All categories

4 years ago

What are the possible units for a spring constant

2 answers:

6 0

[any unit of force] divided by [any unit of length or distance]

can be a unit of spring constant.

Useful spring units can include [newtons per meter] and [pounds per foot].

AleksAgata [21]4 years ago

5 0

Hooke's Law says that F=-kx where k is the spring constant measured in N/m (newtons per meter)

You might be interested in

A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant

djverab [1.8K]

Answer:

0.005 V

Explanation:

We are given that

Initial circumference of circular loop=C=165 cm

Rate of circumference, $\frac{dC}{dt}=12 cm/s$

Magnetic field,B=0.5 T

We have to find the induced emf at time t=9 s

We know that induced amf,E= $\frac{Bd(A)}{dt}$

Area of circular coil,A= $\pi r^2$

$E=B\frac{d(\pi r^2)}{dt}=B(2\pi r)\frac{dr}{dt}$

Circumference of circular coil,C= $2\pi r$

$165=2\pi r$

$r=\frac{165}{2\pi}$

$\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}\times (12)=\frac{6}{\pi} cm/s=\frac{6\times 10^{-2}}{\pi} m/s$

Radius of coil at time t=9 s

$r=\frac{165}{2\pi}-(\frac{6}{\pi}\times 9)=9.08 cm=9.08\times 10^{-2} m$

$1 m=100 cm$

E= $-0.5(2\pi\times 9.08\times 10^{-2})\times \frac{6\times 10^{-2}}{\pi}=-0.005 V$

Magnitude of induced emf=0.005 V

4 0

3 years ago

Read 2 more answers

Consider two diffraction gratings with the same slit separation. The only difference between the two gratings is that one gratin

kobusy [5.1K]

Answer:

True The grid with more slits gives more angle separation increases

True. The grating with 10 slits produces better-defined (narrower) peaks

Explanation:

Such a system can be seen as a diffraction network in this case with different number of lines per unit length, the expression for the constructive interference of a diffraction network is

d sin θ = m λ

where d is the distance between slits or lines, m the order of diffraction and λ the wavelength.

For network with 5 slits

d = 1/5 = 0.2

For the network with 10 slits

d = 1/10 = 0.1

let's calculate the separation (teat) for each one

θ = sin⁻¹ (m λ / d)

for 5 slits

θ₅ = sin⁻¹ (m λ 5)

for 10 slits

θ₁₀ = sin⁻¹ (m λ 10)

we can appreciate that for more slits the angle increases

the intensity of a series of slits is

I = I₀ sin²2 (N d/2) / sin² d/2)

when there are more slits (N) the peaks have greater intensity and are more acute (half width decreases)

let's analyze the claims

False

True The grid with more slits gives more angle separation increases

False

True The expression for the intensity of the diffraction peaks the intensity of the peaks increases with the number of slits as well as their spectral width decreases

False

5 0

4 years ago

DESPERATE WILL GIVE BRAINLIST AND THANKS

Bas_tet [7]

Answer:

true

Explanation:

8 0

3 years ago

What is the escape speed for an electron initially at rest on the surface of a sphere with a radius of 1.3 cm and a uniformly di

Vlad [161]

Answer:

2.37 * 10^4 m/s

Explanation:

Constants :

Mass of electron = 9.11 * 10^(-31) kg

Electric charge of an electron = 1.602 * 10^(-19) C

Parameters given:

Radius of sphere = 1.3cm = 0.013m

Charge of sphere = 2.3 * 10^(−15) C

Using the law of conservation of energy, we have that:

K. E.(initial) + P. E.(initial) = K. E.(final) + P. E.(final)

K. E.(final) = 0, since final velocity is zero and P. E.(final) = 0 since the electron reaches a final distance of infinity.

Hence,

K. E.(initial) = P. E.(initial)

0.5mv^2 = (kqQ)/r

Where k = Coulumbs constant

Q = charge of the sphere.

r = radius of the sphere.

=> 0.5*m*v^2 = (kqQ)/r

0.5 * 9.11 * 10^(-31) * v^2 = (9 * 10^9 * 1.602 * 10^(-19) * 2.3 * 10^(-15))/0.013

4.555 * 10^(-31) * v^2 = 2550.88 * 10^(-25)

=> v^2 = 2550.88 * 10^(-25) / 4.555 * 10^(-31)

v^2 = 560 * 10^6 = 5.60 * 10^8

=> v = 2.37 * 10^4 m/s

4 0

3 years ago

1 A steel ball of mass 5 kg is released from rest at a height of 3 m above the ground. It rolls down the frictionless section AB

KonstantinChe [14]

The speed of the ball when it reached the point B is 7.7 m/s. The angle of inclination of section BC is 11.54⁰.

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

The force between the two mating surfaces and having the relative motion is called the frictional force.

The speed of ball at point B is equal to the relation

v = √(2gh)

Substitute the values of height h = 3m and acceleration due to gravity g =9.8 m/s², we get the speed at B

v =√(2 x 9.8 x 3)

v =7.7 m/s

Thus the speed of ball when it reaches the point B is 7.7 m/s.

Given is the mass of steel ball is 5 kg and the frictional force is 10N.

The frictional force is equal to the product of coefficient of friction and the normal force on the steel ball \.

f = μN

f =μmg

Substitute the values into the equation, we get the coefficient of friction.

10 = μ x 5 x 9.8

μ = 0.2041

Angle of inclination for the section BC is

θ = tan⁻¹(0.2041) = 11.54⁰

Thus, the angle of inclination of section BC is 11.54⁰.

Learn more about frictional force.

brainly.com/question/14662717

#SPJ1

7 0

2 years ago