mu_0n)(\pi d^2)cos\theta}$

$=\frac{1.28\times 10^{-4}\times 4}{(4\pi \times10^{-7}\times385 )\times(\pi\times17\times 10^{-2})^2cos 0^\circ}$

=11.7 A

The current of the solenoid is 11.7 A.

(b)

$I =\frac{4\phi}{(\mu_0\pi n d^2cos\theta)}$

From the above equation it is clear that, the current is inversely proportional to the square of the diameter of a solenoid.

$I\propto \frac1{d^2}$ .

Consider d' be the new diameter of the solenoid .

Since the new diameter of the solenoid is double of the initial diameter.

That is d'= 2d.

$\frac{I}{I'}= \frac{(d')^2}{d^2}$

$\Rightarrow \frac{I}{I'}=\frac{(2d)^2}{d^2}$

$\Rightarrow \frac{I}{I'}=4$

⇒I=4I'

$\Rightarrow I'=\frac{I}{4}$

The new current will be one-fourth of the initial current.

7 0

3 years ago