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BabaBlast [244]
3 years ago
11

A 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to os

cillate. What is the spring’s elastic constant?a) 3.5 N/mb) 5.2 N/mc) 1.9 N/md) 27 N/m
Physics
1 answer:
Bingel [31]3 years ago
4 0

Answer:

k = 3.5 N/m

Explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{m}{k}

\frac{l}{g} = \frac{m}{k}\\\\ k = g\frac{m}{l}

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

k = (9.81\ m/s^2)(\frac{0.125\ kg}{0.35\ m})\\\\

<u>k = 3.5 N/m</u>

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Stratosphere because the atmosphere get denser
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3 years ago
The velocity of a passenger relative to a boat is -vpb. The velocity of the boat relative to the river it is moving on is vbr. T
RideAnS [48]

Answer:

vps = vbr + vrs - vpb

Explanation:

  • If the passenger were at rest, his speed relative to the shore will be identical to the boat's, as follows:
  • vps = vbr + vrs
  • As he is moving in a direction opposite to the boat's, his velocity relative to the shore must be less than if he were at rest, in the same quantity that he was moving opposite to the boat, as follows:
  • vps = vbr+ vrs -vpb
5 0
2 years ago
Assume a change at the source of sound reduces the wavelength of a sound wave in air by a factor of 3.
noname [10]

Explanation:

The speed of a wave is given by :

v=f\lambda ......(1)

(i) Here, the wavelength of a sound wave in air reduces by a factor of 3. Equation (1) becomes :

\lambda=\dfrac{v}{f}

Wavelength and frequency are inversely proportional to each other. So, if wavelength of a sound wave in air reduces by a factor of 3, then the frequency will increases by a factor of 3.

(ii) It remains the same.

8 0
3 years ago
A satellite that goes around the earth once every 24 hours (86,400 s) is called a geosynchronous satellite. If a geosynchronous
Olegator [25]

Answer:

42244138.951 m

Explanation:

G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²

r = Radius of orbit from center of earth

M = Mass of Earth = 5.98 × 10²⁴ kg

m = Mass of Satellite

The satellite revolves around the Earth at a constant speed

Speed = Distance / Time

The distance is the perimeter of the orbit

v=\frac{2\pi \times r}{24\times 3600}

The Centripetal force of the satellite is balanced by the universal gravitational force

m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m

The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m

6 0
3 years ago
A piece of tape is pulled from a spool and lowered toward a 120-mg scrap of paper. Only when the tape comes within 8.0 mm is the
Anton [14]

Answer:

1.176\times 10^{-3} N

Upward

Explanation:

We are given that

Mass of scarp paper,m=120 mg=120\times 10^{-6}kg

1mg=10^{-6} mkg

Distance,d =8 mm=8\times 10^{-3} m

Magnitude of electric force =F_E= w=mg

Where g=9.8 m/s^2

Substitute the values

F_E=120\times 10^{-6}\times 9.8

F_E=1.176\times 10^{-3} N

Gravitational force act in  downward direction.

The electric force acts in opposite direction and magnitude of electric force is equal to gravitational force.

Hence, the direction of electric force is upward.

6 0
3 years ago
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