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padilas [110]
3 years ago
8

Two carts connected by a 0.25 m spring hit a wall, compressing the spring to 0.1 m. The spring constant k is 100 N m . What is t

he elastic potential energy stored from the spring's compression?
Physics
2 answers:
Liula [17]3 years ago
7 0

Answer:

1.125 J

Explanation:

The correct answer is 1.125 J

Anna [14]3 years ago
6 0

Given that,

Initial position of the spring, x₁ = 0.25 m

Final position of the spring, x₂ = 0.1 m

The spring constant of the spring, k = 100 N/m

To find,

The elastic potential energy stored from the spring's compression.

Solution,

The work done in moving a spring from on position to another is stored in the form of elastic potential energy. It can be given as follows :

E=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 100\times (0.1^2-0.25^2)\\\\=2.62\ J

So, the required energy is 2.62 J.

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The Statue of Liberty weighs nearly 205 metric tons. If a person can pull an average of 90 kg, how many people would it take to
charle [14.2K]
3 people what is the math or history?
4 0
3 years ago
A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A pie
-BARSIC- [3]

a) 3.27\cdot 10^{-3} J

b) 11.60\cdot 10^{-3} J

c) 8.33\cdot 10^{-3} J

Explanation:

a)

The energy stored in a capacitor is given by

U=\frac{1}{2}CV^2

where

C is the capacitance of the capacitor

V is the potential difference across the plates of the capacitor

For the capacitor in this problem, before insering the dielectric, we have:

C=13.5 \mu F = 13.5\cdot 10^{-6}F is its capacitance

V = 22.0 V is the potential difference across it

Therefore, the initial energy stored in the capacitor is:

U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J

b)

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

C'=kC

where

k = 3.55 is the dielectric constant of the material

C is the initial capacitance of the capacitor

Therefore, the energy stored now in the capacitor is:

U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2

where:

C=13.5\cdot 10^{-6}F is the initial capacitance

V = 22.0 V is the potential difference across the plate

Substituting, we find:

U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J

C)

The initial energy stored in the capacitor, before the dielectric is inserted, is

U=3.27\cdot 10^{-3} J

The final energy stored in the capacitor, after the dielectric is inserted, is

U'=11.60\cdot 10^{-3} J

Therefore, the change in energy of the capacitor during the insertion is:

\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J

So, the energy of the capacitor has increased by 8.33\cdot 10^{-3} J

8 0
3 years ago
A sample of an ideal gas is heated, and its kelvin temperature doubles. What happens to the average speed of the molecules in th
BabaBlast [244]

A sample of an ideal gas is heated, and its kelvin temperature doubles. The average speed of the molecules in the sample will increases by a factor of  \sqrt{2}

The root-mean square (RMS) velocity is the value of the square root of the sum of the squares of the stacking velocity values divided by the number of values. The RMS velocity is that of a wave through sub-surface layers of different interval velocities along a specific ray path.

Root mean square speed is a statistical measurement of speed.

The root mean square speed can be calculated as : V1 : \sqrt{3 R T / Mo}

if  temperature becomes double

let T1 is initial temperature

So ,  T2 = 2 * T1

now ,

Root mean square speed will be (V2) =  \sqrt{(3 R (2T)) / Mo}

                                                     = \sqrt{2} * \sqrt{3 R T / Mo}

                                                     = \sqrt{2} V1

Thus when temperature becomes double, the root mean square speed increases by a factor of  \sqrt{2}

To learn more about root mean square velocity here

brainly.com/question/13751940

#SPJ4

3 0
1 year ago
A boat with a mass of 1000 kg drifts with the current down a straight section of river parallel to the +x+x axis with a speed of
IrinaK [193]

Answer:

A.

Explanation:

Our values are defined by,

m=1000kg\\v = 2m/s\\a = 2.7 \angle 45\°

We can express also in a vectorial way, then

v=2\hat{i} \rightarrowno values in \hat{j}

Acceleration as follow,

a= 2.7cos45 \hat{i} + 2.7 sin45 \hat{j}

We know that velocity is given by,

V(t) = v_0+at \rightarrow v_0 = 2

We need to calculate for t=3, then

v(3) = 2\hat{i} +(2.7cos45 \hat{i} + 2.7 sin45 \hat{j})*3

v(3) = (2\hat{i}+3*2.7cos45 \hat{i})+(3*2.7 sin45 \hat{j})

v(3) = (2+3*2.7cos45)\hat{i}+(5.7 sin45 )\hat{j}

v(3) = 7.7\hat{i}+5.7\hat{j}

Our mass is 1000Kg, so the momentum is

P = mv

P= 1000(7.7\hat{i}+5.7\hat{j})

P=7700\hat{i}+5700\hat{j}

7 0
4 years ago
How do you find pressure related to temperature
Ivenika [448]
Temperature and pressure are directly proportional to each other. If this helps,please pick me the best. Also,please say thanks.

8 0
3 years ago
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