Circle the larger unit:

A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.58 m. A child appl

lozanna [386]

Answer:

The value is $KE = 259.6 \ J$

Explanation:

From the question we are told that

The weight of the horizontal solid disk is $W = 805 \ N$

The radius of the horizontal solid disk is $r = 1.58 \ m$

The force applied by the child is $F = 49.5 \ N$

The time considered is $t = 2.95 \ s$

Generally the mass of the horizontal solid disk is mathematically represented as

$m_h = \frac{W}{ g}$

=> $m_h = \frac{805}{ 9.8 }$

=> $m_h = 82.14 \ N$

Generally the moment of inertia of the horizontal solid disk is mathematically represented as

$I = \frac{1}{2} * m * r^ 2$

=> $I = \frac{1}{2} * 82.14 * 1.58^ 2$

=> $I = 102.5 \ kg \cdot m^2$

Generally the net torque experienced by the horizontal solid disk is mathematically represented as

$T = I * \alpha = F * r$

=> $\alpha = \frac{ F * r }{ I }$

=> $\alpha = \frac{ 49.5 * 1.58 }{ 102.53 }$

=> $\alpha = 0.7628$

Gnerally from kinematic equation we have that

$w = w_o + \alpha t$

Here $w_o$ is the initial angular velocity velocity of the horizontal solid disk which is $w_o = 0\ rad/s$

So

$w = 0 + 0.7628 * 2.95$

=> $w = 2.2503 \ rad/s$

Generally the kinetic energy is mathematically represented as

$KE = \frac{1}{2} * I * w^2$

=> $KE = \frac{1}{2} * 102.53 * 2.2503 ^2$

=> $KE = 259.6 \ J$

8 0

2 years ago

A longitudinal wave is a combination of a transverse wave and a surface wave.(true or false)

Lisa [10]

Answer: True

Explanation:

8 0

3 years ago

Josh and Jake are both helping to build a brick wall which is 6 meters in height. They each lay 250 bricks, but Josh finishes th

Molodets [167]

<span>Each laid 250 bricks but while Jake was still working, Josh was lounging in the shade. Josh has more power but that power was only on for 3 hours out of 4.5. Obviously Josh could get more done is less time as long as he keeps working. Jake will get the hang of it soon.</span>

5 0

2 years ago

Which statement describes the magnetic field inside a bar magnet?

kakasveta [241]

The statement that describes the magnetic field inside a bar magnet is as follows: it points from north to south.

<h3>What is a bar magnet?</h3>

A bar magnet is a permanent magnet of rectangular shape.

A magnet generally possess a magnetic field, which is a condition in the space around a magnet which there is a detectable magnetic force and the presence of two magnetic poles.

A bar magnet like every other magnet possesses a magnetic field that points from the north pole to the south pole.

Learn more about magnets at: brainly.com/question/13026686

#SPJ1

5 0

2 years ago

In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st

bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

Total time will be the sum of the partial times between stations plus the time stopped at the stations.
Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

$v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s (1)$

Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

$t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)$

Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

$x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m (3)$

In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

$t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)$

We can find the distance traveled while the train was decelerating as follows:

$x_{3} = (v_{1} * t_{3}) + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m (5)$

Finally, we need to know the time traveled at constant speed.
So, we need to find first the distance traveled at the constant speed of 26.4m/s.
This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)

The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

$t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s (7)$

Total time between two stations is simply the sum of the three times we have just found:
t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
tm = t*5 = 131.6 * 5 = 658.2 s (9)
Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

Using all the same premises that for a) we know that the only difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
We can find the distance traveled at constant speed, rewriting (6) as follows:

x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)

The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

$t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s (12)$

Total time between two stations is simply the sum of the three times we have just found:
t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
tm = t*3 = 207.4 * 3 = 622.2 s (14)
Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)

7 0

2 years ago