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Levart [38]
3 years ago
6

An experimenter finds that no photoelectrons are emitted from a particular metal unless the wavelength of light is less than 295

nm. her experiment will require photoelectrons of maximum kinetic energy 2.4 ev. what frequency light should be used to illuminate the metal?
Physics
1 answer:
wariber [46]3 years ago
7 0
If the object, ends up with a positive charge, then it is missing electrons. if it is missing electrons, then it must have been removed form the object during the rubbing process.
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why is a train more difficult to stop then a rolling ball even if they are traveling at the same speed? URGENT
zloy xaker [14]
Think of the formula force=mass x acceleration. even though they have the same acceleration, a train has more mass. is that helpful?
5 0
3 years ago
The position of an object in simple harmonic motion is defined by the function y = (0.50 m) sin (πt/2). Determine the maximum sp
Gnoma [55]

The maximum speed of the object under simple harmonic motion is 0.786 m/s.

The given parameters:

  • Position of the particle, y = 0.5m sin(πt/2)

<h3>Wave equation for simple harmonic motion;</h3>

y = A sin(ωt + Ф)

where;

  • A is the amplitude = 0.5 m
  • ω is the angular speed = π/2

The maximum speed of the object is calculated as follows;

V_{max} = A \omega\\\\V_{max} = 0.5 \times \frac{\pi}{2} = \frac{\pi}{4} \ m/s  = 0.786 \ m/s

Thus, the maximum speed of the object under simple harmonic motion is 0.786 m/s.

Learn more about simple harmonic motion here: brainly.com/question/17315536

5 0
2 years ago
An empty 230 kg elevator accelerates upward
elena-s [515]

Answer:

7.2 as used in the equation

3 0
3 years ago
Refer to the attached image!!!​
dimaraw [331]

The time of motion of the track star is determined as 0.837 s.

<h3>Time of motion of the track star</h3>

The time of motion of the track star is calculated as follows;

T = (2u sinθ)/g

where;

  • T is time of motion
  • g is acceleration due to gravity
  • θ is angle of projection

T = (2 x 12 x sin20)/9.8

T = 0.837 s

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

6 0
2 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
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