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Lelechka [254]
3 years ago
15

A change in which property of light will have no effect on whether or not the photoelectric effect occurs??? Frequently energy i

ntensity and wavelength
Physics
2 answers:
defon3 years ago
8 0

Answer: intensity

Explanation:

MaRussiya [10]3 years ago
5 0
The intensity of the light has no connection with the photoelectric effect.

That's what was so baffling about it before the particle nature of light
was suspected ... a match with a blue flame might stimulate the
photoelectric effect, but a high-power red searchlight couldn't do it.
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A very light ping-pong ball moving east at a speed of 4 m/s collides with a very heavy stationary bowling ball. The Ping-Pong ba
KengaRu [80]

Answer:

They experience the same magnitude impulse

Explanation:

We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:

where is the initial momentum of the ping-poll ball

is the initial momentum of the bowling ball (which is zero, since the ball is stationary)

is the final momentum of the ping-poll ball

is the final momentum of the bowling ball

We can re-arrange the equation as follows or

which means (1) so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.

However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:

(2) therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse:

3 0
3 years ago
A ball is thrown horizontally from the top of a building 37.5 m high. The ball strikes the ground at a point 80.3 m from the bas
trasher [3.6K]

Answer:

t = 2.77 s

Explanation:

The ball in its movement describes a curved line called a semiparabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane:

Equation of movement of the ball in the X axis

X = v₀x*t   Equation  (1)

Equation of movement of the ball in the Y axis

Y = y₀+ v₀y*t -½ g*t² Equation  (2)

Where

X : horizontal position in meters (m)  

Y : vertical  position in meters (m)  

y₀ : initial vertical  position in meters (m)

v₀x :  X-initial speed in m/s  

v₀y :  Y-initial speed in m/s  

g: acceleration due to gravity in m/s²

t : time to position (X,Y)

Data

y₀ = 37.5 m

v₀y = 0

g = 9.8 m/s²

Problem development

The time the ball remains in the air is the same as the ball takes to touch the floor, that is, Y = 0

We apply the Equation (2):

Y = y₀+ (v₀y)*t - (½) g*t²

0 =  37.5 +(0)*t- (1/2)*(g)*t²

0 =  37.5 - (1/2)*(9.8)*t²

(1/2)*(9.8)*t²  = 37.5

t² = (2)(37.5)/(9.8)

t= \sqrt{\frac{2*37.5}{9.8} }

t = 2.77 s

5 0
3 years ago
Please help me guys please​
leonid [27]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Solution is in attachment ~

I hope that you got what you were looking for, and if there's different data then go through the same procedure, using same formula with different values and you will get your answer ~

\mathrm{✌TeeNForeveR✌}

8 0
2 years ago
Give me 2 animals to combine their inherited traits
Kazeer [188]

Answer:

dogs and cats

Explanation:

hope this helps

7 0
3 years ago
Read 2 more answers
Be sure to answer all parts. By what factor does the fraction of collisions with energy equal to or greater than activation ener
qwelly [4]

Answer:

Factor = 8.77

Explanation:

Fraction of collision with energy greater than or equal to the activation energy  can be given by the formula:

f = \exp(\frac{-E_{a} }{RT} )

E_{a}  = Activation Energy\\E_{a}  = 100 kJ /mol\\E_{a}  = 10^{5} J /mol\\

R = 8.314 J/mol.K

When Temperature = 34⁰C = 34 + 273 = 307 K

f_{1}  = \exp(\frac{-10^{5}  }{8.314 * 307} )\\f_{1} = \exp(\frac{-10^{5}  }{2552.398} )\\f_{1}  = \exp(-39.18)\\f_{1} = 964.59 * 10^{-20}

When Temperature =  52⁰C = 52 + 273 = 325 K

f_{2}  = \exp(\frac{-10^{5}  }{8.314 * 325} )\\f_{2}= \exp(\frac{-10^{5}  }{2702.05} )\\f_{2}= \exp(-37.009)\\f_{2} = 8456.6 * 10^{-20}

\frac{f_{2}}{f_{1}}= \frac{8456.6 * 10^{-20} }{964.59 * 10^{-20} }

Factor = 8.77

8 0
3 years ago
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