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Strike441 [17]
4 years ago
14

Describe 3 ways a bicyclist can change velocity.

Physics
2 answers:
umka2103 [35]4 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]4 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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Please help me with this, I will mark as Brianalist if right...​
Solnce55 [7]

Explanation:

To me, it's option 1

Newton third law states that Every action produces an equal and opposite reaction. Hence, given the fact that the engine is on, it should be producing a constant force every time so, the speed should be constant as the force propelling it is the same

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3 years ago
Nuclear power has come under higher scrutiny in recent years. Which statement would make the strongest argument for the use of n
yan [13]

Nuclear power generates large amounts of power with limited production of greenhouse gases.

is the answer

4 0
3 years ago
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How do its eight arms help an octopus obtain food
Andrews [41]
The more arms it has the less of a chance the prey has to swim away.
5 0
3 years ago
A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be
FrozenT [24]

Answer:

\omega=0.31\frac{rad}{s}

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

a_c=\frac{v^2}{r}(1)

Here, r is the radius and v is the tangential speed, which is given by:

v=\omega r(2)

Here \omega is the angular velocity, we replace (2) in (1):

a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r

Recall that r=\frac{d}{2}=\frac{200m}{2}=100m.

Solving for \omega:

\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}

3 0
4 years ago
The stoplight had just changed and a 2200 kg cadillac had entered the intersection, heading north at 2.8 m/s , when it was struc
tekilochka [14]
<span>3. The attempt at a solution So basically what I did was divided into components. x: (3)(2000) = (3000)*v_x y: (v_vw)*(10000) = (3000)*v_y v_x, v_y is the velocity (after collision) in the x and y direction, respectively, of both cars stuck together (since it is an inelastic collision). v_vw is the initial velocity of the Volkswagen. Now what I did was that the angle is 35 degrees north of east. So basically made a triangle and figured that tan(35) = (v_y)/(v_x). This means (v_x)*(tan35) = v_y. Then, I simplified the component equations to get: x: 2 = v_x y: v_vw = 3*v_y Then plugging in for v_y, I got: v_vw = 3(2)(tan35) = 4.2 m/s as the velocity of the volkswagen. However, the answer key says 8.6 m/s. Could someone please help me out? Thanks Phys.org - latest science and technology news stories on Phys.org • Game over? Computer beats human champ in ancient Chinese game • Simplifying solar cells with a new mix of materials • Imaged 'jets' reveal cerium's post-shock inner strength Oct 24, 2012 #2 ehild Homework Helper Gold Member What directions you call x and y? Reference https://www.physicsforums.com/threads/2d-momentum-problem.646613/</span>
7 0
3 years ago
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