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wel
3 years ago
7

Show the equation of simple pendulum to be dimensionally consistent

Physics
1 answer:
nataly862011 [7]3 years ago
6 0
T is in seconds (s) 

<span>2pi is dimensionless </span>

<span>L is in meters (m) </span>

<span>g is in meters per second squared (m/s^2) </span>

<span>so you can write the equation for the period of the simple pendulum in its units... </span>

<span>s=sqrt(m/(m/s^2)) </span>

<span>simplify</span>

<span>s=sqrt(m*s^2*1/m) cancelling the m's </span>

<span>s=sqrt(s^2) </span>

<span>s=s </span>

<span>therefore the dimensions on the left side of the equation are equal to the dimensions on the right side of the equation.</span>
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The lightest car in the world was built in London and had a mass of less than 10 kg. it's maximum speed was 25.0 km/h. Suppose t
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15. If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the
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The final combined velocity after the collision is 20.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two car and of the truck must be conserved before and after the collision.

This means that we can write the following equation:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 = 800 kg is the mass of the sport car

u_1 = 13.0 m/s is the initial velocity of the car (taking its direction as positive  direction)

m_2 = 1200 kg is the mass of the truck

u_2 = 25.0 m/s is the initial velocity of the truck

v is the final combined velocity of the car and the truck, after the collision

Re-arranging the equation and substituting the values, we find the velocity after the collision:

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(800)(13)+(1200)(25)}{800+1200}=20.2 m/s

And the positive sign indicates their final direction is the same as the initial direction of the two vehicles.

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A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener
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Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

Explanation:

Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

The angular speed is calculated as follows;

\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s

(a) the force constant of the spring

\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m

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3 years ago
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