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aleksandr82 [10.1K]
2 years ago
7

elastic wire extend by 1.ocm when a load on 20g range from It, what additional load will it be required Cause the futher extensi

on of 2.0cm​
Physics
1 answer:
Citrus2011 [14]2 years ago
6 0

Answer:

40g

Explanation:

20g range > 1.0cm

Therefore,

40g range > 2.0cm

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A plane travels 350 km south along a straight path with an average velocity of 125 km/h to the south. The plane has a 30 minute
faust18 [17]
Average Velocity = Total Displacement / Total time

1st part of journey,  350 km at velocity 125 km/h

Time = 350 / 125 = 2.8 hours.

2nd part of journey,  220 km at velocity 115 km/h

Time = 220 / 115 = 1.9 hours


Average Velocity = Total Displacement / Total time

                               = (350 + 220) / (2.8 + 1.9)

                                =   570 / 4.7  ≈ 121.3 km/hr

Average Velocity ≈ 121 km/hr due south. 

Option C. 
6 0
3 years ago
Giving brainiest to correct answer.
mixas84 [53]

Answer:

5.33\ m/s

Explanation:

We\ know\ that,\\Momentum=Mass*Velocity\\p=mv\\Hence,\\Lets\ first\ consider\ the\ case\ of\ the\ two\ balls\ 'Before\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Initial\ Velocity\ of\ the\ green\ ball=5\ m/s\\Initial\ Momentum\ of\ the\ green\ ball=5*0.2=1\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Initial\ Velocity\ of\ the\ pink\ ball=2\ m/s\\Initial\ Momentum\ of\ the\ pink\ ball=0.3*2=0.6\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'Before\ Collision'=1+0.6=1.6\ kg\ m/s

Hence,\\Lets\ now\ consider\ the\ case\ of\ the\ two\ balls\ 'After\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Final\ Velocity\ of\ the\ green\ ball=0\ m/s\\Final\ Momentum\ of\ the\ green\ ball=0\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Final\ Velocity\ of\ the\ pink\ ball=v\ m/s\\Final\ Momentum\ of\ the\ pink\ ball=0.3*v=0.3v\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'After\ Collision'=0+0.3v=0.3v\ kg\ m/s

As\ we\ know\ that,\\Through\ the\ law\ of\ conservation\ of\ momentum,\\In\ an\ isolated\ system:\\Total\ Momentum\ Before\ Collision=Total\ Momentum\ After\ Collision\\Hence,\\1.6=0.3v\\v=\frac{1.6}{0.3}=5.33\ m/s

5 0
3 years ago
A pebble is released from rest at a certain height and falls freely, reaching an impact speed of 6 m/s at the floor. Next, the p
Anna71 [15]

Answer:

Explanation:

Let h be the height .

initial velocity in first case u = 0

final velocity v = 6 m /s

acceleration due to gravity g = 9.8 m /s²

v² = u² + 2 g h

6² = 0 + 2 x 9.8 x h

h = 1.837 m .

For second case u = 3 m /s

v² = u² + 2 gh

= 3² + 2 x 1.837 x 9.8

= 9 + 36

= 45 m

v = 6.7 m /s

8 0
3 years ago
A Ferris wheel has diameter of 10 m. It rotates at a uniform rate and makes one revolution in 8.0 seconds. A person weighing 670
Nikolay [14]

Answer:  459.14 N

Explanation:

from the question, we have

diameter = 10 m

radius (r)  = 5 m

weight (Fw) = 670 N

time (t) = 8 seconds

Circular motion has centripetal force and acceleration pointing perpendicular and inwards of the path, therefore we apply the equation below

∑ F = F c =  F w − Fn ..............equation 1

Fn = Fw − Fc = mg − (mv^2 / r) ...................equation 2

substituting the value of v as (2πr / T) we now have

Fn = mg − (m(2πr / T )^2) / r

Fn= mg − (4(π^2)mr / T^2)   ..........equation 3

Fw (mass of the person) = mg

therefore m = Fw / g

                m = 670 / 9.8 = 68.367 kg

now substituting  our values into equation 3

Fn = 670 - ( (4 x (π^2) x 68.367 x 5 ) / 8^2)

Fn = 670 - 210.86

Fn = 459.14 N

4 0
3 years ago
A refrigeration cycle has Qout = 1000 Btu and Wcycle = 300 Btu. Determine the coefficient of performance for the cycle.
DENIUS [597]

Answer:

The coefficient of performance for the cycle is 2.33.

Explanation:

Given that,

Output energy Q_{out}=1000\ Btu

Work done W_{cycle}=300\ Btu

We need to calculate the coefficient of performance

Using formula of  the coefficient of performance

COP=\dftrac{Q_{in}}{W_{cycle}}

We need to calculate the Q_{in}

W_{cycle}=Q_{out}-Q_{in}

Put the value into the formula

300=1000-Q_{in}

Q_{in}=300-1000

Q_{in}=700\ Btu

Now put the value of Q_{in} into the formula of COP

COP=\dfrac{700}{300}

COP=\dfrac{7}{3}=2.33

Hence, The coefficient of performance for the cycle is 2.33.

5 0
3 years ago
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