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2 years ago

A 5.8x10 ^4 watt elevator motor can lift a total weight of 2.1 x10^4 newtons with a maximum constant speed of

2 answers:

8 0

The formula for power is used to solve for the speed. Given are P = 5.8x10^4 W and W = 2.1 x10^4 N = F. Formula: v =P / F this would result to v = 5.8 x10^4 W / 2.1x10^4 N = 2.76 m/s. I hope this helps you

Elodia [21]2 years ago

5 0

Answer : The speed of motor will be, 2.76 m/s

Solution : Given,

Power of motor = $5.8\times 10^4watt$

Force = $2.1\times 10^4N$

Formula used :

$Power=Force\times Velocity\\\\Velocity=\frac{Power}{Force}$

Now put all the given values in this formula, we get the speed of motor.

$Velocity=\frac{Power}{Force}=\frac{5.8\times 10^4watt}{2.1\times 10^4N}=2.76m/s$

Therefore, the speed of motor will be, 2.76 m/s

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A longitudinal wave is observed to be moving along a slinky. Adjacent crests are 2.4 m apart. Exactly 6 crests are observed to m

Umnica [9.8K]

Answer:

Explanation:

8 0

2 years ago

During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

Ray Of Light [21]

Answer:

part a : The dry unit weight is 0.0616  $lb/in^3$ 

part b : The void ratio is 0.77

part c : Degree of Saturation is 0.43

part d : Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

The dry unit weight is 0.0616  $lb/in^3$ 

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

The void ratio is 0.77

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

Degree of Saturation is 0.43

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0

2 years ago

A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the disk is allowed to fall and th

Tju [1.3M]

Answer:

$a = \frac{2}{3}g$

$T = \frac{mg}{3}$

Explanation:

As the disc is unrolling from the thread then at any moment of the time

We have force equation as

$mg - T = ma$

also by torque equation we can say

$TR = I\alpha$

$TR = \frac{1}{2}mR^2(\frac{a}{R})$

$T = \frac{1}{2}ma$

Now we have

$mg - \frac{1}{2}ma = ma$

$mg = \frac{3}{2}ma$

$a = \frac{2}{3}g$

Also from above equation the tension force in the string is

$T = \frac{1}{2}ma$

$T = \frac{mg}{3}$

7 0

2 years ago

An electric clock is hanging on a wall. As you are watching the second hand rotate, the clock's battery stops functioning, and t

Setler [38]

Answer:

B. W is positive and a is negative

Explanation:

As we know that the angular speed of the second clock is in positive direction so as it comes to halt from its initial direction of motion then we have

initial angular velocity is termed as positive angular velocity

$\omega = positive$