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ki77a [65]
2 years ago
6

If it takes Ashley 6.1 seconds to run at an average speed of 5.7 meters per second, what is the distance (in meters) she covers

in that time? Round your answer to one decimal place.​
Physics
1 answer:
Softa [21]2 years ago
3 0

Answer:

34.8m

Explanation:

distance = speed x time

6.1 x 5.7 = 34.77

34.8

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The speed limit sign in indiana reads 60 miles per hour. convert these units to meters per second
cluponka [151]

Answer:

26.822 m/s

Explanation:

60 mi/hr  *  5280 ft/mile *  1 hr / 3600 sec  *  12 in / foot * 1 meter / 39.37 in = <u>26.822 m/s</u>

3 0
1 year ago
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau
mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = \frac{7}{5} (\frac{1}{2} m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

         \frac{y_{sphere}} {y_bolck} = 5/7

3 0
3 years ago
A model train traveling at a constant speed around a circular track has a constant velocity
Mrac [35]
The answer should be yes
7 0
3 years ago
Read 2 more answers
PLEASE PLEASE PLEASE A mad scientist places massive amounts of charge on basketball sized aluminum balls. The charge on the ball
bazaltina [42]

Answer:

4.2 x 10⁷N

Explanation:

Given parameters:

Charge on ball:

             q₁  = 3C

              q₂ = 14C

Distance between balls  = 9000m

Unknown:

Force acting on the two balls

Solution:

The force experienced by the two charges is given by coulombs law. It is mathematically expressed as;

                      F  = \frac{k q_{1} q_{2} }{r^{2} }

where k  = 9 x 10⁹Nm²/C²

           q is the charges

             r is the distance

Input the variables and solve;

                 

        F  = \frac{9 x 10^{9} x 3 x 14 }{9000}  = 4.2 x 10⁷N

8 0
3 years ago
A substance that cannot be broken into simpler elements ia an
Anastaziya [24]
Element. element cannot be simplerfide 
3 0
2 years ago
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