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kolbaska11 [484]
3 years ago
13

A 5.0 kg mass is suspended from a spring. Pulling the mass down by an additional 10 cm takes a force of 20 N. If the mass is the

n released, it will rise up and then come back down. How long will it take for the mass to return to its starting point 10 cm below its equilibrium position?
Physics
1 answer:
Rudik [331]3 years ago
4 0

Answer:

0.99 seconds

Explanation:

The problem depicts a simple harmonic motion.

Now, from Hooke's law,

The spring constant is given as;

k =F/x

where:

x is the displacement of the spring's end from its equilibrium position

F is the restoring force exerted by the spring on that end

From the question, F = 20N while x = 10cm = 0.1m. Thus,

K = 20/0.1 = 200 N/m

Now, time to take for the mass to return to its starting point is a period

The period oscillation of the mass is given as;

T = 2π√m/k

Where m = mass = 5kg

T = 2π√(5/200)

T = 0.99 s

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here's the solution,

we know that,

=》

wave \: speed = wavelength \times frequency

so,

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18 = 3 \times f

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f =  \dfrac{18}{3}

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f = 6

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What is the stopping distance at night for a passenger vehicle traveling at 50 mph?
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An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA.
rewona [7]

Answer:

the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

Explanation:

Given that;

diameter D = 2.0 mm

current I = 1.0 mA

K.E of each proton is 20 MeV

the number density of the protons in the beam = ?

Now, we make use of the relation between current and drift velocity

I = MeAv ⇒ 1 / eAv

The kinetic energy of protons is given by;

K = \frac{1}{2}m_{p}v²

v = √( 2K / m_{p} )

lets relate the cross-sectional area A of the beam to its diameter D;

A = \frac{1}{4}πD²

now, we substitute for v and A

n = I / \frac{1}{4}πeD² ×√( 2K / m_{p} )

n = 4I/π eD² × √(m_{p} / 2K )

so we plug in our values;

n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )

n =  1.98695 × 10¹⁸ × 1.6157967  × 10⁻⁵

n = 3.2 × 10¹³ m⁻³  

Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

8 0
2 years ago
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's a
s2008m [1.1K]

Answer:

A. Power generated by meteor = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Workdone = 981000 J

Power required = 19620 Watts

Note: The question is incomplete. A similar complete question is given below:

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?

Explanation:

A. Power = workdone / time taken

Workdone = Kinetic energy of the meteor

Kinetic energy = mass × velocity² / 2

Mass of meteor = 1.5 g = 0.0015 kg;

Velocity of meteor = 50 km/s = 50000 m/s

Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J

Power generated = 1875000/2.1 = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Work done by elevator against gravity = mass × acceleration due to gravity × height

Work done = 1000 kg × 9.81 m/s² × 100 m

Workdone = 981000 J

Power required = workdone / time

Power = 981000 J / 50 s

Power required = 19620 Watts

Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.

6 0
3 years ago
If a ????=87.5 kgm=87.5 kg person were traveling at ????=0.900????v=0.900c , where ????c is the speed of light, what would be th
Diano4ka-milaya [45]

Answer:

\frac{K.E_r}{K.E}=2.875

Explanation:

Given:

mass, m = 87.5kg

Velocity, V = 0.900c

now,

the relativistic kinetic energy id given as:

K.E_r=(\gamma-1)mc^2 ...........(1)

where,

\gamma = relativistic factor, given as; \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

Now, the classical kinetic energy is given as:

K.E = \frac{1}{2}mv^2    ..........(2)

Dividing the equation (1) by (2) we get

\frac{K.E_r}{K.E}=\frac{(\gamma-1)mc^2}{\frac{1}{2}mv^2}

or

\frac{K.E_r}{K.E}=\frac{(\gamma-1)c^2}{\frac{1}{2}v^2}

substituting the values in the equation we get,

\frac{K.E_r}{K.E}=\frac{(\frac{1}{\sqrt{1-\frac{(0.90c)^2}{c^2}}}-1)c^2}{\frac{1}{2}\times(0.90c)^2}

or

\frac{K.E_r}{K.E}=2.875

5 0
3 years ago
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