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2 years ago

14

Find the time t1 it takes to accelerate the flywheel to ω1 if the angular acceleration is α. express your answer in terms of ω1

and α.

2 answers:

Ivahew [28]2 years ago

6 0

the time t1 it takes to accelerate:

$\large{\boxed{\bold{t_1=\frac{\omega_1 }{\alpha}}}}$

<h3>Further explanation </h3>

The speed of the rotating object is called angular velocity. This value is part of circular motion.

The angular velocity is denoted by w including the vector quantity. The unit used is usually rpm (rotation per minute) or radians / sec

Average angular velocity is the ratio of angular displacement to the time interval

$\rm \omega=\dfrac{\Delta \theta}{\Delta t}$

Circular motion with constant acceleration (angular acceleration denoted by α) can be formulated:

$\rm \omega=\omega o+\alpha t$

ωo = initial angular velocity

ω = angular velocity after t

the flywheel accelerate to ω1, so:

wo = 0 because The flywheel is assumed to be at rest at time t = 0

so the equation becomes:

ω1 = ωo + αt

ω1 = 0 + αt

ω1 = αt

$\large{\boxed{\bold{t=\frac{ \omega_1}{\alpha }}}$

Learn more

the average velocity

brainly.com/question/5248528

resultant velocity

brainly.com/question/4945130

velocity position

brainly.com/question/2005478

VMariaS [17]2 years ago

3 0

T1 = ω1/α.............

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If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it

Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of 27.0 m/s2 is

$\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

Explanation:

From the question we are told that

The mass of the rocket is m = 6300 kg

The velocity at gas is being ejected is u = 2000 m/s

The initial acceleration desired is $a = 27.0 \ m/s$

The time taken for the gas to be ejected is t = 1 s

Generally this desired acceleration is mathematically represented as

$a = \frac{u * \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}$

Here $\frac{\Delta m}{\Delta t }$ is the rate at which gas is being ejected with respect to time

Substituting values

$27 = \frac{2000 * \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}$

=> $170100 -27* \frac{\Delta m}{\Delta t} = 2000 * \frac{\Delta m}{\Delta t}$

=> $170100 = 2027 * \frac{\Delta m}{\Delta t}$

=> $\frac{\Delta m}{\Delta t} = \frac{170100}{2027}$

=> $\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

3 0

2 years ago

The speed of light in a transparent medium is 0.6 times that of its speed in vacuum. Find the refractive index of the medium.

grin007 [14]

<h2>Answer:</h2>

The refractive index is 1.66

<h2>Explanation:</h2>

The speed of light in a transparent medium is 0.6 times that of its speed in vacuum .

Refractive index of medium = speed of light in vacuum / speed of light in medium

So

RI = 1/0.6 = 5/3 or 1.66

3 0

2 years ago

Read 2 more answers

A firecracker breaks up into several pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.

MakcuM [25]

Answer:

The magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.

Explanation:

Hi there!

The total momentum is calculated as the sum of the momenta of the pieces.

The momentum of each piece is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass.

v = velocity.

The momentum is a vector. The 200 g-piece flies along the x-axis then, its momentum will be:

p = (m · v, 0)

p = (0.200 kg · 82.0 m/s, 0)

p = (16.4 kg m/s, 0)

The 300 g-piece flies along the y-axis. Its momentum vector will be:

p =(0, m · v)

p = (0, 0.300 kg · 45.0 m/s)

p = (0, 13.5 kg m/s)

The total momentum is the sum of each momentum:

Total momentum = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)

Total momentum = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)

Total momentum = (16.4 kg m/s, 13.5 kg m/s)

The magnitude of the total momentum is calculated as follows:

$|p| = \sqrt{(16.4 kgm/s)^2+(13.5 kg m/s)^2}= 21.2 kg m/s$

The direction of the momentum vector is calculated using trigonometry:

cos θ = px/p

Where px is the horizontal component of the total momentum and p is the magnitude of the total momentum.

cos θ = 16.4 kg m/s / 21.2 kg m/s

θ = 39.3 (39.5° if we do not round the magnitude of the total momentum)

Then, the magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.

6 0

2 years ago

The thrust of a certain boat’s engine generates a power of 10kW as the boat moves at constant speed 10ms through the water of a

Lunna [17]

Answer:

The change in power is 4400 W.

Explanation:

Given that,

Power = 10 kW

Speed = 10 m/s

Increases speed = 12 m/s

Given equation is,

$F=kv$

We know that,

The power is,

$P=Fv$

Put the value of F into the formula

$P=(kv)v$

$P=kv^2$

$P\propto v^2$

We need to calculate the new power

Using formula for power

$\dfrac{P}{P'}=\dfrac{v^2}{v'^2}$

Put the value into the formula

$\dfrac{10}{P'}=(\dfrac{10}{12})^2$

$P'=(\dfrac{12}{10})^2\times10$

$P'=14.4\ kW$

We need to calculate the change in power

Using formula of change in power

$\Delta P=P'-P$

Put the value into the formula

$\Delta P=14.4-10$

$\Delta P=4.4\ kW$

$\Delta P=4.4\times1000$

$\Delta P=4400\ W$

Hence, The change in power is 4400 W.

6 0

2 years ago

An 80-kg person stands at one end of a 130-kg boat. He then walks to the other end of the boat so that the boat moves 80 cm with

a_sh-v [17]

Explanation:

We assume that length of boat is L cm. And, we suppose to have a coordinate system relative to the bottom of the lake with:

Boat's position initially from x = 0 cm to x = L cm

Let person (S) is initially at x = L cm (right hand end of boat)

and, let boat's center of mass at x = C cm

Therefore, initially center of mass of overall (system's) is at $X_{1}$ where (taking moments about x = 0).

$(80 + 130)X_{1}$

= 130C + 80L ........... (i)

After S walks left to the other end of the boat, the boat moves right 80 cm.

Hence,

Boat's position is now from x = 80 cm to x = (L+ 80) cm

S's position is x = 80 (left end of boat)

Boat's center of mass is at (C + 80).

Therefore, overall (system's) final center of mass is at $X_{2}$ where (taking moments about x = 0).

$(80 + 130)X_{2}$

= $(130)(C + 80) + 80 \times 80$ ............. (ii)

As no external force is acted on the system, the system's center of mass has not moved, so $X_{1} = X_{2}$ .

This means the left hand sides of both equations (i) and (ii) are equal. So, we can equate the right hand sides giving: the equation as follows.

130C + 80L = $(130)(C+80) + 80 \times 80
$

80L = 16800 cm

L = 210 cm

= 2.1 m

Length of boat = 2.1 m

When the person is walking from one end to the other as there was no external force then center of mass of the person-boat system did not move.

6 0

2 years ago