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klasskru [66]
3 years ago
14

When a mass of 29 g is attached to a certain spring, it makes 20 complete vibrations in 3.1 s. What is the spring constant of th

e spring? Answer in units of N/m.
Physics
1 answer:
never [62]3 years ago
4 0

Answer:

The spring constant of the spring is 47.62 N/m

Explanation:

Given that,

Mass that is attached with the spring, m = 29 g = 0.029 kg

The spring makes 20 complete vibrations in 3.1 s. We need to find the spring constant of the spring. We know that the number of oscillations per unit time is called frequency of an object. So,

f=\dfrac{20}{3.1}

f = 6.45 Hz

The frequency of oscillator is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

k is the spring constant

k=4\pi^2f^2m

k=4\pi^2\times (6.45)^2\times 0.029

k = 47.62 N/m

So, the spring constant of the spring is 47.62 N/m. Hence, this is the required solution.

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Answer:

Explanation:

Given

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Negative Pressure is applied to raise the water

Pressure is given by

P=\rho gh

where P=pressure

\rho =Density

h=depth

P=10^3\times 9.8\times 90

P=8.82\times 10^{5}\ N/m^2\approx 8.82\ atm

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(c)The fact of applying a negative pressure of 8.74 below the vapor pressure of water

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3 years ago
According to Newton’s first law of motion, what will an object in motion do when no external force acts on it?
KonstantinChe [14]

By definition, we have to:

Newton's first law states that any object will remain in a state of rest or with a uniform rectilinear motion unless an external force acts on it.

Therefore, according to the first law of Newton, if the object is already in motion and has no force acting on it then, it will remain with a uniform rectilinear motion.

Answer:

The object will remain with a uniform rectilinear movement when the external force does not act on it.

4 0
3 years ago
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The drawing shows 6 point charges arranged in a rectangle. The value of q is 2.83 uC and the distance d is 0.123 m. Find the tot
vova2212 [387]

the total electric potential at location P, which is at the center of the rectangle is 0V.

The charges placed at the corner of the rectangle are same in magnitude but different in charge. hence the total electric potential will be same in  magnitude but different in charge and will be cancelled. Similarly, all the total electric potential will be cancelled and resultant will be zero.

<h3>What is total electric potential?</h3>
  • The amount of labor required to convey a unit of electric charge from a reference point to a given place in an electric field is known as the electric potential (also known as the electric field potential, potential drop, or the electrostatic potential).
  • More specifically, it is the energy per unit charge for a test charge that is negligibly disruptive to the field under discussion. In order to prevent the test charge from gaining kinetic energy or radiating, the travel across the field is also meant to occur with very little acceleration.
  • The electric potential at the reference location is, by definition, zero units. Any point may be used as the reference point, but typically it is earth or a point at infinity.

To learn more about total electric potential with the given link

brainly.com/question/14776328

#SPJ4

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2 years ago
Fill in the blank for four and five answer. Question for number 6.
ehidna [41]
Increase .... decrease .... presumably it's the "best shape" for a body which has been formed by the gravitational force
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2 years ago
A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at
pav-90 [236]

Answer:

Explanation:

Given

mass of drop m=2 kg

height of fall h=1 m

ball leaves the foot with a speed of 18 m/s at an angle of 55^{\circ}

Velocity of ball just before the collision with the floor

u^=2gh

u=\sqrt{2gh}

u=\sqrt{2\times 9.8\times 1}=4.42 m/s

Impulse delivered in Y direction

J_y=m(v\sin (55)-(-u))

J_y=2(18\sin (55)+4.42)

J_y=38.32 kg-m/s

Impulse in x direction

J_x=m\times v\cos (55)

J_x=2\times 4.42\cos (55)=20.646

J_{net}=\sqrt{J_x^2+J_y^2}

J_{net}=\sqrt{(38.32)^2+(20.64)^2}

J_{net}=43.52 kg-m/s

at an angle of \tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}

\phi =tan^{-1}(1.856)

\phi =61.7^{\circ}  

7 0
2 years ago
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