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4 years ago

43.278 kg - 28.1 g use significant figures rule

1 answer:

3 0

You cannot directly subtract the given values because their units are different. To be consistent, let's convert kilograms to grams with the conversion that 1,000 g = 1 kg.

43.278 kg(1,000 g/1 kg) - 28.1 g = 43249.9 grams

In subtraction, you will base the number of significant figures to the least of the given data. Since 28.1 has 3 significant figures, the answer should also contain 3 significant figures. Thus, it is much more convenient to report it in terms of kilograms.

43249.9 grams * 1 kg/ 1,000 g = 43.2 kg

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An experimental rocket designed to land upright falls freely from a height of 2.59 102 m, starting at rest. At a height of 86.9

aleksandr82 [10.1K]

Answer:

The acceleration required by the rocket in order to have a zero speed on touchdown is 19.96m/s²

The rocket's motion for analysis sake is divided into two phases.

Phase 1: the free fall motion of the rocket from the height 2.59*102m to a height 86.9m

Phase 2: the motion of the rocket due to the acceleration of the rocket also from the height 86.9m to the point of touchdown y = 0m.

Explanation:

The initial velocity of the rocket is 0m/s when it started falling from rest under free fall. g = 9.8m/s² t1 is the time taken for phase 1 and t2 is the time taken for phase2.

The final velocity under free fall becomes the initial velocity for the accelerated motion of the rocket in phase 2 and the final velocity or speed in phase 2 is equal to zero.

The detailed step by step solution to the problems can be found in the attachment below.

Thank you and I hope this solution is helpful to you. Good luck.

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3 years ago

A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de

EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/π $r^{2}$

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / π $r^{2}$ ex

E = $\frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6} } = 17.368 * 10^{9} Pa = 17.4 GPa$

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = $\frac{1}{E}$ (бy - v (бx + бz)) = $-\frac{v}{E}$ бx

= $\frac{vFx}{Epir^{2} }$ = $\frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9} } = -0.63 *10^{-6}$

Finally

ey = Δr / r

Δr = ey * r = 10 * $-0.63* 10^{-6} mm = -6.3 * 10^{-6} mm$

Δd = 2Δr = $-12.6 * 10^{-6} mm$

Explanation:

5 0

4 years ago

Is block b speeding up or slowing down once it is set into downward motion and the cat is on block a?

stepan [7]

Answer:

The answer is "Slowing down ".

Explanation:

please find the complete question in the attached file.

In this question, if the block B weight were accounted for by kinetic the friction of frame A, because the blocks pushed at a consistent speed throughout the beginning.

Afterward, on block A, the resistance intensity rises, which allows frames to also be negative, which is defined in the graph, that's why the answer Slowing down is correct.

6 0

3 years ago

How condensation affects the temperature of the area where liquids forms

NISA [10]

Condensation does not effect the temperature of the area, condensation is the product the temperature difference between an object and the area it is in

7 0

3 years ago

Learning Goal:

Rudiy27

Answer:

The questions are not complete so this is the complete questions

1. How much work W does the motor do on the platform during this process?

2. What is the angular velocity ωf of the platform at the end of this process?

3. What is the rotational kinetic energy, Ek, of the platform at the end of the process described above?

4. How long does it take for the motor to do the work done on the platform calculated in Part 1?

5. What is the average power delivered by the motor in the situation above?

6. . Note that the instantaneous power P delivered by the motor is directly proportional to ω, so P increases as the platform spins faster and faster. How does the instantaneous power P•f being delivered by the motor at the time t•f compare to the average power

P(average) calculated in Part e?

Explanation:

Given that,

The torque τ=25Nm

Moment of inertia I =50kgm²

The platform is initially at rest,

ω•i=0 rad/sec

Revolution the torque produce is 12

Then, θ=12 revolution

1 revolution=2πrad

So, θ=24πrad

1. Work done in a rotational motion is give as

W=τ•Δθ

Given that the τ=25Nm and the initial angular displacement is 0rad

The final angular displacement is 24πrad

Δθ =(θ2-θ1)

Δθ=24π-0

Δθ=24πrad

Then,

W=τ•Δθ

W=25(24π)

W=25×24π

W=1884.96J

To 4s.f, W=1885J

2. Final angular velocity ωf

Using the angular equation

ω•f²=ω•i²+2•α•Δθ

We need to get angular acceleration

The torque is given as

τ=I•α

Given that,

I is moment if inertia =50kgm²

τ=25Nm

α=τ/I

α= 25/50

α=0.5rad/s²

Now, using the angular acceleration

ω•f²=ω•i²+2•α•Δθ

ω•f²=0²+2×0.5×24π

ω•f²=0+75.398

ω•f²=75.398

ω•f=√75.398

ω•f=8.68 rad/sec.

3. We need to find rotational Kinetic energy and it is given as

K.E, = ½I•ω²

Given that, I=50kgm² and ω•f=8.68rad/sec

Then,

K.E, =½I•ω²

K.E, =½×50×8.68²

K.E, =1884.96J

To 4s.f,

K.E, =1885J

Which is the same as the work done by the motor.

4. Time taken to complete part 1,

Using the rotational equation

ω•f=ω•i+α•t

Since, ω•f=8.68 rad/sec and ω•i=0

And α=0.5rad/s²

Then,

ω•f=ω•i+α•t

8.68=0+0.5t

8.68=0.5t

Then, t=8.68/0.5

t=17.36secs

5. The average power of rotational motion is given as

P(average) =Workdone/timetaken

Since,

Work done =1884.96J

Time taken =17.36sec

P(average) =Workdone/timetaken

P(average)=1884.96/17.36

P(average)= 108.58Watts

To 4s.f

P(average)=108.6Watts

6. We need to find •, it is given as

• =τ•ωf

Given that, ω•f=8.68rad/sec, τ=25Nm

•=25×8.68

•=217Watts

Then, the ratio of • to P(average) is

Ratio = •/ P(average)

Ratio= 217/108.58

Ratio=1.9985

Then, the ratio is approximately 2

Ratio=2

5 0

4 years ago