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4 years ago

An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20.0°C.

Physics

1 answer:

amm18124 years ago

6 0

Answer:

a) T ’= 0.999 s , b) t = 3596.4 s

Explanation:

The angular velocity of a simple pendulum is

w = √g / L

The angular velocity, frequency and period are related

w = 2π f = 2π / T

2π / T = √ g / L

T = 2π √ L / g

L = T² g / 4π²

L = 1² 9.8 / 4π²

L = 0.248 m

To know the effect of the temperature change let's use the thermal expansion ratios

ΔL = α L ΔT

ΔL = 24 10⁻⁶ 0.248 (-4 - 20)

ΔL = 142.8 10⁻⁶ m

Lf - L = -142. 8 10⁻⁶

Lf = 142.8 10⁻⁶ + 0.248

Lf = 0.2479 m

Let's calculate new period

T ’= 2π √ L / g

T ’= 2π √ (0.2479 / 9.8)

T ’= 0.999 s

We can see that the value of the period is reduced so that the clock is delayed

b) change of time in 1 hour

When the clock is at 20 ° C in one hour it performs 3600 oscillations, for the new period the time of this number of oscillations is

t = 3600 0.999

t = 3596.4 s

Therefore the clock is delayed almost 4 s

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A lumberjack (mass = 103 kg) is standing at rest on one end of a floating log (mass = 277 kg) that is also at rest. The lumberja

Lorico [155]

Answer:

Explanation:

Given

mass of lumber jack $m_L=1000 kg$

mass of floating log $m_f=277 kg$

velocity of lumber jack =3.74 m/s (w.r.t. shore)

conserving momentum

Initial momentum=Final momentum

$0=m_L\times 3.74+m_f\times v$

$v=-\frac{1000\times 3.74}{277}$

$v=-13.501 m/s$

(b)For second Log

Conserving Momentum

Initial momentum of log and Lumber jack=Final momentum of log and lumberjack

$m_L\times 3.74=(m_L+m_f)u$

u is final velocity of lumberjack and log

$u=\frac{m_L}{m_L+m_f}\times 3.74$

$u=\frac{1000}{1000+277}=2.92 m/s$

5 0

4 years ago

A hollow cylinder is given a velocity of 5.3 m/s and rolls up an incline to a height of 2.87 m. If a hollow sphere of the same m

Valentin [98]

Answer:

E = 1/2 M V^2 + 1/2 I ω^2 = 1/2 M V^2 + 1/2 I V^2 / R^2

E = 1/2 M V^2 (1 + I / (M R^2))

For a cylinder I = M R^2

For a sphere I = 2/3 M R^2

E(cylinder) = 1 + 1 = 2 omitting the 1/2 M V^2

E(sphere) = 1 + 2/3 = 1.67

E(cylinder) / E(sphere) = 2 / 1.67 = 1.2

The cylinder initially has 1.20 the energy of the sphere

The PE attained is proportional to the initial KE

H(sphere) = 2.87 / 1/2 = 2.40 m since it has less initial KE

5 0

2 years ago

In a physics lab experiment, a compressed spring launches a 24 g metal ball at a 35o angle above the horizontal. Compressing the

Levart [38]

Answer:

k = 45.95 N/m

Explanation:

First, we will find the launch speed of the ball by using the formula for the horizontal range of the projectile.

$R = \frac{v_{o}^{2}\ Sin\ 2\theta}{g} \\\\v_{o}^{2} = \frac{Rg}{Sin\ 2\theta}\\$

where,

Vo = Launch Speed = ?

R = Horizontal Range = 5.3 m

θ = Launch Angle = 35°

Therefore,

$v_{o}^{2} = \frac{(5.3\ m)(9.81\ m/s^{2})}{Sin\ 2(35^{o})}\\$

v₀² = 55.33 m²/s²

Now, we know that the kinetic energy gain of ball is equal to the potential energy stored by spring:

$Kinetic\ Energy\ Gained\ By\ Ball = Elastic\ Potential\ Energy\ Stored\ in \ Spring\\\frac{1}{2}mv_{o}^{2} = \frac{1}{2}kx^{2}\\\\k = \frac{mv_{o}^{2}}{x^2} \\$

where,

k = spring constant = ?

x = compression = 17 cm = 0.17 m

m = mass of ball = 24 g = 0.024 kg

Therefore,

$k = \frac{(0.024\ kg)(55.33\ m^2/s^2)}{(0.17\ m)^2} \\$

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3 years ago

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boyakko [2]

Answer:

coefficient

Explanation:

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Ksju [112]

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3 years ago