Ask question

Ask question

All categories

3 years ago

13

A 2.00 kg object is moving in a circular path with a radius of 5.00 cm. The object starts from rest and with constant angular ac

celeration, obtains an angular velocity of 6.00 rad/s in 3.00 s. The object then comes to a stop with constant deceleration in 4.00 s. The centripetal component of acceleration of the object at 2.00 s is

1 answer:

Rom4ik [11]3 years ago

3 0

Answer:

0.800 m/s²

Explanation:

First, calculate the angular acceleration:

ω = αt + ω₀

6.00 rad/s = α (3.00 s) + 0 rad/s

α = 2.00 rad/s²

Now calculate the angular velocity at t = 2.00 s:

ω = αt + ω₀

ω = (2.00 rad/s²) (2.00 s) + 0 rad/s

ω = 4.00 rad/s

Calculate the linear velocity:

v = ωr

v = (4.00 rad/s) (0.0500 m)

v = 0.200 m/s

Finally, calculate the centripetal acceleration:

a = v² / r

a = (0.200 m/s)² / (0.0500 m)

a = 0.800 m/s²

You might be interested in

Which of the following items best embodies the physical property of conductivity?

Ostrovityanka [42]

Copper penny is the answer

8 0

3 years ago

Read 2 more answers

Why do hot stars look bluer than cool stars?

brilliants [131]

Answer:

(B) The wavelength that a star radiates the most energy is inversely proportional to the temperature.

Explanation:

As we know that

According to Wien's law wavelength is inverse proportional to the temperature .

λ.T = Constant.

λ.∝ 1 /T

As we know that star radiates wavelength and this wavelength is inverse proportional to the temperature of the star.

The temperature of cool star is cooler than the temperature of hot star that is cool star looks red and hot star looks blue.Cool star have low energy and hot star have high energy.

So option B is correct.

(B) The wavelength that a star radiates the most energy is inversely proportional to the temperature.

5 0

3 years ago

PLEASE HELP ME 45 POINTS

sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

$\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a$ (1)

Mass B

$\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a$ (2)

Where:

$T$ - Tension force in the cord, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

$m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g$

$(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g$

$a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g$

If we know that $m_{A} = 5\,kg$ , $m_{B} = 2\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the acceleration of the masses is:

$a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)$

$a = 4.203\,\frac{m}{s^{2}}$

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

$T = 28.02\,N$

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

$\Delta y = \frac{1}{2}\cdot a\cdot t^{2}$ (3)

Where:

$a$ - Net acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

$\Delta y$ - Covered distance, measured in meters.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $\Delta y = 1.5\,m$ , then the time taken by the system is:

$t = \sqrt{\frac{2\cdot \Delta y}{a} }$

$t = \sqrt{\frac{2\cdot (1.5\,m)}{4.203\,\frac{m}{s^{2}} } }$

$t \approx 0.845\,s$

The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is calculated by the following formula:

$v = a\cdot t$ (4)

Where $v$ is the final speed of the system, measured in meters per second.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $t \approx 0.845\,s$ , then the final speed of the system is:

$v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)$

$v = 3.551\,\frac{m}{s}$

The final speed of the system is 3.551 meters per second.

8 0

3 years ago

Which three quantities can be used to calculate acceleration?

PtichkaEL [24]

D is the correct answer, assuming that this is the special case of classical kinematics at constant acceleration. You can use the equation V = Vo + at, where Vo is the initial velocity, V is the final velocity, and t is the time elapsed. In D, all three of these values are given, so you simply solve for a, the acceleration.
A and C are clearly incorrect, as mass and force (in terms of projectile motion) have no effect on an object's motion. B is incorrect because it is not useful to know the position or distance traveled, unless it will help you find displacement. Even then, you would not have enough information to use a kinematics equation to find a.

4 0

3 years ago

Which of the moon's properties prevents it from being pulled inward by Earth?

allsm [11]

The answer is Inertia

4 0

3 years ago

Read 2 more answers