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3 years ago

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A(n) 591 kg elevator starts from rest. It moves upward for 4.79 s with a constant acceleration until it reaches its cruising spe

ed of 2.17 m/s. The acceleration of gravity is 9.8 m/s 2 . Find the average power delivered by the elevator motor during the period of this acceleration. Answer in units of kW

1 answer:

Cloud [144]3 years ago

3 0

Answer:

Power delivered will be 10755.87 W

Explanation:

We have given mass of elevator m = 591 kg

Time t = 4.79 sec

As the elevator starts from rest so initial velocity u = 0m/sec

Final velocity v = 2.17 m/sec

From first equation of motion

v=u+at. here v is final velocity, u is initial velocity and t is time

So $2.17=0+a\times 4.79$

$a=0.453m/sec^2$

Height is given by $h=ut+\frac{1}{2}at^2=0\times 4.79+\frac{1}{2}\times 0.453\times 4.79^2=5.196m$

As elevator moves upward and acceleration due to gravity is downward so net acceleration due to gravity is downward so net acceleration = g-a=9.8-0.453 = 9.347 $m/sec^2$

Now work done $W=m(g-a)h=951\times (9.8-0.453)\times 5.196=51520.626J$

We know that power $P=\frac{work\ done}{time}=\frac{51520.626}{4.79}=10755.871W$

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n explosion breaks an object initially at rest into two pieces, one of which has 2.0 times the mass of the other. If 8400 J of k

ANTONII [103]

Answer:

The heavier piece acquired 2800 J kinetic energy

Explanation:

From the principle of conservation of linear momentum:

0 = M₁v₁ - M₂v₂

M₁v₁ = M₂v₂

let the second piece be the heavier mass, then

M₁v₁ = (2M₁)v₂

v₁ = 2v₂ and v₂ = ¹/₂ v₁

From the principle of conservation of kinetic energy:

¹/₂ K.E₁ + ¹/₂ K.E₂ = 8400 J

¹/₂ M₁(v₁)² + ¹/₂ (2M₁)(¹/₂v₁)² = 8400

¹/₂ M₁(v₁)² + ¹/₄M₁(v₁)² = 8400

K.E₁ + ¹/₂K.E₁ = 8400

Now, we determine K.E₁ and note that K.E₂ = ¹/₂K.E₁

1.5 K.E₁ = 8400

K.E₁ = 8400/1.5

K.E₁ = 5600 J

K.E₂ = ¹/₂K.E₁ = 0.5*5600 J = 2800 J

Therefore, the heavier piece acquired 2800 J kinetic energy

3 0

3 years ago

Read 2 more answers

A(n) 1.3 kg mass sliding on a frictionless surface has a velocity of 7.1 m/s east when it undergoes a one-dimensional elastic co

Oxana [17]

Answer: 2.12 kg

Explanation:

Since the 1.3 kg object moves to the west after the collision, the other object will move to the east after the collision.

In an elastic collision, the relative velocity after the collision is the opposite of the relative velocity before the collision. Since the 1.3 kg object’s velocity before the collision is 6.7 m/s greater than the other object, after the collision, its velocity will be 6.7 m/s less than the other object. To determine the other object’s velocity, use the following equation.

v = 1.7 – 7.1 = -5.4 m/s

The negative sign means it is moving eastward. Let’s use this number is a momentum equation to determine its mass.

Initial momentum = 1.3 * 7.1 = 9.23 east

For the 1.3 object, final momentum = 1.3 * 1.7 = 2.21 west

To determine the final momentum of the other object, add these two numbers.

Final momentum = 11.44 east

To determine its mass, use the following equation.

m * 5.4 = 11.44

m = 11.44 ÷ 5.4 = 2.12 kg

To make sure that kinetic energy is conserved, let’s round this number to 2 kg and determine the final kinetic energies.

For the 1.3 object, KE = 1/2 * 1/3* 1.7^2 = 0.48

For the 2 kg object, KE = 1/2* 2 * 5.4^2 = 29.64

Total final KE = 29.64

Initial KE = 0.5* 1.3 * 7.1^2 = 32.77

Since I rounded the mass up to 2kg, this proves that kinetic energy is conserved and the mass is correct!

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4 years ago

If you need 40.0 Nm of torque in order to loosen a nut on a wn

KonstantinChe [14]

Answer:

0.301 m

Explanation:

Torque = Force × Radius

τ = Fr

40.0 Nm = 133 N × r

r = 0.301 m

The mechanic must apply the force 0.301 m from the nut.

6 0

3 years ago

Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo

ZanzabumX [31]

a) 893 N

b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

Explanation:

a)

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

$F_{net}=ma$

where

m is Bolt's mass

a is the acceleration

In the first 0.890 s of motion, we have

m = 94.0 kg (Bolt's mass)

$a=9.50 m/s^2$ (acceleration)

So, the net force is

$F_{net}=(94.0)(9.50)=893 N$

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).

b)

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

$v=u+at$

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

In the first phase of Bolt's race we have:

u = 0 m/s (he starts from rest)

$a=9.50 m/s^2$ (acceleration)

t = 0.890 s (duration of the first phase)

Solving for v,

$v=0+(9.50)(0.890)=8.5 m/s$

c)

First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

$W=K_f - K_i = \frac{1}{2}mv^2-0$

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

$K_i = 0 J$ is the initial kinetic energy

So the work done is

$W=\frac{1}{2}(94.0)(8.5)^2=3396 J$

The power expended is given by

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

where

t = 0.890 s is the time elapsed

Substituting,

$P=\frac{3396}{0.890}=3816 W$

d)

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

$F_1=893 N$

We know that the average force for the whole race is

$F_{avg}=820 N$

Which can be rewritten as

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}$

And solving for $F_2$ , we find the average force exerted by Bolt on the ground during the second phase:

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N$

The net force exerted by Bolt during the second phase can be written as

$F_{net}=F_2-D$ (1)

where D is the air drag.

The net force can also be rewritten as

$F_{net}=ma$

where

$a=\frac{v-u}{t}$ is the acceleration in the second phase, with

u = 8.5 m/s is the initial speed

v = 12.4 m/s is the final speed

t = 8.69 t is the time elapsed

Substituting,

$a=\frac{12.4-8.5}{8.69}=0.45 m/s^2$

So we can now find the average drag force from (1):

$D=F_2-F_{net}=F_2-ma=812.5 - (94.0)(0.45)=770.2 N$

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

$\Delta E=W=Ds$

where

d is Bolt's displacement in the second part, which can be found by using suvat equation:

$s=\frac{v^2-u^2}{2a}=\frac{12.4^2-8.5^2}{2(0.45)}=90.6 m$

And so,

$\Delta E=Ds=(770.2)(90.6)=69780 J$

e)

The power that Bolt must expend just to voercome the drag force is given by

$P=\frac{\Delta E}{t}$

where

$\Delta E$ is the increase in internal energy due to the air drag

t is the time elapsed

Here we have:

$\Delta E=69780 J$

t = 8.69 s is the time elapsed

Substituting,

$P=\frac{69780}{8.69}=8030 W$

And we see that it is about twice larger than the power calculated in part c.

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First of all, there are not <u>just</u> two applications that are solely applicable to the electron beam welding process. There are MANY.

Please visit out website at the URL below and you can click the "View Application" button under each listed Industry segment to view case studies of commonly EB welded applications.

https://www.ptreb.com/electron-beam-welding-applications

And for more general information on our welding process, we have an informational section you can peruse as well:

https://www.ptreb.com/electron-beam-welding-information

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