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andrezito [222]
3 years ago
5

A block of mass m attached to a horizontally mounted spring with spring constant ???? undergoes simple harmonic motion on a fric

tionless surface. How would the maximum speed of the block be affected if the spring constant was increased by a factor of 4 while holding the amplitude of oscillation constant?
(A) It would increase by a factor of 4.
(B) It would decrease by a factor of 12.
(C) It would decrease by a factor of 14.
(D) It would remain unchanged.
(E) It would increase by a factor of 2.
Physics
1 answer:
Firdavs [7]3 years ago
8 0

Answer:

(E) It would increase by a factor of 2.

Explanation:

Maximum speed (v) is given by:

v = \omega *A

Where ω is the angular frequency and A is the amplitude.

The angular frequency of a spring-mass system is given by:

\omega = \sqrt{\frac{k}{m}}

Where k is the spring constant and m is the mass of the block attached to the spring.

Therefore, maximum speed can be written as:

v= A\sqrt{\frac{k}{m}}\\

Since mass and amplitude are constant values, a relationship between maximum velocity and spring constant can be defined as:

v =C \sqrt{k}

There is a linear relationship between maximum velocity and the square root of the spring constant. Therefore if the spring constant is increased by a factor of 4, the velocity is increased by a factor of the square root of 4, which is 2.

The answer is (E) It would increase by a factor of 2.

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mr_godi [17]
As the magnet is moved inside a coil of wire, the number of lines of magnetic field passing through the coil changes. Faraday stated that : it is the change in the number of field lines passing through the the coil of wire that induces emf in the loop. Specifically, it is the rate of change in the number of magnetic field lines passing through the loop that determines the induced emf. There is a term called magnetic flux same as electric flux, this magnetic flux can be a measure of the number of field lines passing through a surface. It is given by ( Φ=ΣB. dA. Where B is magnetic field and dA is small elementary area). The induced emf is given by (ξ = dΦ/dt). This equation states that THE MAGNITUDE OF THE INDUCED CURRENT IN A CIRCUIT IS EQUAL TO THE RATE AT WHICH THE MAGNETIC FLUX THROUGH THE CIRCUIT IS CHANGING WITH TIME. So more rapid you move the coil, more will be the change in flux and hence more emf will be produced. So option D is the correct answer. I hope this long description will help you out.
5 0
3 years ago
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A solid sphere of radius r carries charge q distributed uniformly throughout its volume. find the potential difference from the
erastova [34]

Answer: The potential difference from the sphere's surface to its center is zero.

Explanation:

The expression for the potential difference due to charge is as follows;

\Delta V=\frac{q}{4\pi \epsilon _{0}r}

Here, \Delta V is the potential difference, q is the charge, r is the position and \epsilon _{0}} is the absolute permittivity of free space.

In the given problem, a solid sphere of radius r carries charge q distributed uniformly throughout its volume.

To find the potential difference from the sphere's surface to its center. Put r=0 in the expression of the potential difference.

\Delta V=\frac{q}{4\pi \epsilon _{0}(0)}

\Delta V=0

Therefore, the  potential difference from the sphere's surface to its center is zero.

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PLEASE HELP!! What needs to happen for solid water to change to liquid water?
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Answer:

The answer is D.

Explanation:

I took the test.

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2 years ago
Assume both snowballs are thrown with the same initial speed 27.2 m/s. The first snowball is thrown at an angle of 75◦ above the
Leona [35]

Answer:

15 deg

Explanation:

Assume both snowballs are thrown with the same initial speed 27.2 m/s. The first snowball is thrown at an angle of 75◦ above the horizontal. At what angle should you throw the second snowball to make it hit the same point as the first? The acceleration of gravity is 9.8 m/s 2 . Answer in units of ◦ .

Given:

For first ball, θ1 = 75◦

initial velocity for both the balls, u = 27.2 m/s

for second ball, θ2 = ?

since distance covered by both the balls is same.

Therefore,..

R1=(u^{2} sin2\alpha _{1}) /g[/tex]

the range for the first ball

the range for the second ball

R2=(u^{2} sin2\alpha _{2}) /g[/tex]

(u^{2} sin2\alpha _{2}) /g[/tex]=(u^{2} sin2\alpha _{1}) /g[/tex]

sin2\alpha _{2})=sin2\alpha _{1})

2\alpha _{2}=sin^-1(sin2\alpha _{1})

\alpha _{2}=1/2sin^-1(sin2\alpha _{1})

\alpha _{2}=

15 deg

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4 years ago
Rosný bod závisí od ?
WITCHER [35]

Answer:

what did u say and what language are you speaking in

8 0
3 years ago
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