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3 years ago

12

A bicycle rider traveling east at 10 km/hr sees a blue car pass her appearing to travel west at 50 km/hr. What is the blue car's

apparent speed for an observer riding in a red car traveling 15 km/hr west relative to the Earth? g

1 answer:

BigorU [14]3 years ago

8 0

Answer:

$v_{bR} = 25 km/h$ towards west

Explanation:

As we know that the speed of the blue car as appear to the bicycle rider is given as

$v_{bc} = 50 km/h$ towards west

also it is given that bicycle is moving at speed of 10 km/h towards East

so here we have

$v_{bc} = v_b - v_c$

so we have

$v_b = v_{bc} + v_c$

$v_b = -50 + 10 = 40 km/h$ towards west

now speed of the red car is given as 15 km/h towards west

so here the relative speed of blue car with respect to red car is given as

$v_{bR} = v_b - v_R$

$v_{bR} = 40 - 15 = 25 km/h$ towards west

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The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu

Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

$E=\phi + K$

where

$E=\frac{hc}{\lambda}$ is the energy of the incident light, with h being the Planck constant, c being the speed of light, and $\lambda$ being the wavelength

$\phi$ is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

$\lambda=190 nm=1.9\cdot 10^{-7}m$ , so the energy of the incident light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J$

Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: $\phi = 4.5 eV$

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

$E=\phi+K=4.5 eV+2.7 eV=7.2 eV$

Then the copper is replaced with sodium, which has work function of

$\phi = 2.3 eV$

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

$K=E-\phi = 7.2 eV-2.3 eV=4.9 eV$

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4 years ago

In physics what does 7.56 × 5.746 equal ?

lapo4ka [179]

Answer:

43.43

Explanation:

5.746 x 7.56 = 43.43976

As the least number of desimal is two so our awnser should contain two digits after the decimal point.

Ans: 43.43.

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4 years ago

The half life of uranium-235 is 4. 5 billion years. If 0. 5 half-lives have elapsed, how many years have gone by?

igor_vitrenko [27]

Answer:

2.25 billion years

Explanation:

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2 years ago

A girl exerts a horizontal force of 109 N on a crate with a mass of 31.2 kg. HINT (a) If the crate doesn't move, what's the magn

vesna_86 [32]

Answer:

(a) Magnitude of static friction force is 109 N

(b) Minimum possible value of static friction is 0.356

Solution:

As per the question;

Horizontal force exerted by the girl, F = 109 N

Mass of the crate, m = 31.2 kg

Now,

(a) To calculate the magnitude of static friction force:

Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:

F = f = 109 N

(b) The maximum possible force of friction between the floor and the crate is given by:

$f_{m} = \mu_{s}N$

where

N = Normal reaction = mg

Thus

$f_{m} = \mu_{s}mg$

For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.

$f\leq f_{m}$

$f \leq \mu_{s}mg$

$109 \leq \mu_{s}\times 31.2\times 9.8$

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Which of the following scenarios will cause the current to stop flowing within a circuit?

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3 years ago

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