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3 years ago

What is the drawback to use period of pendulum as time standard

1 answer:

3 0

The time period of a simple pendulum depends upon its length and value of'g'at any place. The period of the pendulum of fixed length

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A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length ℓ and of negligible

mr Goodwill [35]

Answer:

$mv l$

$\frac{M }{(M + m)}$

Explanation:

Complete question statement is as follows :

A wooden block of mass M resting on a friction less, horizontal surface is attached to a rigid rod of length ℓ and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.

(a) What is the angular momentum of the bullet–block system about a vertical axis through the pivot? (Use any variable or symbol stated above as necessary.)

(b) What fraction of the original kinetic energy of the bullet is converted into internal energy in the system during the collision? (Use any variable or symbol stated above as necessary.)

$m$ = mass of the bullet

$v$ = velocity of the bullet before collision

$r$ = distance of the line of motion of bullet from pivot = $l$

$L$ = Angular momentum of the bullet-block system

Angular momentum of the bullet-block system is given as

$L = m v r$

$L = mv l$

$V$ = final velocity of bullet block combination

Using conservation of momentum

Angular momentum of bullet block combination = Angular momentum of bullet

$(M + m) V l = m v l\\V =\frac{mv}{(M + m)}$

$K_{o}$ = Initial kinetic energy of the bullet

Initial kinetic energy of the bullet is given as

$K_{o} = (0.5) m v^{2}$

$K_{f}$ = Final kinetic energy of bullet block combination

Final kinetic energy of bullet block combination is given as

$K_{f} = (0.5) (M + m) V^{2}$

Fraction of original kinetic energylost is given as

Fraction = $\frac{(K_{o} - K_{f})}{K_{o}} = \frac{((0.5) m v^{2} - (0.5) (M + m) V^{2})}{(0.5) m v^{2}}$

Fraction = $\frac{(m v^{2} - (M + m) (\frac{mv}{(M + m)})^{2})}{m v^{2}} = \frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}$

Fraction = $\frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}\\ \frac{M }{(M + m)}$

6 0

3 years ago

Newton's Third Law of Motion relates to action and reaction. Which of the following scenarios accurately names the correct actio

Tomtit [17]

Newton’s Third Law of Motion states that for every action there is an equal and opposite reaction. So look for a scenario in which something had force applied upon it and the reaction is a force in the opposite direction of the same size.

5 0

2 years ago

A bullet is shot vertically upward with an initial velocity of 128 ft/s. The bullet's height after t seconds is y(t) = 128t - 16

saul85 [17]

The height of the bullet when the velocity is zero is 256 ft.

<h3>Height of the bullet when the velocity is zero </h3>

The height of the bullet when the velocity is zero is determined by taking derivative of the function as shown below;

$v = \frac{dy}{dt} = 128 -32t \\\\when \ v \ is \ zero\\\\v = 0\\\\128 - 32t = 0\\\\32t = 128\\\\t = \frac{128}{32} \\\\t = 4 \ s$

The height of the bullet at this time is calculated as follows;

$y(4) = 128(4) - 16(4)^2\\\\y(4) = 256 \ ft$

Learn more about height of projectiles here: brainly.com/question/10008919

6 0

3 years ago

Can an object with constant acceleration reverse its direction of travel? Can it reverse its direction twice? Explain.

Neko [114]

Answer:yes

Explanation:The constan acceleration means that it wont stop moving but if you kick it a different direction then it will change direction

6 0

3 years ago

One speed skater starts across a frozen lake at an average speed of 8 m/s. Ten seconds later, a second speed skater starts from

Luba_88 [7]

Answer:

40 s

Explanation:

After 10 seconds, the first skater would have a 8m/s * 10s = 80 m head start

Let t be the number of seconds after the second skater starts will the second skater overtake the first skater

The distance traveled by the first skater after t seconds is

$s_1 = v_1t = 8t$

Similarly the distance traveled by the 2nd skater after t seconds is

$s_2 = v_2t = 10t$

Since the 2nd skater catches up to the 1st one after 80 m behind, the distance traveled by the 2nd one must be 80m greater than the distance of the 1st skater

$s_2 = s_1 + 80$

We can substitute $s_1 = 8t, s_2 = 10t$

$10t = 8t + 80$

$2t = 80$

$t = 80 / 2 = 40 s$

7 0

3 years ago