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2 years ago

A pulley with the radius of 10.0 cm was connected to a motor with a massless

Physics

1 answer:

kogti [31]2 years ago

8 0

Answer:

(i) -556 rad/s²

(ii) 17900 revolutions

(iii) 11250 meters

(iv) -55.6 m/s²

(v) 18 seconds

Explanation:

(i) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

α = (10000 − 15000) / 9

α ≈ -556 rad/s²

(ii) Constant acceleration equation:

θ = θ₀ + ω₀ t + ½ αt²

θ = 0 + (15000) (9) + ½ (-556) (9)²

θ = 112500 radians

θ ≈ 17900 revolutions

(iii) Linear displacement equals radius times angular displacement:

s = rθ

s = (0.100 m) (112500 radians)

s = 11250 meters

(iv) Linear acceleration equals radius times angular acceleration:

a = rα

a = (0.100 m) (-556 rad/s²)

a = -55.6 m/s²

(v) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

-556 = (0 − 15000) / t

t = 27

t − 9 = 18 seconds

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Two firefighters are trying to break through a door. One firefighter is heavy, and the other is light. If they run at the same s

frez [133]

Answer:

The Heavier Firefighter

Explanation:

Generally, more massive objects will have more intertia than less massive objects. As such it takes more force to halt a more massive object if its moving at the same speed as a smaller object. This can also be thought of in the context of Newton's second law. The more force needed to accelerate an object means the more force the object will have.

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2 years ago

An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has s

gizmo_the_mogwai [7]

Answer:

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Explanation:

Part A)

heat required to melt the aluminium is given by

$Q = ms\Delta T + mL$

here we have

$Q = 40(950)(680 - 32) + 40(450 \times 10^3)$

$Q = 24624 kJ + 18000 kJ$

$Q = 42624 kJ$

Since this is the amount of aluminium per hour

so power required to melt is given by

$P = \frac{Q}{t}$

$P = \frac{42624}{3600} kW$

$P = 11.84 kW$

Since the efficiency is 85% so actual power required will be

$P = \frac{11.84}{0.85} = 13.93 kW$

Part B)

Total energy consumed by the furnace for 30 hours

$Energy = power \times time$

$Energy = 13.93 kW\times 30 h$

$Energy = 417.9 kWh$

now the total cost of energy consumption is given as

$R = P \times 20 \frac{Cents}{kWh}$

$R = 417.9 kWh\times 20 \frac{cents}{kWh}$

$R = 8357.6 Cents$

3 0

3 years ago

When the distance between two charges is halved, the electrical force between them?

Llana [10]

If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
$F= k \frac{q_1 q_2}{r^2}$
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
$r'= \frac{r}{2}$
the magnitude of the force changes as follows:
$F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k \frac{q_1 q_2}{r^2}=4 F$
so, the force increases by a factor 4.

3 0

3 years ago

At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

$\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

$\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

5 0

2 years ago

Which property of table salt is also a property of other ionic compounds

Jet001 [13]

s alluded to in the other answers, salt refers to any ionic compound that doesn't have “oxides” in it. Table salt is sodium chloride. Going down the periodic table, the first column contains lithium, sodium, potassium, rubidium, cesium, and francium. This group (alkali metals) of atoms (and their corresponding positive ions) gets larger in the order shown above. Therefore, their ionic bonds with chloride (or any nonmetal) gets smaller. The trend of their corresponding compounds is a decreasing hardness, decreasing melting point, decreasing boiling point, and decreasing thermal stability. These are the major periodic trends of these corresponding compounds. Other metal ions generally have higher positive charges on them. This makes the ionic bonds considerably larger and you can probably surmise most of their corresponding properties listed above. However, the details of their lattice structures may cause the overall trend to vary.

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3 years ago

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