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2 years ago

I need to calculate a coefficient of linear expansion given the following

1 answer:

4 0

The coefficient of linear expansion is

<em>(change of length, feet) / (original length, feet)</em> per °F.

Since the numerator and denominator of the fraction have the same
unit, the coefficient turns out to be just a number, with no unit.

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Calculate the time period of simple pendulum whose length is 98.2cm

alekssr [168]

The time period resulting in oscillations will be 1.986 seconds.

<h3>What is the period of oscillation?</h3>

The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.

The time period of the oscillation is;

$\rm T = 2 \pi\sqrt{\frac{L}{g}} \\\\ \rm T = 2 \times 3.14 \times \sqrt{ \frac{98.2 \ \times 10^{-2} \ m}{9.81 \ m/s^2}} \\\\ T= 1.986 \ sec$

Hence the time period resulting oscillations will be 1.986 seconds.

To learn more about the time period of oscillation refer to the link;

brainly.com/question/20070798

#SPJ1

5 0

1 year ago

Help me please I only have a few minutes to get this doneplease and thank q

DIA [1.3K]

A: particles are more spread out in gas

6 0

2 years ago

A 5 Kg bowling ball is thrown at a stationary 1.6 Kg bowling pin at 5 m/s. If the final velocity of the ball is 2.5 m/s. The fin

Sergio039 [100]

Answer: hope this helps

Explanation:

7 0

2 years ago

A solid cylinder with a mass of 2.72 kg and a radius of 0.083 m starts from rest at a height of 4.20 m and rolls down a 88.7 ◦ s

Bingel [31]

Explanation:

According to the law of conservation of energy ,

Potential energy = kinetic energy

$mgh = \frac{1}{2} \times mv^{2} + \frac{1}{2} \times I \times \omega^{2}$

I = $\frac{mr^{2}}{2}$

$\omega = \frac{v}{r}$

$mgh = [\frac{1}{2} \times mv^{2}] + [\frac{1}{2} \times (\frac{mr^{2}}{2}) \frac{v^{2}}{r^{2}}]$

mgh = $[\frac{1}{2} \times mv^{2}] + [\frac{1}{4} \times mv^{2}]$

$g \times h = \frac{3}{4} \times v^{2}$

$9.8 \times 4.2 = \frac{3}{4} \times v^{2}$

v = 7.4 m/s

thus, we can conclude that the translational speed of the cylinder when it leaves the incline is 7.4 m/s.

5 0

2 years ago

A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of

adelina 88 [10]

Answer:

x = 11.23 m

Explanation:

For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.

Let's reduce to SI system units

θ = 155 rev (2pi rad / rev) = 310π rad

α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²

Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)

w² = w₀² + 2 α θ

w =√ 2 α θ

w = √(2 4pi 310pi)

w = 156.45 rad / s

The relationship between angular and linear velocity

v = w r

v = 156.45 0.175

v = 27.38 m / s

In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive

y = $v_{oy}$ t - ½ g t²

As it leaves the highest point its speed is horizontal

y = 0 - ½ g t²

t = √ (-2y / g)

t = √ (-2 (-0.820) /9.8)

t = 0.41 s

With this time we calculate the horizontal distance, because the constant horizontal speed

x = vox t

x = 27.38 0.41

x = 11.23 m

5 0

3 years ago