All categories

2 years ago

If an object has a density of .9g/ml and a volume of 12ml what is the mass of the object

Physics

1 answer:

BaLLatris [955]2 years ago

3 0

Answer: 10.8 g

Explanation: Density = mass/ volume

So, Mass = density * volume = 0.9 g/ml * 12 ml = 10.8g

You might be interested in

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J

Explanation:

We have given the player turntable initially rotating at speed of $33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec$

Now speed is reduced by 75 %

So final speed $\frac{3.49\times 75}{100}=2.6175rad/sec$

Time t = 5.5 sec

From first equation of motion we know that '

$\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2$

(a) Now final velocity $\omega =0rad/sec$

So time t to come in rest $t=\frac{0-3.49}{-0.158}=22sec$

(b) The work done in coming rest is given by

$\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J$

4 0

2 years ago

Please help it’s do soon

viktelen [127]

When y’all are all over the place and I wanna is a time to come sit in my

8 0

2 years ago

Write SHORT examples of Gravitational potential energy and Elastic energy

Nikolay [14]

Dropping a bouncy ball and stretching a rubber ban.

6 0

2 years ago

Read 2 more answers

A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exe

Karolina [17]

Answer:

The velocity is 40 ft/sec.

Explanation:

Given that,

Force = 3200 lb

Angle = 30°

Speed = 64 ft/s

The resistive force with magnitude proportional to the square of the speed,

$F_{r}=kv^2$

Where, k = 1 lb s²/ft²

We need to calculate the velocity

Using balance equation

$F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}$

Put the value into the formula

$3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}$

Put the value of k

$3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}$

$1600-v^2=m\dfrac{d^2v}{dt^2}$

At terminal velocity $\dfrac{d^2v}{dt^2}=0$

So, $1600-v^2=0$

$v=\sqrt{1600}$

$v=40\ ft/sec$

Hence, The velocity is 40 ft/sec.

4 0

3 years ago

At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir

Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

$\omega = \omega_{o} + \alpha \cdot t$

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

Where:

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

$\omega_{o}$ , $\omega$ - Initial and final angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

$\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}$

$\theta-\theta_{o} = 134.88\,rad$

The final angular angular speed can be found by the equation:

$\omega = \omega_{o} + \alpha \cdot t$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)$

$\omega = 87.4\,\frac{rad}{s}$

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

$\Delta \theta = 134.88\,rad + 435\,rad$

$\Delta \theta = 569.88\,rad$

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

The angular acceleration is now cleared:

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$