PLEASE HELP

A small object with mass 1.30 kg is mounted on one end of arod

Julli [10]

Answer:

(a) $I_{system} = 1.014\ kg.m^{2}$

(b) $\tau = 0.0179\ N-m$

Solution:

As per the question:

Mass of the object, m = 1.30 kg

Length of the rod, L = 0.780 m

Angular speed, $\omega = 5010\ rev/min$

Now,

(a) To calculate the rotational inertia of the system about the axis of rotation:

Since, the rod is mass less, the moment of inertia of the rotating system and that of the object about the rotation axis will be equal:

$I_{system} = ML^{2} = 1.30\times (0.780)^{2} = 0.791\ kg.m^{2}$

(b) To calculate the applied torque required for the system to rotate at constant speed:

Drag Force, F = $2.30\times 10^{- 2}\ N$

$\tau = FLsin\theta 90 = 2.30\times 10^{- 2}\times 0.780\times 1 = 0.0179\ N-m$

3 0

3 years ago

The process of _________________ replaces a faulty gene with a normal working gene.

ch4aika [34]

DNA transcription. ....

4 0

4 years ago

Read 2 more answers

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

4 years ago

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi

Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

$m_{m}$ = mass of metal = 400 g

$c_{m}$ = specific heat of metal = ?

$T_{mi}$ = initial temperature of metal = 100 °C

$m_{a}$ = mass of aluminum cup = 100 g

$c_{a}$ = specific heat of aluminum cup = 900.0 J/kg ∙ K

$T_{ai}$ = initial temperature of aluminum cup = 15 °C

$m_{w}$ = mass of water = 500 g

$c_{w}$ = specific heat of water = 4186 J/kg ∙ K

$T_{wi}$ = initial temperature of water = 15 °C

$T$ = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

$m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}$

7 0

4 years ago

Receiver maxima problem. When the receiver moves through one cycle, how many maxima of the standing wave pattern does the receiv

shusha [124]

Answer:

Two.

Explanation:

A standing wave or stationary wave are the type of wave in which it oscillates with time but the amplitude does not move.

One complete wave consists of two loops.
It means that when the receiver moves through one cycle, two maxima of standing wave pattern the receiver pass through.

8 0

3 years ago