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4 years ago

What is hooke's law? does it apply to elastic materials or to inelastic materials?

1 answer:

6 0

When you talk about Hooke's law, it always have to do something with springs. Hooke's Law, from Robert Hooke, saw a relation between the force applied to the spring and the extension of its length. The equation is: F = kx, where k is the spring constant and x is the displacement of the original and stretched lengths. In other words, x is the length of deformation. Hence, the object must be elastic to come up with a displacement or deformation, in the first place. Then, the Hooke's Law is only applicable to elastic materials.

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Determine the force Santa needs to accelerate a 25g model of the helpful guy in the blue polo shirt to 200,000.00 m/s2 assume th

liubo4ka [24]

Answer:

The force needed will be 5000[N]

Explanation:

Using the Newton's second law we have:

$F = m*a\\where\\F = force [N]\\m = mass [kg]\\a= acceleration [m/s^2]\\Replacing we have\\F = 0.025[kg]*200000[m/s^2]\\F = 5000 [N]$

4 0

3 years ago

A solenoidal coil with 21 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 22.0 cm long

Thepotemich [5.8K]

Answer:

(a)The average of magnetic flux trough each turns is $3.63 \times 10^{-8}$ wb.

(b) The magnitude of mutual inductance of the two solenoid is $1.596 \times 10^{-5}$ H.

(c)The magnitude of emf induced in the outer solenoid by changing current in the inner solenoid is 0.027132 v .

Explanation:

Given that,

The number of turns of the solenoid=N₁= 350

The diameter of the solenoid=D= 2.20 cm=0.022

The length of the solenoid is = $l$ = 22.0 cm=0.22 m

The second solenoid with $N_2$ = 21 turns which is wound around the solenoid at its center.

The current in the inner solenoid is $I_1$ = 0.150 A and is increasing at a rate of 1700 A/s i.e $\frac{dI_1}{dt}= 1700$ A/s

(a)

The formula of magnetic field of a solenoid is

$B=\mu_0 nI=\frac{\mu_0 N_1I_1}{l}$

$=\frac{(4\pi \times 10^{-7} T.m/A) (350)(0.15 \ A)}{0.220 \ m}$

$\approx 3.0 \times 10^{-4}$ T

So, the magnetic flux of each turns

$\phi = BA=B \pi (\frac d2)^2$

$=(3.0\times 10^{-4}\ T)\pi (\frac {0.022}{2} \ m)^2$

$=1.14 \times 10^{-7}$ wb

The average of magnetic flux trough each turns is $3.63 \times 10^{-8}$ wb.

(b)

Since the both coil wound tightly. So, the magnitude magnetic flux though each turns of both coils is same.

The mutual inductance of the two solenoid is

$M=\frac{N_2\phi}{I_1}$

$=\frac{(21)(1.14 \times 10^{-7}\ wb)}{0.15\ A}$

$=1.596 \times 10^{-5}$ H

(c)

The formula of emf is

$\varepsilon =-M\frac{dI_1}{dt}$

$=-(1.596\times 10^{-5}\ H) (1700 \ A/s)$

$=27.132 \times 10^{-3}$ v

=0.027132 v

The emf induced in the outer solenoid by changing current in the inner solenoid is 0.027132 v .

4 0

3 years ago

Which scientist continued the work of another while studying the elliptical path of planets? Name both scientists. Kepler advanc

Korolek [52]

This question is so vague as to render it useless.

I suspect the answer they're looking for is Kepler advanced Copernicus's work as is stated. but the way science works is that anyone who has contributed will have their work built on. for example, Newton advanced Kepler's work on the elliptical path of planets as Einstein advanced Newton's work.

3 0

4 years ago

Read 2 more answers

A beta particle is identical to _____ the nucleus of a helium atom an electron a proton a neutron

svp [43]

A beta particle is identical to : an electron

Both of beta particle and electron are high in energy and move in high speed.
hope this helps

5 0

4 years ago

Read 2 more answers

Convert the number from scientific into standard notation: 5.9 x 10-2

guapka [62]

Move the decimal point to:
Left : (if the exponent of ten is a negative number -) ... OUR CASE HERE (-2)
or to
Right : (if the exponent is positive +).

You should move the point as many times as the exponent indicates.
Do not write the power of ten anymore.

So, standard form is:
Two points to the left {Exponent of Ten is Negative (-2)}
0.059 ... (without the 10)

6 0

3 years ago