If the mass of the object is doubled and the speed is halved then kinetic energy will change by a factor of:

Look at the diagram of a solar eclipse.

lubasha [3.4K]

Answer:

at location 4

Explanation:

it is because the solar eclipse happens when the moon is covering the sun

3 0

1 year ago

Points of maximum amplitude on a standing wave are called

kupik [55]

Answer:

Anti nodes

Explanation:

There are mainly two points in amplitude of standing wave one is where amplitude is zero and one is where amplitude is maximum

The point where the magnitude of standing wave is zero is called nodes and the points where magnitude of amplitude in standing wave is maximum is called anti nodes

So anti nodes will be answer

6 0

2 years ago

Determine the displacement and distance covered by a man if he walks 10 m north, turns east and walks 20 m, and then turns right

NemiM [27]

Answer:

20m

Explanation:

The two tens cancel each other out, as they are in opposite directions. Now we only care about the 20m, which if we have no 10's, will end up 20m away.

7 0

2 years ago

I really need help can anyone help

sergiy2304 [10]

Answer:

percentage w a slug preference (start) - 35% (0.35)

percentage w a fish preference (start) - 65% (0.65)

percentage w a slug preference (end) - 25% (0.25)

percentage w a fish preference (end) - 75% (0.75)

hope this helps kind stranger

5 0

1 year ago

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.

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