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3 years ago

if a spring stores 5j of energy when it is compressed by0.5m,what is the spring constant of the spring?

Physics

1 answer:

Sindrei [870]3 years ago

5 0

Answer:

K = 40N/m

Explanation:

According to Hooke’s law

Energy = 1/2 kx^2

K is the spring constant

x = displacement

From the question

Energy =5J

x = 0.5m

K = ?

Using the above formula,

We have

5 = 1/2 x k x (0.5)^2

5 = 1/2 x k x (0.5 x 0.5)

5 = 1/2 x k x 0.25

5 = 0.25k/ 2

Cross multiply

5 x 2 = 0.25k

10 = 0.25k

Divide both sides by 0.25 to get k

10/0.25 = 0.25k/ 0.25

40 = k

K = 40N/m

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What is the new pressure of 150 ml of a gas that is compressed to 50 ml when the original pressure was 3.0 atm and the temperatu

Firlakuza [10]

Answer: 9.0 atm

Explanation:

To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

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where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

$P_1=3.0atm\\V_1=150mL\\P_2=?mmHg\\V_2=50mL$

Putting values in above equation, we get:

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4 years ago

An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t

Gelneren [198K]

Answer:

$1650\:\mathrm{m}$

Explanation:

We can use the following kinematics equations to solve this problem:

$v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x$ .

Using the first one to solve for acceleration:

$132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}$ .

Now we can use the second equation to solve for the distance travelled by the airplane:

$132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}$ (three significant figures).

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3 years ago

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i

Elis [28]

Four electrons are placed at the corner of a square

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So here it is given as

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here

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$r = \frac{a}{\sqrt2}$

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now the net potential is given as

$V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}$

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now potential energy of alpha particle at this position

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Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

$r_1 = \frac{a}{2} = 5 nm$

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now potential is given as

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now final potential energy is given as

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Now work done in this process is given as

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$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

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3 years ago

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DerKrebs [107]

Answer:

The point on the rim

Explanation:

All the points on the disk travels at the same angular speed $\omega$ , since they cover the same angular displacement in the same time. Instead, the tangential speed of a point on the disk is given by

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where

$\omega$ is the angular speed

r is the distance of the point from the centre of the disk

As we can see, the tangential speed is directly proportional to the distance from the centre: so the point on the rim, having a larger r than the point halway between the rim and the axis, will have a larger tangential speed, and therefore will travel a greater distance in a given time.

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