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3 years ago

A motorcycle is capable of accelerating at 5.1 m/s starting feom rest how far can it travel im 1.5 seconds

Physics

1 answer:

saul85 [17]3 years ago

6 0

5.1 m/ s

~Multiply 1.5 on both sides since you’re trying to finding out how far it can travel in 1.5 seconds

5.1 x 1.5= 7.65

s x 1.5= 1.5

So a motorcycle can travel 7.65 m in 1.5 seconds.

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A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

alexgriva [62]

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = $I_t$

Rotational inertia ( $I_r$ ) of the record = $0.61\times I_t$

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as $\omega _i\ ,\ \omega_f$ respectively.

Note :

Angular momentum $(L)$ = Product of moment of inertia $(I)$ and angular velocity $(\omega)$ .

Lets say,

⇒ initial angular momentum = final angular momentum

⇒ $L_i=L_f$

⇒ $(I_t)\times \omega_i = (I_t+I_r)\times \omega_f$

⇒ $\omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i)$ ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ $K_i=\frac{I_t\times \omega_i^2}{2}$ ⇒ $K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}$

Their ratios:

⇒ $\frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }$

⇒ $\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}$

Plugging the values of $\omega _f^2$ as $\omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2$ from equation (i) in the ratios of the Kinetic energies.

⇒ $\frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}$

Now,

The Kinetic energy lost in fraction can be written as:

⇒ $\frac{K_f-K_i}{K_i}$

Now re-arranging the terms.

$\frac{K_f-K_i}{K_i} =(\frac{K_f}{K_i} -1)= \frac{I_t}{I_t+I_r} -1=\frac{I_t-I_t-I_r}{I_t+I_r} =\frac{-I_r}{(I_t+I_r)}$

Plugging the values of $I_r$ and $I_t$ .

⇒ $\frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788$

To find the percentage we have to multiply it with $100$ and here negative means for loss of Kinetic energy.

⇒ $\frac{K_f}{K_i} = =-0.3788\times 100= 37.88$

So the percentage of the initial Kinetic energy lost is 37.88

4 0

3 years ago

It decomposes in a first-order reaction with a rate constant of 14 s–1. how long would it take for an initial concentration of 0

VikaD [51]

The rate constant of a reaction can be computed by the ratio of the changes in the concentration and time take taken for it to decompose. Thus, if the rate constant is given to be 14 M/s, we have

$rate = \frac{-(C_{new} - C_{old})}{t}$

where C are the concentration values and t is the time taken for it to decompose.

$14 = \frac{-(0.02 - 0.06)}{t}$
$t = 0.003 s$

Thus, it will take 0.003 s for it to decompose.
Answer: 0.003 s

4 0

3 years ago

the cycle of the moon through its phases, or the synodic month, is a. 21 days long. b. 27 1/3 days long. c. 29 1/2 days long. d.

olasank [31]

It takes 29.5 days long

7 0

3 years ago

An artist working on a piece of metal in his forging studio plunges the hot metal into oil in order to harden it. The metal piec

lutik1710 [3]

Answer:

The temperature of the metal is $T_m = 376.8 ^o C$