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3 years ago

A spring has a spring constant of 115 N/m. How much energy is stored in the

Physics

2 answers:

noname [10]3 years ago

8 0

Explanation:

Elastic energy stored in a spring is:

EE = ½ kx²

EE = ½ (115 N/m) (0.15 m)²

EE = 1.3 J

andrey2020 [161]3 years ago

6 0

<u>Answer:</u> The energy stored in the spring is 1.33 J

<u>Explanation:</u>

To calculate the energy stored in the spring, we use the equation:

$E=\frac{1}{2}kx^2$

where,

E = energy stored in the spring

k = spring constant = 115 N/m

x = compression of the spring = 0.15 m

Putting values in above equation, we get:

$E=\frac{1}{2}\times 115\times (0.15)^2\\\\E=1.33J$

Hence, the energy stored in the spring is 1.33 J

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In deserts the temperatures can reach extreme values in deep of summer and winter. A steel bridge is being constructed and it is

ryzh [129]

Answer:

The length of the bridge during the hottest part of summer is $L_s = 1235.925 m$

Explanation:

From the question we are told that

The length of the steel bridge is $L = 1234.567m$

The temperature for this length is $T_1 = 233.15K$

The temperature at summer $T_2 = + 140.0F = \frac{140 - 32}{180} *100 + 273= 333 K$

The coefficient of linear expansion is $\alpha = 11.0123*10^{-6} K^{-1}$

Generally the change in length of the steel bridge is mathematically represented as

$\Delta L = \alpha L \Delta T$

Substituting value

$\Delta L = 11.0123*10^{-6} * 1234.567 (333-233.15)$

$\Delta L = 1.3575 \ m$

The length of the bridge in summer is mathematically evaluated as

$L_s = L + \Delta L$

Substituting values

$L_s = 1234.567 + 1.3575$

$L_s = 1235.925 m$

3 0

3 years ago

Example 2 (5 points) Researchers at a university want to know if higher levels of nitrogen in fertilizer will increase the produ

OlgaM077 [116]

The experiment was well designed.

<h3>What is an experiment?</h3>

An experiment is used to determine cause and effect relationships. We know that in this case, the research is aimed at being able to obtain the relationship that exists between the amount of nitrogen in the soil and the production of tomatoes per plant.

As we could see, the experiment includes a control that serves as a baseline for determining the validity of the results. Further more, the sunlight, water and soil were held constant. The experiment was even repeated further south to verify the results.

Based on a careful review of the experiment, the experiment was well designed.

Learn more about experiment:brainly.com/question/11256472

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8 0

2 years ago

How does rolling an object affect its kinetic energy?

ad-work [718]

Answer:

it does happen, search in Google

3 0

3 years ago

Read 2 more answers

Baseball scouts often use a radar gun to measure the speed of a pitch. One particular model of radar gun emits a microwave signa

Firdavs [7]

Since the Units presented are not in the International System we will proceed to convert them. We know that,

$1 mi/h = 0.447 m/s$

So the speed in SI would be

$V=95mi/h(\frac{0.447m/s}{1mi/h})$

$V=42.465 m/s$

The change in frequency when the wave is reflected is

$f'=f(1+\frac{V}{c})$

Or we can rearrange the equation as

$f' = f + f\frac{V}{c}$

f' = Apparent frequency

f = Original Frequency

c = Speed of light

$f'-f = f\frac{V}{c}$

$\Delta f = f\frac{V}{c}$

Replacing,

$\Delta f = (10.525*10^9)(\frac{42.465}{3*10^8})$

$\Delta f =1489.8 Hz$

Since the waves are reflected, hence the change in frequency at the gun is equal to twice the change in frequency

$\Delta f_T = 2 \Delta f$

$\Delta f_T = 2(1489.8Hz)$

$\Delta f_T = 2979.63Hz$

Therefore the increase in frequency is 2979.63Hz

4 0

4 years ago

An electron (charge −e=−1.60×10−19C−e=−1.60×10−19C) is at rest at a distance 1.00×10−10 m 1.00×10−10m from the center of a nucle

shtirl [24]

Answer:

$-9.45\times10^{-16} \text{ J}$

Explanation:

According to Coulomb's law, the force of attraction between two point charges, $q_1$ and $q_2$ , separated by a distance $d$ is given by

$F = k\dfrac{q_1q_2}{d^2}$

$k$ is a constant with a value of $9\times10^9\text{ F/m}$ .

When we substitute the values from the question,

$F = (9\times10^9\text{ F/m})\dfrac{(-1.60\times10^{-19} \text{ C})\times(+1.31\times10^{-17}\text{ C})}{(1.00\times10^{-10}\text{ m})^2} = -1.89\times10^{-6} \text{ N}$

This value is negative because it is in a direction towards the positive charge.

The work done in moving the electron from the nucleus is

$W = F\times r$

$W = (-1.89\times 10^{-6} \text{ N})\times(5.00\times10^{-10}\text{ m}) = -9.45\times10^{-16} \text{ J}$

This is negative because work is done on the electron, not by it.

3 0

3 years ago