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2 years ago

Number these from least (1) to most (5) mass ?

Physics

2 answers:

elena-14-01-66 [18.8K]2 years ago

6 0

1. a feather

2. a baseball

3.a small car

4. a truck

5 a large train

alisha [4.7K]2 years ago

5 0

Answer:

1. A feather

2. A baseball

3. A small car

4. A truck

5. A large train

Explanation:

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A hula hoop is rolling along the ground with a translational speed of 26 ft/s. It rolls up a hill that is 16 ft high. Determine

nikklg [1K]

Answer:12.8 ft/s

Explanation:

Given

Speed of hoop $v=26\ ft/s$

height of top $h=16\ ft$

Initial energy at bottom is

$E_b=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2$

Where m=mass of hoop

I=moment of inertia of hoop

$\omega$ =angular velocity

for pure rolling $v=\omega R$

$I=mR^2$

$E_b=\frac{1}{2}mv^2+\frac{1}{2}mR^2\times (\frac{v}{R})^2$

$E_b=mv^2=m(26)^2=676m$

Energy required to reach at top

$E_T=mgh=m\times 32.2\times 16$

$E_T=512.2m$

Thus 512.2 m is converted energy is spent to raise the potential energy of hoop and remaining is in the form of kinetic and rotational energy

$\Delta E=676m-512.2m=163.8m$

Therefore

$163.8 m=mv^2$

$v=\sqrt{163.8}$

$v=12.798\approx 12.8\ ft/s$

7 0

3 years ago

What type of phase change occurs when water vapor in the atmosphere changes to liquid water in clouds?

Mekhanik [1.2K]

Answer:

condensation, the process of changing from a gas to a liquid

3 0

3 years ago

The flywheel of a steam engine runs with a constant angular speed of 113 $rev/min$. When steam is shut off, the friction of the

Oksana_A [137]

Answer:

α = - 1.883 rev/min²

Explanation:

Given

ωin = 113 rev/min

ωfin = 0 rev/min

t = 1.0 h = 60 min

α = ?

we can use the following equation

ωfin = ωin + α*t ⇒ α = (ωfin - ωin) / t

⇒ α = (0 rev/min - 113 rev/min) / (60 min)

⇒ α = - 1.883 rev/min²

6 0

3 years ago

MamaMia's Pizza purchases its pizza delivery boxes from a printing supplier. MamaMia's delivers on-average 200 pizzas each month

Ira Lisetskai [31]

Answer:

a) 138 units

b) 17 units

c) 17 units

d) Total Cost = $353.35

Explanation:

Given:

Average pizzas delivered = 200

Charge of inventory holding = 30% of cost

Lead time = 7 days

Now,

a) Economic Order Quantity = $\sqrt\frac{2\times\textup{Annual Demand}\times\textup{Cost per Order}}{\textup{Carrying cost}}$

also,

Annual Demand = 200 × 12 = 2400

Cost per Order = Cost of Box + Processing Costs

= 30 cents + $10

= $10.30

and, Carrying Cost = $\frac{\textup{Total Inventory Cost}}{\textup{total annual demand}}$

= $\frac{\textup{Total Cost per order}\times\textup{Annual demand}\times\frac{25}{100}}{\textup{Annual demand}}$

= $\frac{\$10.30\times2400}\times\frac{25}{100}}{2400}$

= $2.575

Therefore,

Economic Order Quantity = $\sqrt\frac{2\times\textup{2400}\times\textup{10.30}}{\textup{2.575}}$

= 138.56 ≈ 138 units

b) Reorder Point

= (average daily unit sales × the lead time in days) + safety stock

= ( $\frac{200}{30}\times7$

= 46.67 ≈ 47 units

c) Number of orders per year = $\frac{\textup{Annual Demand}}{\textup{Economic order quantity}}$

= $\frac{\textup{2400}}{\textup{138}}$

= 17.39 ≈ 17 units

d) Total Annual Cost (Total Inventory Cost)

= Ordering Cost + Holding Cost

Now,

The ordering Cost = Cost per Order × Total Number of orders per year

= $10.30 × 17

= $175.1

and,

Holding Cost = Average Inventory Held × Carrying Cost per unit

Average Inventory Held = $\frac{\textup{0+138}}{\textup{2}}$ = 69

Carrying Cost per unit = $2.575

Holding Cost = 69 × $2.575 =

$177.675

Therefore,

Total Cost = Ordering Cost + carrying cost

= $175.1 + $177.675 = $353.35

5 0

3 years ago

Read 2 more answers

A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication

koban [17]

Answer:

r = 4.24x10⁴ km.

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>

From equation (1), r₁ is:

$r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}$

$r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}$

$r_{1} = 4.24 \cdot 10^{4} km$

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0

3 years ago