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4 years ago

2 thousand+12 hundreds

Physics

2 answers:

Firlakuza [10]4 years ago

6 0

Answer:

3,200

Explanation:

3,200

Umnica [9.8K]4 years ago

4 0

Answer:

3200 is the answer have a good day

Explanation:

hope this helps? :))

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A horizontal vinyl record of mass 0.10 kg and radius 0.10 m rotates freely about a vertical axis through its center with an angu

Tresset [83]

Answer:

The angular speed of the record is 3.36 rad/s.

Explanation:

Given that,

Mass of record= 0.10 kg

Radius = 0.10 m

Angular speed = 4.7 rad/s

Moment of inertia $I=5.0\times10^{-4}\ kgm^2$

Mass of putty = 0.020 kg

We need to calculate the angular speed

Using law of conservation of momentum

$L_{i}=L_{f}$

$I\omega_{i}=(I+mr^2)\omega_{f}$

$\omega_{f}=\dfrac{I\omega_{i}}{(I+mr^2)}$

Put the value into the formula

$\omega_{f}=\dfrac{5.0\times10^{-4}\times4.7}{5.0\times10^{-4}+0.020\times(0.10)^2}$

$\omega_{f}=3.36\ rad/s$

Hence, The angular speed of the record is 3.36 rad/s.

3 0

4 years ago

A tank is 6 m long, 4 m wide, 5 m high, and contains kerosene with density 820 kg/m3 to a depth of 4.5 m. (Use 9.8 m/s2 for the

Mazyrski [523]

The pressure at the bottom of the tank is given by:

P = ρgh

P = pressure, ρ = fluid density, g = gravitational acceleration, h = depth

Given values:

ρ = 820kg/m³, g = 9.8m/s², h = 4.5m

Plug in and solve for P:

P = 820(9.8)(4.5)

P = 36000Pa

5 0

3 years ago

100g converted to kg .

natka813 [3]

Answer: 0.1 kilogram

4 0

3 years ago

Which term best describes a test used to answer a question?

Anastasy [175]

A) experiment. Is the answer.

hypothesis is the educated guess about what the result of the experiment is before conducting the experiment.

Observation is what you see and record during the experiment.

3 0

4 years ago

A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result

vivado [14]

Answer:

$21.6\ \text{kg m}^2$

$3.672\ \text{Nm}$

$54.66\ \text{revolutions}$

Explanation:

$\tau$ = Torque = 36.5 Nm

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2$

Torque is given by

$\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2$

The wheel's moment of inertia is $21.6\ \text{kg m}^2$

t = 60.6 s

$\omega_i$ = 10.3 rad/s

$\omega_f$ = 0

$\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2$

Frictional torque is given by

$\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}$

The magnitude of the torque caused by friction is $3.672\ \text{Nm}$

Speeding up

$\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}$

Slowing down

$\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}$

Total number of revolutions

$\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}$

$\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}$

The total number of revolutions the wheel goes through is $54.66\ \text{revolutions}$ .

3 0

3 years ago