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Naily [24]
3 years ago
13

What is the mass of metal if it has a density of 12.459 hg/cm^3 and displaces 28.7cm^3 of water ​

Physics
1 answer:
natima [27]3 years ago
7 0

Answer:

357.6g

Explanation:

Given parameters:

Density  = 12.459g/cm³

Volume of metal  = 28.7cm³

Unknown:

Mass of metal = ?

Solution:

The density of a substance is its mass per unit volume.

To find the mass;

     Mass of metal  = density x volume

Now insert the parameters and solve;

    Mass of metal  = 12.459 x 28.7  = 357.6g

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Explanation:

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Answer:

0.2 m

Explanation:

PHASE 1

First, we calculate the distance the tongue moved in the first 20 ms (0.02 secs). We use one of Newton's equations of linear motion:

s = ut + \frac{1}{2}at^2

where u = initial velocity = 0 m/s

a = acceleration = 250 m/s^2

t = time = 0.02 s

Therefore:

s = 0 + \frac{1}{2} * 250 * (0.02)^2\\\\\\s = 0.05 m

PHASE 2

Then, for the next 30 ms (0.03 secs), we use the formula:

distance = speed * time

This speed is the same as the final velocity of the tongue after the first 20 ms.

This can be obtained by using the formula:

v = u + at\\\\=> v = 0 + (250 * 0.02)\\\\\\v = 5 m/s

Therefore:

distance = 5 * 0.03 = 0.15 m

Therefore, the total distance moved by the tongue in the 50 ms interval is:

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3 years ago
which of the following objects is considered negatively charged A - one with excess electrons on the surface B - one with excess
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A spinning wheel has a rotational inertia of 2 kg•m². It has an angular velocity of 6.0 rad/s. An average counterclockwise torqu
ira [324]

Answer:

-20.0 kg m^2/s

Explanation:

The angular momentum of an object in rotation is given by

L=I \omega

where

I is the moment of inertia

\omega is the angular speed

In this problem, initially we have

I=2 kg m^2 is the moment of inertia of the wheel

\omega_i = 6.0 rad/s is the initial angular speed

So the initial angular momentum is

L_i = I\omega_i = (2)(6.0)=12 kg m^2/s

Later, a counterclockwise torque of

\tau=-5.0 Nm is applied

So the angular acceleration of the wheel is:

\alpha = \frac{\tau}{I}=\frac{-5.0}{2}=-2.5 rad/s^2 in the direction opposite to the initial rotation.

As a result, the final angular velocity of the wheel will be:

\omega_f = \omega_i + \alpha t

where

t = 4.0 is the time interval

Solving,

\omega_f = +6.0 +(-2.5)(4.0)=-4 rad/s

which means that now the wheel is rotating in the counterclockwise direction.

Therefore, the new angular momentum of the wheel is:

L_f = I\omega_f =(2)(-4.0)=-8.0 kg m^2/s

So, the change in angular momentum is:

\Delta L=L_f - L_i = (-8.0-(12))=-20.0 kg m/s^2

7 0
3 years ago
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