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zmey [24]
3 years ago
15

A 59kg child starting from rest slides down a water slide with a vertical height of 5.0m. what is the child's speed halfway down

the slide's vertical distance
Physics
1 answer:
KIM [24]3 years ago
6 0
<span>EP (potential energy) = mgy -> (59)(9.8)(-5) = -2,891
   EP + EK (kinetic energy) = 0; but rearranging it for EK makes it EK = -EP, such that EK = 2891 when plugged in.
   EK = 0.5mv^2, but can also be v = sqrt(2EK/m).
   Plugging that in for sqrt((2 * 2891)/59), we get 9.9 m/s^2 with respect to significant figures.</span>
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Lana71 [14]
Circumference C=2πr 
<span>C=2π(1.5x10^8)=9.42x10^8 </span>

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<span>V=distance/time </span>

<span>V=(9.42x10^8)/8760=107534.2km/hr </span>
8 0
3 years ago
A sailboat is traveling to the right when a gust of wind causes the boat to accelerate leftward at 2.5​​​​m​​/s2 for 4s. After t
Ber [7]

Answer:

+7.0 m/s

Explanation:

Let's take rightward as positive direction.

So in this problem we have:

a = -2.5 m/s^2 acceleration due to the wind (negative because it is leftward)

t = 4 s time interval

v = -3.0 m/s is the final velocity (negative because it is leftward)

We can use the following equation:

v = u + at

Where u is the initial velocity

We want to find u, so if we rearrange the equation we find:

u = v - at = (-3.0 m/s) - (-2.5 m/s^2)(4 s)=+7.0 m/s

and the positive sign means the initial direction was rightward.

6 0
4 years ago
JOSEPH JOGS FROM END A TO OTHER END B OF A straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back
zzz [600]
(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second 

<span>(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second</span>
3 0
3 years ago
IIIIIIIIII neeeeeed helppppp again?
Juliette [100K]

Answer:

The Principle of Progression

(I searched it up since I never learned this)

Explanation:

The principle of progression states that a person should start slowly and increase exercise gradually. Since Mandy is just getting started on her exercise routine, she should begin with a few workouts over a large span of time, then work her way up so she can do more workouts in a shorter span of time.

7 0
3 years ago
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25
mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0
3 years ago
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