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Vikki [24]
3 years ago
15

What is the speed u of the object at the height of (1/2)hmax? Express your answer in terms of v and g. Use three significant fig

ures in the numeric coefficient.
Physics
1 answer:
max2010maxim [7]3 years ago
3 0

The speed u of the object at the height of (1/2) hmax is 0.707 v_{i}.

<u>Explanation:</u>

At a height that is half of the maximum height, the object will have only half of its initial kinetic energy, the rest half is converted to potential energy at this point. So,

                              1 / 2 m v^{2} _{i} = m (v_{mid})^{2} }

                                  v_{mid} = \frac{v}{\sqrt{2} }

                                v_{mid} =   0.707 v_{i}.

The speed u of the object at the height of (1/2) hmax is 0.707 v_{i}.

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It can be either C or B

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But also, it depends on what kind of filter it is, if the filter is like a screen filter then it will just come out in blue with the slightly different colors of again purple but in a darker tone then the actual eye can see.

Or it can be just C again cause the filter can be a film but that's a bit too far and to complex for right now so I believe it is B
8 0
3 years ago
Cheetahs can accelerate to a speed of 20.0 m/s in 2.50 s and can continue to accelerate to reach a top speed of 29.5 m/s. Assume
insens350 [35]

Answer:

boy if you dont get t f outa here

4 0
3 years ago
Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing?
Oksanka [162]

Answer:

1.58 Hz

Explanation:

The frequency of the simple pendulum is given by

f = 1/T

 = 1/2π√g/l  

In this problem, I = 10.0 cm = 0.1 m  

f = 1/2π√9.8/0.1

=  1.58 Hz

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What <br> material gives the soil its high fertility?
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Humus

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8 0
3 years ago
A railroad freight car rolls on a track at 3.50 m/s toward two identical coupled freight cars, which are rolling in the same dir
ladessa [460]

Answer:

Explanation:

Hello! To solve this problem we must be clear about the concept of energy conservation, and kinetic energy with the following sentence

The kinetic energy of the two cars (v = 1.2m / S) plus the kinetic energy of the third car (v = 3.5m / S) must be equal to the kinetic energy of the three cars together.

The kinetic energy is calculated by the following equation.

E=0.5mV^2

m= mass of the cars=26500kg

V=speed

E=kinetic energy

taking into account the above, the following equation is inferred

1=  the cars are separated

2= the cars are togheter

E1=E2

E1=0.5mV1^2+0.5mV1^2+0.5m(Va)^2

where

m= mass of each car

V1= 1.2m/s

Va=3.5,m/S

E2=0.5(3)(m)V^2

m= mass of each car

V=speed (in m/s) of the three coupled cars after the first couples with the other two

Solving

0.5mV1^2+0.5mV1^2+0.5m(Va)^2=0.5(3)(m)V^2

V1^2+V1^2+(Va)^2=(3)V^2.\\2V1^2+(Va)^2=(3)V^2\\V^2=\frac{2V1^2+(Va)^2}{3} \\

V=\sqrt{\frac{2V1^2+(Va)^2}{3}} \\V=\sqrt{\frac{2(1.2)^2+(3.5)^2}{3}} \\\\V=2.245m/s

the speed  of the three coupled cars after the first couples with the other two is 2.245m/s

7 0
3 years ago
Read 2 more answers
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