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soldier1979 [14.2K]
3 years ago
5

Describe three ways a bicyclist can change velocity.

Physics
2 answers:
olga55 [171]3 years ago
6 0

Answer:

changing their speed, changing direction, or when both speed and direction change

Explanation:

changing their speed,

changing direction,

or when both speed and direction change

Sloan [31]3 years ago
3 0
Three ways of bicyclist can change velocity is changing their speed changing the direction or when both speed and direction change
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Which of the following is an inelastic collision?
Luda [366]
I'm pretty sure B would be your answer. 
7 0
4 years ago
Emma is riding on a train. The train is moving at 50 m/s. Emma walks down the aisle at 1 m/s relative to the train in the same d
anzhelika [568]

low speed means non relativistic.

the velocities relative to an observer outside the train are added.

51 m/s.

Were ita light wave, rather than Emma, the speed wold not depend on the speed of the train. Though that may sound surprising, I think it's true. Special relativity says more about this.

Special relativity "shows up" when the speeds get very high indeed.

6 0
4 years ago
In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst
dem82 [27]

Answer:

r_\alpha=16cm

Explanation:

The radius of the circumference described by a particle in a cyclotron is given by:

r=\frac{mv}{qB}(1)

m is the particle's mass, v is the speed of the particle, q is the particle's charge and B is the magnitude of the magnetic field.

Kinetic energy is defined as:

K=\frac{mv^2}{2}=\frac{m^2v^2}{2m}\\

Solving this for mv:

mv=\sqrt{2mK}(2)

Replacing (2) in (1):

r=\frac{\sqrt{2mK}}{qB}

For protons, we have:

r_p=\frac{\sqrt{2m_pK}}{eB}(3)

For alpha particles, we have:

r_\alpha=\frac{\sqrt{2m_\alpha K}}{(2e)B}(4)

Dividing (4) in (3):

\frac{r_\alpha}{r_p}=\frac{\frac{\sqrt{2m_\alpha K}}{(2e)B}}{\frac{\sqrt{2m_p K}}{(e)B}}\\r_\alpha=\frac{r_p}{2}\sqrt{\frac{m_\alpha}{m_p}}\\r_\alpha=\frac{16cm}{2}\sqrt{\frac{6.64*10^{-27}kg}{1.67*10^{-27}kg}}\\r_\alpha=\frac{16cm}{2}(\sqrt{3.98})\\\\r_\alpha=\frac{16cm}{2}(2)\\r_\alpha=16cm

4 0
4 years ago
An insulating cup contains 200 grams of water at 25 ∘C. Some ice cubes at 0 ∘C is placed in the water. The system comes to equil
Nataly [62]

Answer:

The amount of ice added in gram is 32.77g

Explanation:

This problem bothers on the heat capacity of materials

Given data

Mass of water Mw= 200g

Temperature of water θw= 25°c

Temperature of ice θice= 0°c

Equilibrium Temperature θe= 12°c

Mass of ice Mi=???

The specific heat of ice Ci= 2090 J/(kg ∘C)

specific heat of water Cw = 4186 J/(kg ∘C)

latent heat of the ice to water transition Li= 3.33 x10^5 J/kg

heat heat loss by water = heat gained by ice

N/B let us understand something, heat gained by ice is in two phases

Heat require to melt ice at 0°C to water at 0°C

And the heat required to take water from 0°C to equilibrium temperature

Hence

MwCwΔθ=MiLi +MiCiΔθ

Substituting our data we have

200*4186*(25-12)=Mi*3.3x10^5+

Mi*2090(12-0)

837200*13=Mi*3.3x10^5+Mi*2090

10883600=332090Mi

Mi=10883600/332090

Mi= 32.77g

4 0
4 years ago
An object of mass M oscillates on the end of a spring. To double the period, replace the object with one of mass: (a) 2M. (b)M/2
harkovskaia [24]

Answer:

(c) 4M

Explanation:

The system is a loaded spring. The period of a loaded spring is given by

T = 2\pi\sqrt{\dfrac{m}{k}}

<em>m</em> is the mass and <em>k</em> is the spring constant.

It follows that, since <em>k</em> is constant,

T\propto\sqrt{m}

\dfrac{T}{\sqrt{m}} = C where <em>C</em>  represents a constant.

\dfrac{T_1}{\sqrt{m_1}} = \dfrac{T_2}{\sqrt{m_2}}

m_2 = m_1\left(\dfrac{T_2}{T_1}\right)^2

When the period is doubled, T_2 = 2T_1.

m_2 = m_1\left(\dfrac{2T_1}{T_1}\right)^2 = 4m_1

Hence, the mass is replaced by 4M.

7 0
4 years ago
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