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A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of

lidiya [134]

Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

K = 18.4N/m

At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

V = √(59.966)

V = 7.74m/s

8 0

2 years ago

Read 2 more answers

A top-fuel dragster starts from rest and has a constant acceleration of 42.0 m/s2. What are (a) the final velocity of the dragst

disa [49]

Answer:

a) Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) The displacement of the dragster at the end of 1.8 s = 68.04 m

d) The displacement of the dragster at the end of 3.6 s = 272.16 m

Explanation:

a) We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 1.8 s

Substituting

v = u + at

v = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 3.6 s

Substituting

v = u + at

v = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 1.8 s

Substituting

s= ut + 0.5 at²

s = 0 x 1.8 + 0.5 x 42 x 1.8²

s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 3.6 s

Substituting

s= ut + 0.5 at²

s = 0 x 3.6 + 0.5 x 42 x 3.6²

s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

3 0

2 years ago

Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The long

nordsb [41]

Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

For the shorter cylindrical resistor

Length = L

Diameter = D

Resistance = R1

For the longer cylindrical resistor

Length = 8L

Diameter = 4D

Resistance = R2

So;

We all know that the resistance of a given material can be determined by using the formula :

$R = \dfrac{\rho L }{A}$

where;

A = πr²

$R = \dfrac{\rho L }{\pi r ^2}$

For the shorter cylindrical resistor ; we have:

$R = \dfrac{\rho L }{\pi r ^2}$

since 2 r = D

$R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}$

$R = \dfrac{ 4 \rho L }{\pi \ D ^2}$

For the longer cylindrical resistor ; we have:

$R = \dfrac{\rho L }{\pi r ^2}$

since 2 r = D

$R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}$

$R = \dfrac{32\rho L }{\pi \ (4 D) ^2}$

$R = \dfrac{2\rho L }{\pi \ (D) ^2}$

Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

$\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D ^2}}{ \dfrac{2\rho L }{\pi \ (D) ^2}}$

$\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D ^2}}* { \dfrac {\pi \ (D) ^2} {2\rho L}}$

$\dfrac{R_s}{R_L} =2$

${R_s}=2{R_L}$

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

7 0

2 years ago

n an experiment of a simple pendulum, measurements show that the pendulum has length m, mass kg, and period s. Take m/s2 . i. Us

barxatty [35]

Answer:

The answer is " $(1.265 \pm 0.010) \ s \ and \ 0.709 \%$ "

Explanation:

In point i:

$T_{theo}= 2\pi \sqrt{\frac{l}{g}}$

$=2\pi\sqrt{\frac{0.397}{9.8}}\\\\= 1.265 \ s$

If error in the theoretical time period :

$\frac{\Delta T_{theo}}{T_theo} = \frac{1}{2} \frac{\Delta l }{l}\\\\\Delta T_{theo} = 1.265 \times \frac{1}{2} \times \frac{0.006}{0.397}$

$= 0.010 \ s$

$T_{theo} = (1.265 \pm 0.010) \ s$

In point ii:

$\% \ difference = \frac{|T_{exp} -T_{theo}|}{\frac{T_{exp}+T_{theo}}{2}} \times 100$

<h3> $= \frac{1.274 -1.265}{\frac{1.274+1.265}{2}} \times 100\\\\=0.709 \%$ </h3>

5 0

2 years ago

The valence electron occur in what part of the atom

Burka [1]

<span>The valence electrons occur in the outer shell of the atom. </span>

5 0

3 years ago