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Strike441 [17]
3 years ago
14

Describe 3 ways a bicyclist can change velocity.

Physics
2 answers:
umka2103 [35]3 years ago
8 0
She can pedal harder to increase the speed of the bicycle. She can apply the brakes to decrease the speed of the bicycle. And she can turn the handle bars to change the bicycle's Direction.
ivann1987 [24]3 years ago
4 0
1. Applying more force and energy, speeding up
2. Releasing potential and kinetic energy by going down a slope, speeding up
3. Applying less force to slow down and let friction overcome the bicyclist
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6. Mass affects how fast an object falls.<br> True<br> False
dimulka [17.4K]

The statement: Mass affects how fast an object falls is true.

6 0
3 years ago
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A spring with k = 500 N/m stores 704 J. How far is it extended from the equilibrium position
kaheart [24]

Given that:

k = 500 n/m,

work (W) = 704 J

spring extension (x) = ?

         we know that,

                      Work = (1/2) k x²

                          704 = (1/2) × 500 × x²

                            x = 1.67 m

A spring stretched for 1.67 m distance.

4 0
3 years ago
Which of the following has potential but not kinetic energy?
gavmur [86]

The answer would be option D "a ball sitting on a shelf." Potential energy is the amount of energy a object has while it's at rest.. (or not moving) Kinetic energy is how much energy a object is while it's moving. So in this case it's option D because a ball sitting on a shelf isn't moving therefore it has potential energy. It's not option A because thats a example of kinetic energy since how the roller coaster is moving. It's not option B because it's kinetic energy because the bike is moving. It's also not option C because it's kinetic energy because the bird is moving.


Hope this helps!

7 0
3 years ago
Joshua was driving to a friend’s house to study. During his trip, he started on pavement. At one point, he hit an ice patch on t
Tems11 [23]

Answer:

b. Friction decreased when he went from pavement to ice and then increased two more times.

Explanation:

Frictional force depends on the normal force of the surface and a friction coefficient.

F_{f} = -\mu N

Since we're talking about the same car, the value of N will remain constant whereas μ will represent the change in the frictional coefficient of the surface. Now we consider the different surfaces, cars will slide in an icy road which means that the frictional coefficient is smaller than the pavement.

After Joshua returns to the pavement road, the resulting frictional force increases and will do so one more time when he reaches the gravel road. Gravel roads have greater frictional coefficients than pavement roads which means the frictional force will increase a second time.

7 0
3 years ago
Read 2 more answers
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
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