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2 years ago

A flowerpot weighing 2 kilograms, what will be its speed at the end of 5 seconds?

1 answer:

8 0

That depends on what its speed was at the beginning of 5 seconds,
and on what exactly happened to it.

If it got pushed off of the window sill and fell to the ground below,
then the answer would depend on how high the window sill is.

If the flower pot kept falling for the whole 5 seconds, then its speed
would be <em>49 meters per second</em> at the end of the 5 seconds. That's
because here on Earth, gravity increases the speed of falling things
by 9.8 meters per second for every second they fall.

Naturally, this answer is only true if the window sill is more than
122.5 meters above the ground ... about 402 feet, or 40 floors.
If it's lower than that, then the flower pot hits the ground in less
than 5 seconds, and its speed at the end of 5 seconds is zero.

By the way ...

-- It makes no difference how much the flower pot weighs.
All objects fall with the same acceleration, and hit the ground
at the same time.

-- 5 kilograms is a mass, not a weight.

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Natalija [7]

Answer:

I think its (A)The specific heat of an object explains how easily it changes temperatures.

Explanation:

6 0

2 years ago

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

Margaret [11]

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:

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

8 0

2 years ago

One end of a horizontal spring with force constant 130.0 Ni'm is attached to a vertical wall. A 4.00-kg block sitting on the flo

tekilochka [14]

Answer:

Explanation:

When the spring is compressed by .80 m , restoring force by spring on block

= 130 x .80

= 104 N , acting away from wall

External force = 82 N , acting towards wall

Force of friction acting towards wall = μmg

= .4 x 4 x 9.8

= 15.68 N

Net force away from wall

= 104 -15.68 - 82

= 6.32 N

Acceleration

= 6.32 / 4

= 1.58 m / s²

It will be away from wall

Energy released by compressed spring = 1/2 k x²

= .5 x 130 x .8²

= 41.6 J

Energy lost in friction

= μmg x .8

= .4 x 4 x 9.8 x .8

= 12.544 J

Energy available to block

= 41.6 - 12.544 J

= 29 J

Kinetic energy of block = 29

1/2 x 4 x v² = 29

v = 3.8 m / s

This will b speed of block as soon as spring relaxes. (x = 0 )

4 0

2 years ago

Which of the following lists the composition of the Earth’s outer core?

iris [78.8K]

Well,

The outer core of the Earth is mostly composed of iron and nickel.

The correct option is C.

5 0

2 years ago

an object of mass 20kg is lifted to a 25m building. how much potential energy is stored on a mass?(take g=10m/s²)

Goshia [24]

Answer:

The answer is 5000 Joules

4 0

2 years ago