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Vera_Pavlovna [14]
3 years ago
13

5. If the average velocity of a duck is zero in a given

Physics
1 answer:
Ronch [10]3 years ago
5 0
If the average velocity of the duck is zero, it means that the duck's location at the end of the time interval was the same as at the beginning of the interval, but says nothing about the duck's motion during that time.
For instance, the duck could have waddled around in a circle 20 times; as long as it wound up at the starting point, the displacement and average velocity is zero.
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Phương trình trạng thái tổng quát của khí lí tưởng diễn tả là​
tatyana61 [14]

Expla

pV/T=const goodluke

6 0
2 years ago
If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star
torisob [31]
There's no such thing as "stationary in space".  But if the distance
between the Earth and some stars is not changing, then (A) w<span>avelengths
measured here would match the actual wavelengths emitted from these
stars. </span><span>

</span><span>If a star is moving toward us in space, then (A) Wavelengths measured
would be shorter than the actual wavelengths emitted from that star.

</span>In order to decide what's actually happening, and how that star is moving, 
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7 0
3 years ago
A large crane consists of a 20 m, 3,000 kg arm that extends horizontally on top of a vertical tower. The arm extends 15 m toward
Anni [7]

Answer:

m=18000kg

Explanation:

From the question we are told that:

Crane Length l=20m

Crane Mass m_a=3000kg

Arm extension at lifting end l_l=15m

Arm extension at counter weight end l_c=5m

Load M_l=5000kg

Generally the equation for Torque Balance is mathematically given by

T_1 *l_c-(m_a*g) *l_c-(T_2)*l_l=0

mg*5 *-(3000*9.8) *5-(5000*9.8)*15=0

m=18000kg

7 0
3 years ago
A flywheel in a motor is spinning at 590 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75
Gnom [1K]

Answer:

Explanation:

Hello,

Let's get the data for this question before proceeding to solve the problems.

Mass of flywheel = 40kg

Speed of flywheel = 590rpm

Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.

Time = 30s = 0.5 min

During the power off, the flywheel made 230 complete revolutions.

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + ω₂) / 2] × 0.5

But ∇θ = 230 revolutions

∇θ/t = (530 + ω₂) / 2

230 / 0.5 = (530 + ω₂) / 2

Solve for ω₂

460 = 295 + 0.5ω₂

ω₂ = 330rpm

a)

ω₂ = ω₁ + αt

but α = ?

α = (ω₂ - ω₁) / t

α = (330 - 590) / 0.5

α = -260 / 0.5

α = -520rev/min

b)

ω₂ = ω₁ + αt

0 = 590 +(-520)t

520t = 590

solve for t

t = 590 / 520

t = 1.13min

60 seconds = 1min

X seconds = 1.13min

x = (60 × 1.13) / 1

x = 68seconds

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + 0) / 2] × 1.13

∇θ = 333.35 rev/min

8 0
3 years ago
Consider an electrical transformer has 10 loops on its primary coil and 20 loops on its secondary coil. What is the voltage in t
Arlecino [84]

Answer:

240 V

Explanation:

number of turns in primary coil, Np = 10

Number of loops in secondary coil, Ns = 20

Voltage in primary coil, Vp = 120 V

Let the voltage in secondary coil is Vs.

So, Vs / Vp = Ns / Np

Vs / 120 = 20 / 10

Vs / 120  = 2

Vs = 240 V

Thus, the voltage in secondary coil is 240 Volt.

4 0
3 years ago
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