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viktelen [127]
3 years ago
11

I HAVE 5 MINUTES!!!!!! A block oscillating on the end of a spring moves from its position of maximum spring stretch to maximum s

pring compression in 0.25 s. Determine the period and frequency of this motion​
Physics
1 answer:
Verdich [7]3 years ago
4 0

Therefore, if the block moves from its position of maximum spring stretch to maximum spring compression in 0.25 s, the time required for a full cycle is twice as much; T = 0.5 s.

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The minimum energy needed to eject an electron from a sodium atom is 4.41 x 10-19 j. what is the maximum wavelength of light, in
Minchanka [31]
The energy of an electron as it is ejected from the atom can be calculated from the product of the Planck's constant and the frequency of the light energy. We can calculate the wavelength from the frequency we can calculate. We do as follows:

E = hv
 4.41 x 10-19  = 6.62607004 × 10<span>-34 (v)
v = 6.66x10^14 /s

wavelength = speed of light / frequency
</span>
wavelength = 3x10^8 / 6.66x10^14
wavelength = 4.51x10^-7 m = 450.75 nm
5 0
3 years ago
Can someone see if it's right? thank you. ​
bearhunter [10]

Answer:

it's right you did a great job

8 0
2 years ago
A car going 22 m/s increases its speed to pass a truck. Five seconds later the car is going 35 m/s. Calculate the acceleration o
jonny [76]

Answer:

I don't know

Explanation:

4 0
3 years ago
Read 2 more answers
A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d
prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance  

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

V_{t}= V_{o}e^{\frac{-t}{R*C} }

Voltage in this case is the energy dissipated so

E_{t}= E_{o}e^{\frac{-t}{R*C} }

\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }

\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }

Using the equation to find capacitance

ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }

C= 5.32x10^{-6} F

C= 5.32 uF because u is the symbol for micro that is equal to 10^{-6}

8 0
3 years ago
A system releases 255 cal of heat to the surroundings while delivering 428 cal of work. what is the change in internal energy of
vichka [17]

Answer:-683 cal

Explanation:

Given

Heat released by system Q=-255 cal

as heat released is taken as negative and vice-versa

Work done by system W=428 cal

From First law of thermodynamics

Q=\Delta +W , where \Delta U=change in internal Energy

\Delta U=-255-428=-683

7 0
3 years ago
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