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Rudiy27
2 years ago
15

What is the final speed of an object that starts from rest and accelerates uniformly at 4.0 meters per second2 over a distance o

f 8.0 meters? 1. 8.0 m/s 2. 16 m/s 3. 32 m/s 4. 64 m/s
Physics
1 answer:
jonny [76]2 years ago
4 0
Considering that the acceleration is uniform a=4 (m/s^2) we apply the equation
v^2=v0^2+2as
with zero initial speed 
v^2=2as
and we obtain the speed
v^2 =2*8*4 =64 (m/s)^2
Thus v=8 (m/s)

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Arrange the movement/act/organization in ascending order of occurrence. Energy Supply and Environmental Coordination Act Nature
nadya68 [22]

Answer:

PLATO!!!!! THESE ARE RIGHT!!! OTHERS ARE NOT CORRECT. I JUST GOT IT RIGHT ON PLATO

Explanation:

1. federal water pollution control act

2.nature conservancy

3. clean air act

4. water quality act

5. endangered species preservation act

6. clean water act

7. energy supply and environmental coordination act

8. eastern wilderness act

9. toxic substance act

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5 0
2 years ago
Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4
kati45 [8]

Answer:

velocity  = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = \frac{\pi }{4} × (4.50×10^{-2})²  × velocity

solve it we get

velocity  = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = \frac{\pi }{4} × 3 ×  (4.50×10^{-2})²  × velocity

velocity = 52.4 m/s

5 0
2 years ago
At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig
Elena-2011 [213]

Complete Question

  The complete Question is shown on the first uploaded image

Answer:

a

The torque acting on the particle is  \tau = 48t \r k

b

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

      The position of the particle is   \= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j

       The mass of the particle is m = 3.0 kg

        The time is  t

   

The torque acting on  the particle is mathematically represented as

           \tau = \frac{ d \r l }{dt}

where \r l is change in angular momentum which is mathematically represented as

       \r l = m (\r r \ \ X  \ \ \r v)

Where X mean cross- product

   \r v is the velocity which is mathematically represented as

           \r v = \frac{d \r r }{dt}

Substituting for  \r r

           \r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]

           \r v =  8t \r i - (2 + 12 t) \r j

Now the cross product of \r r \ and \ \r v is  mathematically evaluated as    

          \r r  \  \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]

                       = 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k

                      \r r \ \  X \ \ \r v = 8t^2 \r k

So the angular momentum becomes

       \r l = m (8t^2 \r k)

Substituting for m

      \r l = 3 *  (8t^2 \r k)

      \r l =24t^2  \r k

Substituting into equation for torque

       \tau = \frac{d}{dt} [24t^2 \r k]

       \tau = 48t \r k

The magnitude of the angular momentum can be evaluated mathematically as

        |\r l| = \sqrt{(24 t^2) ^2}

        |\r l| = 24 t^2

From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also

4 0
2 years ago
. A girl runs and jumps horizontally off a platform 10m above a pool with a speed of 4.0m/s. As soon as she leaves the platform,
faust18 [17]

Answer:

2.39 revolutions

Explanation:

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t^2 = 2

t = \sqrt{2} = 1.414 s

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\theta = \alpha t^2/2

\theta = 15 * 2/2 = 15 rad

As each revolution is 2π, the total number of revolution that she could make is: 15 / 2π = 2.39 rev

3 0
2 years ago
In a given city, the permissible limit of CO (carbon monoxide) in the air is 100 parts per million (ppm). The city monitors the
Roman55 [17]
The city monitors the steady rise of CO from various sources annually. In the year "C: 2019"<span> (rounded off to the nearest integer) will the CO level exceed the permissible limit.

If this isn't the answer, let me know and i'll figure out what it is. But I believe this is it.           :) </span>
8 0
2 years ago
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