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1 year ago

what is the force constant, in newtons per meter, needed to produce a period of 0.45 s for a 0.011-kg mass on the spring?

Physics

1 answer:

topjm [15]1 year ago

8 0

The force constant is 2.145 N/m.

<h3>What is spring constant?</h3>

The spring constant is the force required to stretch or compress a spring divided by the distance traveled by the spring. It is used to determine whether a spring is stable or unstable.

K is the proportionality constant, also known as the 'spring constant.' Hooke's law (F = -kx) specifies stiffness and strength via the k variable. The greater the value of k, the greater the force required to stretch an object to a given length.

Using the relation;

T = 2π√m/k

T = time period = 0.45 s

m = mass of object in kilograms = 0.011kg

k = spring constant

To find k based on the formula,

k = 4 × (3.142)^2 × 0.011 / (0.45 )^2

k = 2.145 N/m

Therefore the force constant is 2.145 N/m.

To learn more about force refer to :

brainly.com/question/12785175

#SPJ4

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ycow [4]

A rotational movement.

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3 years ago

Para que exista una fuerza se necesita cuando menos…

kozerog [31]

Answer:

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8 0

3 years ago

The magnetic field perpendicular to a single 16.7-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to z

sammy [17]

Answer:

1.24 C

Explanation:

We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.

The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m

So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.

Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.

So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²

So, 4iρD/d² = 0.750πD²/4Δt.

iΔt = 0.750πD²/4 ÷ 4iρD/d²

iΔt = 0.750πD²d²/16ρ.

So the charge Q = iΔt

= 0.750π(Dd)²/16ρ

= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)

= 123.76 × 10⁻² C

= 1.2376 C

≅ 1.24 C

4 0

3 years ago

A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten

babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

$P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2$

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0

3 years ago

We dont got the force so i got no clue what to do

konstantin123 [22]

Answer:

work = 1275.3 J

Explanation:

work = (force)(distance)cosø ------- force = ma

=(mass*acceleration)(distance)cosø

=(20*9.81)(6.5)cos0

=1275.3J

nite that the angle of cosine is the difference between the angle of force and the distance. in this case, the force and the distance are in the same direction. :)

3 0

3 years ago