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3 years ago

What are 5 examples from daily life in which you see periodic motion caused by a spring or cord?

1 answer:

8 0

<span>A wrist watch has a little weight on spring that it uses to count the seconds.
A guitar sting is a vibrating cord, but the wieght of the cord (and the length and tension) determine the frequency of the vibration.
A car goes over a bump and has a dampened oscillation due to the springs and shock absorbers.

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Standing still at the top of the hill at point a,what kind of energy( potential and/or kinetic) would the snowboarder have ? How

Alja [10]

He would have potential, he hasn’t gone down the mountain yet so he remains potential

5 0

3 years ago

If an object weighs 40N, calculate the mass (gravity = 9.8m/s^2)

Elan Coil [88]

Answer:

4.08 kg

Explanation:

We can apply the Newton's second law of motion to find the mass of this object:

$F=ma$ ,

where $m$ is mass, and $a$ is acceleration, which is substituted by $g$ here. Now, plugging given numbers in the equation, we have

$40=m\cdot9.8$ .

Solving for mass, we have

$m=\frac{40}{9.8}\\=4.08 \;\text{kg}.$

4 0

4 years ago

Which of the following statements best describes the general pattern of composition among the four jovian planets?A) Jupiter is

kompoz [17]

<h2>Answer: Jupiter and Saturn have compositions that are fairly different from the compositions of Uranus and Neptune.</h2>

Jovian Planets (means "similar to Jupiter") is the name given to the "gaseous giant" planets of our solar system. Although they share many similarities, they have certain differences.

For example, Jupiter and Saturn are classified as "gas giants", while Uranus and Neptune are "ice giants".

This is also due to the fact that Uranus and Neptune have higher concentrations of <u>methane</u> (hence its bluish color) and some heavier elements such as oxygen, carbon, nitrogen, and sulfur. On the other hand, Jupiter and Saturn tend to be orange-white in appearance due to the mixture of <u>hydrogen</u> that gives off a red aspect.

7 0

4 years ago

Derive the velocity of an electron in a hydrogen atom using Bohr's model. The electron has a mass of m, charge of (-e) and an or

Ahat [919]

Answer:

$v=\frac{Ze^2}{2 \epsilon_0\times n\times h}$

$v=\frac {2.185\times 10^6}{n}\ m/s$

Explanation:

According to Bohr's Theory,

The force of attraction acting between the electron and the nucleus is equal to the centrifugal force acting on the electron.

Thus,

$\frac{Ze^2}{4\pi \epsilon_0 r^2}=m_e\frac{v^2}{r}$ ......1

Where, Z is the atomic number or the number of protons

r is the atomic radius

v is the velocity of the electron

$m_e$ is the mass of the electron

Also,

Accoriding to Bohr, the angular momentum is quantized. He states that the angular momemtum is equal to the integral multiple of $\frac {h}{2\times \pi}$ .

$m_e vr=n\times \frac {h}{2\times \pi}$ ....2

solving r from equation 2, we get that:

$r=n\times \frac {h}{2\times \pi\times m_e v}$

Putting in 1 , we get that:

$v=\frac{Ze^2}{2 \epsilon_0\times n\times h}$

Applying values for hydrogen atom,

Z = 1

Mass of the electron ( $m_e$ ) is 9.1093×10⁻³¹ kg

Charge of electron (e) is 1.60217662 × 10⁻¹⁹ C

$\epsilon_0$ = 8.854×10⁻¹² C² N⁻¹ m⁻²

h is Plank's constant having value = 6.626×10⁻³⁴ m² kg / s

We get that:

$v=\frac {2.185\times 10^6}{n}\ m/s$

4 0

4 years ago

The end diastolic volume of a heart is 140 mL Assume that it is a sphere. At end diastole, the intraventricular pressure is 7mmI

Vera_Pavlovna [14]

Answer:

Explanation:

We know that, V = 140 mL = 0.00014 m3

Assume that it is a sphere. so, we have

V = (4/3) \pir3

r3 = (0.00014 m3) (3) / (4) (3.14)

r = \sqrt[3]{}\sqrt[3]{}3\sqrt{}3.34 x 10-5 m3

r = 1.93 x 10-7 m

(a) The wall tension at end diastole will be given as :

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end diastole = 7 mmHg = 933.2 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (933.2 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 8.18 x 10-3 N

(b) The wall tension at the end of isovolumetric contraction will be given as :

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end of isovolumetric contraction = 80 mmHg = 10665.7 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (10665.7 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 9.35 x 10-2 N

(d) The wall stress from A and B which will be given as :

we know that, \sigma = T / w

For part A, we have

\sigmaA = (8.18 x 10-3 N) / (0.011 m)

\sigmaA = 0.743 N/m

For part B, we have

\sigmaB = (9.35 x 10-2 N) / (0.011 m)

\sigmaB = 8.5 N/m

4 0

3 years ago