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1 year ago

In Hooke's law, Fspring=kΔx , what does the Fspring stand for?

Physics

2 answers:

Mrrafil [7]1 year ago

6 0

Answer:

The amount of force acting on the spring.

Explanation:

Here, Fspring= kΔx is a equation denoting the amount of force acting on the spring.

where, k = spring constant.

Δx= change in the length of the spring.

so, for every Δx change in spring length, kΔx force acts on the spring.

Agata [3.3K]1 year ago

6 0

Answer:

The amount of force acting on the spring.

Explanation:

Here,

Fspring= kΔx is a equation denoting the amount of force acting on the spring.

where, k = spring constant.

Δx= change in the length of the spring.

so, for every Δx change in spring length, kΔx force acts on the spring.

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The top floor of the Ostankino TV Tower in Moscow is located at a height of 360.4 m. Assume a stone is dropped from this top flo

Ulleksa [173]

Answer:

$x=360.4-4.905t^2$

8.57181 s

84.0894561 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Let distance from ground be x

From equation of motion we have

$s=ut+\frac{1}{2}at^2$

Here, distance covered while the stone is falling will be $360.4-x$

$360.4-x=ut+\frac{1}{2}at^2\\\Rightarrow 360.4-x=\frac{1}{2}9.81t^2\\\Rightarrow 360.4-x=4.905t^2\\\Rightarrow x=360.4-4.905t^2$

The equation is $x=360.4-4.905t^2$

At the ground x = 0

$0=360.4-4.905t^2\\\Rightarrow t=\sqrt{\dfrac{-360.4}{-4.905}}\\\Rightarrow t=8.57181\ s$

The time taken by the stone to fall to the ground is 8.57181 s

$v=u+at\\\Rightarrow v=0+9.81\times 8.57181\\\Rightarrow v=84.0894561\ m/s$

The velocity of the stone when it reaches the ground is 84.0894561 m/s

8 0

2 years ago

An application of this principle is that a line mounted on transparent slide casts the same diffraction pattern as a dark film w

kifflom [539]

Explanation:

It is given that,

The distance between the first spot and the central minimum is, s = 0.007 cm

Length, l = 12 m

Wavelength, $\lambda=6\times 10^{-7}\ m$

We need to find the width of the hair. Using the condition of diffraction pattern as :

$s=\dfrac{m\lambda l}{d}$ , d is the width of the hair

$d=\dfrac{m\lambda l}{s}$

$d=\dfrac{1\times 6\times 10^{-7}\times 12}{0.007}$

d = 0.00102

$d=1.02\times 10^{-3}\ m$

So, the width of the hair is $1.02\times 10^{-3}\ m$ . Hence, this is the required solution.

8 0

2 years ago

A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial

Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

$t=\frac{2usin\theta}{g}$

Here θ = 30° and t = 20 s

a) $t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s$

Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

$H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m$

Maximum altitude = 490.5 m

c) Range of projectile is given by

$R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m$

Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

We have equation of motion s = ut + 0.5 at²

Horizontal motion

u = 169.91 m/s

a = 0 m/s²

t = 15 s

Substituting

s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

Vertical motion

u = 98.1 m/s

a = -9.81 m/s²

t = 15 s

Substituting

s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

$\texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m$

Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

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2 years ago

Janice is unsure about her future career path. She has grown up on her family farm, but she is also interested in medicine. Jani

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Answer:

d not joining FRA and joining HOSA INSTEAD

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2 years ago

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Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol

klasskru [66]

The current flowing in silicon bar is 2.02 $\times$ 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

J = (1.6 $\times$ 10^-19) / 0.001 (104 $\times$ 1200 + 1016 $\times$ 500)

J = 1012480 $\times$ 10^-16 A / m^2.

J = 1.01 $\times$ 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 $\times$ 10^-9 $\times$ 0.002 = 2.02 $\times$ 10^-12

So, the current flowing in silicon bar is 2.02 $\times$ 10^-12 A.

6 0

2 years ago