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3 years ago

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A spring is 20cm long is stretched to 25cm by a load of 50N. What will be its length when stretched by 100N. assuming that the e

lastic limit is not reached

1 answer:

IgorLugansk [536]3 years ago

6 0

Answer:

Final Length = 30 cm

Explanation:

The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:

F = kΔx

where,

F = Force applied

k = spring constant

Δx = change in length of spring

First, we find the spring constant of the spring. For this purpose, we have the following data:

F = 50 N

Δx = change in length = 25 cm - 20 cm = 5 cm = 0.05 m

Therefore,

50 N = k(0.05 m)

k = 50 N/0.05 m

k = 1000 N/m

Now, we find the change in its length for F = 100 N:

100 N = (1000 N/m)Δx

Δx = (100 N)/(1000 N/m)

Δx = 0.1 m = 10 cm

but,

Δx = Final Length - Initial Length

10 cm = Final Length - 20 cm

Final Length = 10 cm + 20 cm

<u>Final Length = 30 cm</u>

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A uniform rod of length L is pivoted at L/4 from one end. It is pulled to one side through a very small angle and allowed to osc

ludmilkaskok [199]

Answer:

T= 4.24sec

Explanation:

We are going to use the formula below to calculate.

$T=2\pi \sqrt{\frac{L}{g} }$

Where T is period

L is length of rod

g is acceleration due to gravity = $9.8m/s^{2}$

From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this $L_{O}$

$L_{O} = 3/4 * 5.95m$

= 4.4625m

thus $T=2\pi \sqrt{\frac{L_{O} }{g} }$

$T=2\pi \sqrt{\frac{4.4625 }{9.8} }$

T= 4.24sec

8 0

4 years ago

The first ionization energy of a hydrogen atom is 2.18 aj (attojoules). what is the frequency and wavelength, in nanometers, of

jasenka [17]

1) Frequency: $3.29\cdot 10^{15}Hz$

the energy of the photon absorbed must be equal to the ionization enegy of the atom, which is

$E=2.18 aJ=2.18\cdot 10^{-18} J$

The energy of a photon is given by

$E=hf$

where $h=6.63\cdot 10^{-34}Js$ is the Planck's constant. By using the energy written above and by re-arranging thsi formula, we can calculate the frequency of the photon:

$f=\frac{E}{h}=\frac{2.18\cdot 10^{-18} J}{6.63\cdot 10^{-34} Js}=3.29\cdot 10^{15} Hz$

2) Wavelength: 91.2 nm

The wavelength of the photon can be found from its frequency, by using the following relationship:

$\lambda=\frac{c}{f}$

where $c=3\cdot 10^8 m/s$ is the speed of light and f is the frequency. Substituting the frequency, we find

$\lambda=\frac{3\cdot 10^8 m/s}{3.29\cdot 10^{15}Hz}=9.12\cdot 10^{-8} m=91.2 nm$

5 0

3 years ago

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

4 years ago

Plz do it all i will give brainlest and thanks to best answer.

maxonik [38]

Its a, metal is a good conductor of heat so yea
Hope this helps :)

4 0

3 years ago

Read 2 more answers

A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?

ankoles [38]

Given required solution

M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
=12m
so W=12*100
W=1200J

4 0

3 years ago