2 years ago

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Physics

1 answer:

GenaCL600 [577]2 years ago

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Answer: Kjebhk das dajsd ajdn wadm awkdlwandeA fefme ef

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A rod of length 0.82 m, rotating with an angular speed, 4.2 rad/s, about axes that pass perpendicularly through one end, has a m

mariarad [96]

Answer:

Explanation:

KE = ½Iω²

ΚΕ = ½(mL²/3)ω²

ΚΕ = ½(0.63(0.82²)/3)4.2²

ΚΕ = 1.24541928

KE = 1.2 J

5 0

2 years ago

You're carrying a 3.2-m-long, 24kg pole to a construction site when you decide to stop for a rest. You place one end of the pole

Aleksandr [31]

Answer:

Tension of 132N

Explanation:

We need to apply Summatory of Force to find the tension in the hand.

We define te tensión in the hand as $F_2$ and the Tension in fence post as $F_1$ , then

$\sum F = 24(9.8)$

$F_1 + F_2= 24(9.8)$

We apply summatory of moments then

$F_2*1.25 = F_1*1.6$

Where the Force 2 is 1.25m from the center of summatory,

We can note that,

$1.6 m - 0.35m=1.25m$

We have two equation and two incognites, then replacing (1) in (2)

$1.6(235.2 -F_2) = 1.25F_2$

$376.32 = F2(1.6+1.25)$

$F_2= \frac{376.32}{2.85}$

$F_2 =132 N$

5 0

3 years ago

A heat engine does 9200 J of work per cycle while absorbing 22.0 kcal of heat from a hightemperature reservoir. What is the effi

Morgarella [4.7K]

Answer: 9.9%

Explanation: efficiency = (work output /work input) × 100

Note that, 1 kilocalorie = 4184 joules, hence 22kcal = 22× 4184 = 92048 joules.

Work output = 9200 j and work input = 92048 j

Efficiency = (9200/92048) × 100 = 0.099 × 100 = 9.9%

4 0

3 years ago

It took Lightning McGreen 2.5 hours to travel

Law Incorporation [45]

Answer:

240km/h

Explanation:

speed=distance/time

speed=600/2.5

=240km/h

4 0

2 years ago

Read 2 more answers

A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the

vladimir2022 [97]

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k

The Pressure-Volume relation is PV = C

T = C for isothermal process

Calculating for the work done in isothermal process

W = P₁V₁ $ln[\frac{P_{1} }{P_{2} }]$

= mRT₁ $ln[\frac{P_{1} }{P_{2} }]$ [∵pV = mRT]

= (5) (0.287) (272.039) $ln[\frac{2.0}{1.0}]$

= 270.588 kJ

Since the process is isothermal, Internal energy change is zero

ΔU = $mc_{v}(T_{2} - T_{1} ) = 0$

From 1st law of thermodynamics

Q = ΔU + W

= 0 + 270.588

= 270.588 kJ

4 0

3 years ago