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tresset_1 [31]
3 years ago
8

A 5.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical

oscillations having a period of 3.50 s. Find the force constant of the spring.
Physics
1 answer:
iris [78.8K]3 years ago
4 0

Answer:

17.71N/m

Explanation:

The period of the spring is expressed according to the expression;

T = 2 \pi \sqrt{\frac{m}{k} } \\

m is the mass of the object

k is the force constant

Given

m = 5.50kg

T = 3.50s

Substitute into the formula;

T = 2 \pi \sqrt{\frac{m}{k} } \\3.5 = 2 (3.14) \sqrt{\frac{5.5}{k} } \\3.5 = 6.28 \sqrt{\frac{5.5}{k} } \\\frac{3.5}{6.28} =  \sqrt{\frac{5.5}{k} } \\0.557 = \sqrt{\frac{5.5}{k} } \\square \ both \ sides\\0.557^2 = (\sqrt{\frac{5.5}{k} })^2 \\0.3106 = \frac{5,5}{k}\\k = \frac{5.5}{0.3106}\\k =  17.71N/m

Hence the force constant of the spring is 17.71N/m

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Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st
Black_prince [1.1K]

Answer:

Part A:

E_{midpoint}=0

Part B:

E_{center}=2711.7558 N/C

Explanation:

Part A:

Formula of Electric Field Strength:

E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}

Where:

x is the distance from the ring

R is the radius of the ring

\epsilon is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C

Electric Field Strength at center of left ring is same as that of right ring.

E_{center}=2711.7558 N/C

5 0
3 years ago
The California sea lion is capable of making extremely fast, tight turns while swimming underwater. In one study, scientists obs
anygoal [31]

Answer:

Acceleration of Sea Lion is 4.41 g

This is 49% of maximum jet acceleration given as a = 9g

Explanation:

As we know that the radius of the circular loop is given as

R = 0.37 m

The speed of the fish is given as

v = 4 m/s

Now the centripetal acceleration of the sea lion is given as

a_c = \frac{v^2}{R}

a_c = \frac{4^2}{0.37}

a_c = 43.2 m/s^2

as we know that

g = 9.8 m/s^2

so we have

a = 4.41 g

Now Percentage of this acceleration wrt maximum jet acceleration is given as

P = \frac{4.41 g}{9g} \times 100

P = 49%

6 0
3 years ago
A scientist discovers a fossil of an animal and places it in the fossil record. The organism’s bones are similar to the bones of
ZanzabumX [31]

Answer:

I think 3

Explanation:

That makes sense

3 0
2 years ago
A. What are the three longest wavelengths for standing waves on a 240-cm-long string that is fixed at both ends?
vovangra [49]

Answer:

a) the three longest wavelengths = 4.8m, 2.4m, 1.6m

b) what is the frequency of the third-longest wavelength = 75Hz

Explanation:

The steps and appropriate formula and substitution is as shown in the attached file.

5 0
3 years ago
What is the full meaning of (i.p.s.m.n)​
Anna11 [10]

Answer:

Intraductal Papillary Mucinous Neoplasm

6 0
3 years ago
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