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Vikki [24]
2 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]2 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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A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the ) bottom of a hill, starting from rest. If
Semenov [28]

Answer:

(A) = 3.57 m

Explanation:

from the question we are given the following:

diameter (d) = 3.2 m

mass (m) == 42 kg

angular speed (ω) = 4.27 rad/s

from the conservation of energy

mgh = 0.5 mv^{2} + 0.5Iω^{2} ...equation 1

where

Inertia (I) = 0.5mr^{2}

ω = \frac{v}{r}

equation 1 now becomes

mgh = 0.5 mv^{2} + 0.5(0.5mr^{2})(\frac{v}{r})^{2}

gh = 0.5 v^{2} + 0.5(0.5)(v)^{2}

4gh = 2v^{2} + v^{2}

h = 3v^{2} ÷ 4 g .... equation 2

  from ω = \frac{v}{r}

 v  = ωr  = 4.27 x (3.2 ÷ 2)

v = 6.8 m/s

now substituting the value of v into equation 2

h = 3v^{2} ÷ 4 g

h = 3 x (6.8)^{2} ÷ (4 x 9.8)

h = 3.57 m

8 0
2 years ago
A mine car, whose mass is 440kg, rolls at a speed of 0.50m/s on ahorizontal track, as the drawing shows. A 150kg chunk of coalha
ella [17]

Answer: 0.56 m/s

Explanation:

hello, there is 25° inclination angle for the chute in the drawing. Thankfully, I know this problem. The conservation of momentum.

so there are X and Y components for the momentum in this problem. The Y component is not conserved as when the coal gets in the cart, the normal force exerted by the surface reduces it to 0.

Now, the X component is definitely conserved here.

so you have the momentum of the cart which is 440*0.5 added to the momentum of the chunk which is 150*0.8*cos(25°), that is the momentum before the coupling between the objects. Afterwards both objects will have the same velocity, so we write the equation like this:

440*0.5 + 150*0.8*cos(25) = 440*v_{final} + 150*v_{final} \\ => 220+120*cos(25) = (440+150)v_{final} => v_{final} = \frac{220+120*cos(25)}{590}  = 0.56 m/s

3 0
2 years ago
A 12-volt battery causes 0.60 ampere to flow through a circuit that contains a lamp and a resistor connected in parallel. The la
san4es73 [151]
The answer to this question is A
5 0
1 year ago
A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.
Aleonysh [2.5K]

Answer:  53.31\° East of North

Explanation:

We have the following data:

Speed of the wind from East to West: 6.68 m/s

Speed of the bee relative to the air:  8.33 m/s

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}

sin \theta=\frac{6.68 m/s}{8.33 m/s}

Clearing \theta:

\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})

\theta=53.31\°

6 0
2 years ago
A dragster crosses the finish line with a velocity of 140m/s . Assuming the vehicle maintained a constant acceleration from star
Mumz [18]

Answer:

Answer is attached

5 0
2 years ago
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