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3 years ago

How is acceleration impacted when mass remains the same but the amount of force is decreased?

1 answer:

3 0

The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

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Plz help me I’m a desperate little boy

Neporo4naja [7]

It is the first one. But if you have this question you are not a little boy

5 0

3 years ago

What are possible units for impulse? Check all that apply.

Neporo4naja [7]

Units of impulse: N • s, kg • meters per second

Explanation:

Impulse is defined in two ways:

Impulse is defined as the product between the force exerted in a collision and the duration of the collision:

$I=F\Delta t$

where

F is the force

$\Delta t$ is the time interval

Since the force is measured in Newtons (N) and the time is measured in seconds (s), the units for the impulse are

$[I] = [N][s]$

So,

N • s

Impulse is also defined as the change in momentum experienced by an object:

$I=\Delta p$

where the change in momentum is given by

$\Delta p = m\Delta v$

where m is the mass and $\Delta v$ is the change in velocity.

The mass is measured in kilograms (kg) while the change in velocity is measured in metres per second (m/s), therefore the units for impulse are

$[I]=[kg][m/s]$

so,

kg • meters per second

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

6 0

3 years ago

Read 2 more answers

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.

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