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3 years ago

An ideal spring with spring constant k is hung from the ceiling and a block of mass

Physics

1 answer:

xz_007 [3.2K]3 years ago

6 0

Using the idea that the total energy of the system can be thought to be zero at the starting point and remains zero.

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I’ll mark you as brinlist please help.

Ede4ka [16]

Answer:

245 divided by 5.14=47.6653696 or 47.66

Explanation:

8 0

3 years ago

Read 2 more answers

During an earthquake, _______ travels through the Earth's interior as _______ waves.

nordsb [41]

Answer:

During an earthquake, seismic waves travels through the Earth's interior as body or p waves.

Explanation:

If neither of the bold words look familiar from your lesson feel free to ignore this answer

4 0

3 years ago

A spring with spring constant 450 N/m is stretched by 12 cm. What distance is required to double the amount of potential energy

snow_lady [41]

Answer:

The distance required = 16.97 cm

Explanation:

Hook's Law

From Hook's law, the potential energy stored in a stretched spring

E = 1/2ke² ......................... Equation 1

making e the subject of the equation,

e = √(2E/k)........................ Equation 2

Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.

Given: k = 450 N/m, e = 12 cm = 0.12 m.

E = 1/2(450)(0.12)²

E = 225(0.12)²

E = 3.24 J.

When the potential energy is doubled,

I.e E = 2×3.24

E = 6.48 J.

Substituting into equation 2,

e = √(2×6.48/450)

e = √0.0288

e = 0.1697 m

e = 16.97 cm

Thus the distance required = 16.97 cm

6 0

3 years ago

The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

Margaret [11]

Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.

From the above we see that F ∝ mm' and $F\alpha \frac{1}{r^{2} }$

Let the orbital radius of planet A is $r_{1}$ = r and mass of planet is $m_{1}$ .

Let the mass of central star is m .

Hence the gravitational force for planet A is $f_{1} =G \frac{m_{1}*m }{r^{2} }$

For planet B the orbital radius $r_{2} =2r_{1}$ and mass $m_{2} = 3 m_{1}$

Hence the gravitational force $f_{2} =G\frac{m m_{2} }{r^{2} }$

$f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }$

$= \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }$

Hence the ratio is $\frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2} }{Gmm_{1}/r_{1} ^2 }$

$=\frac{3}{4}$ [ ans]

3 0

4 years ago

Read 2 more answers

What is the resistance of : A) A 1.70 m long copper wire that is 0.700 mm in diameter? B) A 20.0 cm long piece of carbon with a

astra-53 [7]

Answer:

(I). The resistance of the copper wire is 0.0742 Ω.

(II). The resistance of the carbon piece is 1.75 Ω.

Explanation:

Given that,

Length of copper wire = 1.70 m

Diameter = 0.700 mm

Length of carbon piece = 20.0 cm

Cross section area $A = (2.00\times10^{-3})^2\ m$

(I). We need to calculate the area of copper wire

Using formula of area

$A=\pi r^2$

$A=3.14\times(\dfrac{0.700\times10^{-3}}{2})^2$

We need to calculate the resistance

Using formula of resistance

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.68\times10^{-8}\times1.70}{3.14\times(\dfrac{0.700\times10^{-3}}{2})^2}$

$R=0.0742\ \Omega$

(II). We need to calculate the resistance

Using formula of resistance

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{3.5\times10^{-5}\times20\times10^{-2}}{(2.00\times10^{-3})^2}$

$R=1.75\ \Omega$

Hence, (I). The resistance of the copper wire is 0.0742 Ω.

(II). The resistance of the carbon piece is 1.75 Ω.

8 0

3 years ago