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julsineya [31]
2 years ago
15

Can you answer this math homework? Please!

Physics
1 answer:
steposvetlana [31]2 years ago
5 0
Using the count data and observational data you acquired, calculate the number of CFUs in the original sample
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Photoelectrons with a maximum speed of 7.00 · 105 m/s are ejected from a surface in the presence of light with a frequency of 8.
lord [1]
This is just testing your ability to recall that kinetic energy is given by: 

<span>k.e. = ½mv² </span>

<span>where m is the mass and v is the velocity of the particle. </span>

<span>The frequency of the light is redundant information. </span>

<span>Here, you are given m = 9.1 * 10^-31 kg and v = 7.00 * 10^5 m/s. </span>
<span>Just plug in the values: </span>

<span>k.e. = ½ * 9.1 * 10^-31 * (7.00 * 10^5)² </span>
<span>k.e. = 2.23 * 10^-19 J 
so it will be d:2.2*10^-19 J</span>
4 0
3 years ago
34.6 cL= (blank) hL convert
shepuryov [24]

<u>Answer</u>

0.00346 hL

<u>Explanation</u>

cL means Centilitre while hL means Hectolitre.

10,000 cL = 1 hL

∴ 34.6 cL = 34.6/10,000  hL

                = <em>0.00346 hL</em>

3 0
3 years ago
4. Who was the proponent of the Neo-classicism? a) Claude Debussy b) Joseph Maurice Ravel c) Igor Stravinsky d) Arnold Schoenber
nirvana33 [79]

Who was the proponent of the Neo-classicism?

a) Claude Debussy

b) Joseph Maurice Ravel

c) Igor Stravinsky

d) Arnold Schoenberg

4 0
2 years ago
A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)
bezimeni [28]

Answer:

205 V

V_{R} = 2.05 V

Explanation:

L = Inductance in Henries, (H)  = 0.500 H

resistor is of 93 Ω so R = 93 Ω

The voltage across the inductor is

V_{L} = - IwLsin(wt)

w = 500 rad/s

IwL = 11.0 V

Current:

I = 11.0 V / wL

 = 11.0 V / 500 rad/s (0.500 H)

 = 11.0 / 250

I = 0.044 A

Now

V_{R} = IR

    = (0.044 A) (93 Ω)

V_{R} = 4.092 V

Deriving formula for voltage across the resistor

The derivative of sin is cos

V_{R} = V_{R} cos (wt)

Putting V_{R} = 4.092 V and w = 500 rad/s

V_{R} = V_{R} cos (wt)

    = (4.092 V) (cos(500 rad/s )t)

So the voltage across the resistor at 2.09 x 10-3 s is which means

t = 2.09 x 10⁻³

V_{R} = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))

    =  (4.092 V) (cos (500 rads/s)(0.00209))

    = (4.092 V) (cos(1.045))

    = (4.092 V)(0.501902)

    = 2.053783

V_{R} = 2.05 V

8 0
2 years ago
Kinetic energy of an object is equal to
Gnom [1K]
See
K.E=1/2(mass*velocity²)
so option B is the correct answer.
Brainliest pls :-)
8 0
2 years ago
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