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kvasek
5 days ago
8

When you ride a bicycle at constant speed, nearly all the energy you expend goes into the work you do against the drag force of

the air. Model a cyclist as having cross-section area 0.40 m2 and, because the human body is not aerodynamically shaped, a drag coefficient of 0.90.
What is the cyclist's power output while riding at a steady 7.3 m/s?
Physics
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Four wires are made of the same highly resistive material, cut to the same length, and connected in series. Wire 1 has resistanc
ValentinkaMS [3465]

Answer:

V_2 = \frac{R_2 V_0}{R_1+R_2+R_3+R_4}

Explanation:

The four wires are arranged in series: this setup indicates that the same current passes through them, while the total voltage from the battery, V0, equals the combined voltages across each resistor:

V_0=V_1+V_2+V_3+V_4

Additionally, the total resistance for this series configuration is

R_{eq}=R_1+R_2+R_3+R_4

Using Ohm's law, we can determine the voltage V2 across wire 2:

V_2 = R_2 I (1)

Here, I represents the entire current flowing through the circuit, which is calculated as:

I=\frac{V_0}{R_{eq}}=\frac{V_0}{R_1+R_2+R_3+R_4}

By substituting into equation (1), we derive V2:

V_2 = \frac{R_2 V_0}{R_1+R_2+R_3+R_4}

8 0
2 months ago
The energy difference between the 5d and the 6s sublevels in gold accounts for its color. If this energy difference is about 2.7
Yuliya22 [3333]
The light's wavelength absorbed during the transition is 459 nm. Energy difference between the 5-d and the 6-s sub-levels in gold is expressed as ΔE. Let the wavelength associated with the electron's transition from the 5-d to the 6-s state be λ. The relationship that describes the connection between energy and wavelength is defined as: E = hc/λ, where E stands for photon energy, h represents Planck's constant, c is the speed of light, and λ denotes the wavelength of the photon. Therefore, the absorption wavelength in this transition stands at 459 nm.
8 0
2 months ago
Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e
Keith_Richards [3271]

Answer:

The radius is r = 4.434 *10^{-5} \ m

Explanation:

The problem states that

    The magnetic field is  B = 90 mT = 90*10^{-3} \ T

     The electron kinetic energy is  KE = 1.4 eV = 1.4 * (1.60*10^{-19}) =2.24*10^{-19} \ J

In general, for a collision to happen, the centripetal force on the electron in its orbit must equal the magnetic force acting on it  

   This can be mathematically expressed as

   \frac{mv^2}{r} = qvB

=>    r = \frac{m* v}{q * B}

Where  m denotes the electron’s mass, which has a value of m = 9.1 *10^{-31} \ kg  

             v signifies the escape velocity, mathematically represented as

                v = \sqrt{\frac{2 * KE}{m} }

Thus,

       r = \frac{m}{qB} * \sqrt{\frac{2 * KE}{m} }

     applying indices

    r = \frac{\sqrt{2 * KE * m} }{qB}

substituting these values

   

       

r = \frac{\sqrt{2 * 2.24*10^{-19}* 9.1 *10^{-31}} }{ 1.60 *10^{-19}* 90*10^{-3}}

       

r = 4.434 *10^{-5} \ m

     

6 0
2 months ago
Two infinite parallel surfaces carry uniform charge densities of 0.20 nC/m2 and -0.60 nC/m2. What is the magnitude of the electr
Yuliya22 [3333]

Answer:

The electric field strength, E = 45.19 N/C

Explanation:

It is indicated that,

Surface charge density on the first surface, \sigma_1=0.2\ nC/m^2=0.2\times 10^{-9}\ C/m^2

Surface charge density on the second surface, \sigma=-0.6\ nC/m^2=-0.6\times 10^{-9}\ C/m^2

The electric field at a location between the two surfaces can be calculated as:

E=\dfrac{\sigma}{2\epsilon_o}

E=\dfrac{\sigma1-\sigma_2}{2\epsilon_o}

E=\dfrac{0.2\times 10^{-9}-(-0.6\times 10^{-9})}{2\times 8.85\times 10^{-12}}

Consequently, E = 45.19 N/C

Therefore, the electric field's magnitude at a point between both surfaces is 45.19 N/C.

6 0
2 months ago
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