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aliya0001 [1]
3 years ago
7

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is

then pulled down 7.5 cm and released. The ball makes 31 oscillations in 15 s seconds.
1) What is its the mass of the ball?
Express your answer using two significant figures.
m = _____ g

2) What is its maximum speed?
Express your answer using two significant figures.

= _____ cm/s
Physics
1 answer:
Tpy6a [65]3 years ago
3 0

Answer:

1)m=89.01 g

2)V(max) = 97.3 cm/s

Explanation:

Given that

K= 15 N/m

The maximum amplitude ,A=7.5 cm = 0.075 m

Given that 31 oscillations in 15 seconds ,this means that frequency f

f=\dfrac{31}{15}

f=2.066 Hz

lets take mass of the ball = m kg

f=\dfrac{1}{2\pi}\times \sqrt{\dfrac{K}{m}}

m=\dfrac{1}{4\times \pi^2}\times{\dfrac{K}{f^2}}

m=\dfrac{1}{4\times \pi^2}\times{\dfrac{15}{2.066^2}}

m=0.08901 kg

m=89.01 g

The maximum speed

V(max)= ω x A

ω = 2π f= 2 x π x 2.066 = 12.98 rad/s

V(max) = 12.98 x 0.075 =0.973 m/s

V(max) = 97.3 cm/s

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Answer:

atomic concentration = 2 atoms/unit cell

lattice parameter: a= 3.22 x 10⁻¹⁰ m

atomic radius: r= 1.39 x 10⁻¹⁰m

Explanation:

The atomic concentration is the number of atoms that can fit into a unit cell. It is a known number for each unit cell crystal structure. For a BCC (body-centered cube) crystal structure, atomic concentration is 2 atoms/unit cell because there are a 1/8 part of an atom in each corner of the cube (1/8 x 8= 1 atom) and 1 central atom in the central position of the cube ⇒ n= 1 atom + 1 atom= 2 atoms/unit cell

In order to calculate the lattice parameter a, we introduce the atomic mass 95.94 g/mol and the density 10.22 g/cm³ in the expression for the volume of the cube:

Vc= a³= \frac{(95.94 g/mol) x (2 atoms/unit cell)}{(10.2 g/cm^{3}) x (6.023 x 10^{23} atoms/mol)  }

a³= 3.12 x 10⁻²³ m³

⇒ a = ∛(3.12 x 10⁻²³ m³) = 3.22 x 10⁻¹⁰m

Once we know the lattice parameter a, we can calculate the atomic radius r by using the expression of a for a BCC structure:

a= \frac{4r}{\sqrt{3}}

⇒ r= a x √3/4= (3.22 x 10⁻¹⁰ m) x √3/4 = 1.39 x 10⁻¹⁰ m

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Please help <br>problems 2a.,2b.,3a.,and 3b.​
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Answer:

2a) x = 32 [mil/h]; 2b) t = 0.5[h]; 3a) t = 2.5 [h]; 3b) x = 185[mil]

Explanation:

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We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 32 [\frac{mil}{h}] \\t=time = 1 [h]\\x=v*t\\x=32[\frac{mil}{h} ]*1[h]\\x=32[mil}

2b)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

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3a)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

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3b)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

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Answer:

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We need to find the magnitude of the average value of the emf induced in the loop. Due to change in magnetic field, an emf is induced in the loop which is given by :

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