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Art [367]
3 years ago
15

Accelerated motion is represented by a _____ line on a nonlinear distance-time graph.

Physics
2 answers:
Natasha_Volkova [10]3 years ago
4 0

Accelerated motion could be changing speed, or changing direction, or both.

If it's only the direction changing and not the speed, then you'll never see it on a distance/time graph.

If the speed is changing, then a <em>velocity</em>/time graph can be any kind of a line except horizontal or vertical, and a <u><em>distance</em></u>/time graph must be <em>curved</em>.

Delicious77 [7]3 years ago
4 0

Answer:

Accelerated motion is represented by a <u>CURVED</u> line on a nonlinear distance-time graph.

Explanation:

For uniform acceleration we know that

\frac{dv}{dt} = a

now integrating both sides we will have

\int dv = \int adt

now after integration we will have

v - v_o = at

now we have

v = v_o + at

again we know that

\frac{dx}{dt} = v_o + at

again by integration both sides

x - x_0 = v_o t + \frac{1}{2}at^2

x = x_0 + v_0t + \frac{1}{2}at^2

now since this equation is a quadratic equation so here distance time graph must be a curved graph

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Crews at the International Space Station are researching the effects of the weightlessness of space on ________.
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A: Human Body

C is wrong because they don’t have the tools to test it on another planet

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3 years ago
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Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration
LenaWriter [7]

Answer:

c.5.6m/s^2

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

Acceleration,a=\frac{v-u}{t}

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Then, we get

Acceleration, a=\frac{28-1}{4.8}=\frac{27}{4.8} m/s^2

Acceleration=a=5.6 m/s^2

Hence, the acceleration of cheetah=5.6m/s^2

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You leave on a 450 miles trip in order to attend a meeting that will start 10.8 hours after you begin your trip. Along the way y
Lorico [155]

Answer:2.6 h

Explanation:

Given

Total Trip distance=450 miles

Meeting starts after 10.8 hours

safe Fastest speed is  55 mi/h

so if he drives all the to the meeting with max speed then it takes =\frac{450}{55}=8.181 h

and total allowable time is 10.8

Therefore longest time he can spend over dinner is 10.8-8.181 \approx 2.6 hours

5 0
3 years ago
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