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Olin [163]
3 years ago
12

A 465 g block slides along a frictionless surface at a speed of 0.35 m/s. It runs into a horizontal massless spring with spring

constant 54 N/m that extends outward from a wall. It compresses the spring, then is pushed back in the opposite direction by the spring, eventually losing contact with the spring.
a) How long does the block remain in contact with the spring?
b) How would your answer change if the block's initial speed was doubled?
Physics
1 answer:
katovenus [111]3 years ago
7 0

Answer:

a) The duration, during which the block remain in contact with the spring is 0.29 s

b) The period of the simple harmonic oscillatory motion depends only on the mass and spring constant, therefore when the speed is doubled, the duration of contact remains the same as 0.29 s.

Explanation:

Mass of the block = 465 g

Surface speed = 0.35 m/s

Spring constant , k = 54 N/m

T = 2\times \pi \times \sqrt{\frac{m}{K} } = 2\times \pi \times \sqrt{\frac{0.465}{54} }  = 0.58 s

a) Since the period for the oscillatory motion is 0.58 s, then the time when the block and spring remain in contact is T/2 = 0.29 s

b) When the speed is doubled, we have

T = 2\times \pi \times \sqrt{\frac{m}{K} }

Therefore, since T is only dependent on the mass, m and the  spring constant, K, then the time it takes when the speed is doubled remain as

T /2 = 0.29 s

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Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?

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<span>Solution:

A) A charge q </span>under an electric field of intensity E will experience a force F  equal to:

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E= \frac{F}{q}= \frac{18\cdot 10^{-9}N}{2.5\cdot 10^{-9}C}=7.2 V/m

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B) The proton charge is equal to

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Therefore, the magnitude of the force acting on the proton will be

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3 years ago
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Input the value of bullet in the formulae;
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Which means that th Energy of the ball is more than the bullet.
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3 years ago
When a cube is inscribed in a sphere of radius r, the length Lof a side of the cube is . If a positive point charge Qis placed a
Nana76 [90]

Answer:

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Explanation:

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where E is the electric field and A is the surface area

Let's shut down the electric field with Gauss's law

       Фi = ∫ E .dA = q_{int} / ε₀

the Gaussian surface is a sphere so its area is

        A = 4 π r²

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we substitute

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To calculate the flow on the two surfaces

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* Cube

Let's find the side value of the cube inscribed inside the sphere.

In this case the radius of the sphere is half the diagonal of the cube

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we substitute

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         Ф_cube /Ф_sphere = 3 /π

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