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Shkiper50 [21]
3 years ago
7

Calculate by how much a spring with spring constant 1.2N/m has been squashed when it has stored 76J,

Physics
1 answer:
fgiga [73]3 years ago
7 0

Answer:

11.25 m

Explanation:

From the question given above, the following data were obtained:

Spring constant (K) = 1.2 N/m

Energy (E) = 76 J

Compression (e) =?

Energy stored in a spring is given by the following equation:

E = ½Ke²

Where:

E => Energy stored in the spring.

K => spring constant

e => Extention or compression

With the above formula, we can obtain how much the spring has been squashed as follow:

Spring constant (K) = 1.2 N/m

Energy (E) = 76 J

Compression (e) =?

E = ½Ke²

76 = ½ × 1.2 × e²

76 = 0.6 × e²

Divide both side by 0.6

e² = 76 / 0.6

Take the square root of both side.

e = √(76 / 0.6)

e = 11.25 m

Thus, the spring has been squashed by 11.25 m

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Sure: ths is called protyping and lets yu get a sende fo the effeciveness tof the cahnge and the cost of the change. 

<span>A - just like the scientific principle syu what to know what other know or have learned. Example would it be silly to build a nuclear power de-salinatiztion plant when a dam in the mountains wuld dothe savme thng and perhaps have the advatage of using local labor and preveinting floods and givng of hydro eletic power. </span>

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8 0
3 years ago
Can someone answer this.
klemol [59]

Take left to be the negative direction, and right to be positive. That means the forces on the squirrel would be -2N and 3N.

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4 0
4 years ago
The tub of a washer goes into its spin-dry cy-cle, starting from rest and reaching an angularspeed of 3.1 rev/s in 10.7 s.At thi
neonofarm [45]

Answer:

35 revolutions

Explanation:

t = Time taken

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

\theta = Number of rotation

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{3-0}{10.7}\\\Rightarrow \alpha=0.28037\ rev/s^2

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{3.1^2-0^2}{2\times 0.28037}\\\Rightarrow \theta=17.13806\ rev

Number of revolutions in the 10.7 seconds is 17.13806

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-3.1}{11.2}\\\Rightarrow a=-0.27678\ rev/s^2

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-3.1^2}{2\times -0.27678}\\\Rightarrow \theta=17.36035\ rev

Number of revolutions in the 11.2 seconds is 17.36035

Total total number of revolutions in the 21.9 second interval is 17.13806+17.36035 = 34.49841 = 35 revolutions

3 0
4 years ago
1. On which date does the Antarctic Circle have 24 hours of daylight? A. June 21
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5 0
4 years ago
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A rectangular wire loop is pulled out of a region of uniform magnetic field B at a constant speed v. What is true about the indu
Paul [167]

Answer:

There is a constant emf induced in the loop.

Explanation:

In the uniform magnetic field suppose the rectangular wire loop of length L and width b is moved out with a uniform velocity v. suppose any instance x length of the loop is out of the magnetic field and L-x length is inside the loop.

Area of loop outside the field = b(L-x)

we know that flux φ= BA

B= magnitude of magnetic field , A=  area

and emf \epsilon= \frac{d\phi}{dt}

\epsilon=B\frac{dA}{dt}

\epsilon=B\frac{db(L-x)}{dt}

\epsilon=Bb\frac{d(L-x)}{dt}

B,b and L are constant and dx/dt = v

⇒ε = -Bbv

which is a constant hence There is a constant emf induced in the loop.

5 0
4 years ago
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