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2 years ago

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An object has an angular velocity of 1.0 rad and an angular acceleration of 0.5 rad/s. Is the speed of its rotation increasing o

r decreasing?

1 answer:

alexdok [17]2 years ago

6 0

Answer:

increasing

Explanation:

Given that,

Initial angular speed =1 rad/s

Angular acceleration = 0.5 rad/s²

We know that the angular acceleration is given by :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

Where

$\omega_f$ city and t is time is the final angualr velo

As the angular acceleration of the object is positive. It means that the final speed of the object is more than that initial. As a result, the speed of its rotation is increasing.

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You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

1 year ago

Read 2 more answers

Một nười thả 420g chì ở nhiệt độ 100°C và 260g nước ở nhiệt độ 58°C làm cho nước nóng lên tới 60°C . Cho nhiệt dung riêng của nư

vlabodo [156]

Hey what language is this?

6 0

2 years ago

How long will it take a ball to roll 10 meters along the floor at a speed of 0.5 m/s?

IRINA_888 [86]

Answer:

20 seconds.

Explanation:

The following data were obtained from the question:

Distance = 10 m

Speed = 0.5 m/s

Time =...?

The speed of an object is simply defined as the distance travelled by the object per unit time. Mathematically, it is expressed as:

Speed = Distance /time

With the above formula, we can obtain the time taken for the ball to travel a distance of 10 m as shown below:

Distance = 10 m

Speed = 0.5 m/s

Time =...?

Speed = Distance /time

0.5 = 10/time

Cross multiply

0.5 × time = 10

Divide both side by 0.5

Time = 10/0.5

Time = 20 secs.

Therefore, it will take 20 seconds for the ball to travel a distance of 10 m.

6 0

3 years ago

What velocity does a 2kg mass have when its kinetic energy is 16 J

alexandr402 [8]

We can use the equation for kinetic energy, K=1/2mv².
Your given variables are already in the correct units, so we can just plug in the variables and solve for v.

K = 1/2mv²
16 = 1/2(2)v²
16 = (1)v²
√16 = v
v = 4 m/s

Therefore, the velocity of a 2 kg mass with 16 J of kinetic energy is 4 m/s.
Hope this is helpful!

7 0

2 years ago

Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving

Iteru [2.4K]

Explanation:

Let $m_p$ is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.

The magnetic force is balanced by the centripetal force acting on the proton as :

$\dfrac{mv^2}{r}=qvB$

r is the radius of path,

$r=\dfrac{mv}{qB}$

Time period is given by :

$T=\dfrac{2\pi r}{v}$

$T=\dfrac{2\pi m_p}{qB}$

Frequency of proton is given by :

$f=\dfrac{1}{T}=\dfrac{qB}{2\pi m_p}$

The wavelength of radiation is given by :

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{2\pi m_pc}{qB}$

So, the wavelength of radiation produced by a proton is $\dfrac{2\pi m_pc}{qB}$ . Hence, this is the required solution.

3 0

2 years ago