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3 years ago

7

If you apply an unbalanced force to this object it will ______. Mention all that will apply: change velocity, accelerate, change

position

1 answer:

Papessa [141]3 years ago

6 0

Answer:

All of them: change velocity, accelerate, change position

Explanation:

We can answer this question by using Newton's second law:

F = ma

where

F is the net force on the object

m is the mass of the object

a is the acceleration

We notice that when there is an unbalanced force on the object, $F\neq 0$ , and therefore

$a\neq 0$

whcih means that the object will accelerate.

But acceleration is the rate of change of velocity, v:

$a=\frac{\Delta v}{\Delta t}$

And so,

$\Delta v \neq 0$

which means that the object will change velocity.

If the object is changing velocity, this means that it is also moving: therefore, the position of the object must be changing, so also the option "change position" is correct.

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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

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Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

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3 years ago

A student used a tuning fork of frequency 320 Hz and observed that the speed of sound was 339 m/s. Calculate the wavelength of t

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$\lambda = \frac{v}{f}$

Where,

v = Velocity

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According to our values the frequency (f) is 320Hz and the speed (v) is 339m / s.

Replacing in the given equation we have to,

$\lambda = \frac{v}{f}\\\lambda = \frac{339}{320}\\\lambda = 1.059m\approx 1.06m$

Therefore the wavelength of this sound wave is 1.06m

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3 years ago

A hoop, a solid cylinder, a spherical shell, and a solid sphere are placed at rest at the top of an inclined plane. All the obje

iogann1982 [59]

Answer:

Solid sphere will reach first

Explanation:

When an object is released from the top of inclined plane

Then in that case we can use energy conservation to find the final speed at the bottom of the inclined plane

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now we have

$I = mk^2$

here k = radius of gyration of object

also for pure rolling we have

$v = R\omega$

so now we will have

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(\frac{v^2}{R^2})$

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$v^2 = \frac{2gh}{1 + \frac{k^2}{R^2}}$

so we will say that more the value of radius of gyration then less velocity of the object at the bottom

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So it will take more time for the object to reach the bottom which will have more radius of gyration

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$mk^2 = mR^2$

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$mk^2 = \frac{2}{3}mR^2$

$k = \sqrt{\frac{2}{3}} R$

For solid sphere

$mk^2 = \frac{2}{5}mR^2$

$k = \sqrt{\frac{2}{5}} R$

So maximum value of radius of gyration is for hoop and minimum value is for solid sphere

so solid sphere will reach the bottom at first

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3 years ago