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Ahat [919]
3 years ago
13

How to find friction force given mass and acceleration?

Physics
1 answer:
qwelly [4]3 years ago
8 0
You do m times a equal ff
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The molar mass of the sample is equal to the summation of the molar mass of the elementas multiplied by the abundance of the elements by mole. In this case, copper has an abundance of 93.69 percent while zinc has 6.31 percent. In this case, the average molecular weight is 63.67 g/mol
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Does exists the friction force in space?If yes,tell an full example
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All are examples of electric forces except _________.
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The correct answer is A
6 0
3 years ago
Can you answer this math homework? Please!
kap26 [50]

\large \mathfrak{Solution : }

9. An object which is in circular motion (moving along a circle) is said to be accelerating because it changes it's direction constantly even if it is moving with a constant speed. cuz acceleration is change in either magnitude or direction of an object with respect to time.

therefore, it's still acceleration as change in direction with time.

10. Average speed of an object can be calculated by dividing the total distance covered by an object by time taken to cover that distance.

i.e

  • \boxed{speed =  \dfrac{distance}{time} }

it can be re- arranged to find the distance as :

  • \boxed{distance = speed \times time}

  • time =  \dfrac{distance}{speed}

11. speed = 20 m/s : conversion into km/h

distance covered : 4 km = 4000 m

  • time =  \dfrac{distance}{speed}

  • t =  \dfrac{4000}{20}

  • t =  200 \:  \: sec

time taken = 200 seconds

12. let's use the first equation of motion to find the acceleration :

  • v = u + at

  • 50 = 80 + 120a

  • 50 - 80 = 120a

  • a =  \dfrac{ - 30}{120}

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3 0
3 years ago
In a photoelectric effect experiment, electrons emerge from a copper surface with a maximum kinetic energy of 1.10 eV when light
Sladkaya [172]

Answer:A) 220 nm

Explanation:

Given

Maximum Kinetic Energy K.E.=1.10 eV

Work Function W=4.65 eV

from Einstein Equation

h\mu =W+K.E.

h\cdot \frac{c}{\lambda }=W+K.E.

h\cdot \frac{c}{\lambda }=4.65+1.10

6.626\times 10^{-34}\cdot \frac{3\times 10^8}{\lambda }=5.75

1 eV=1.6\times 10^{-19} J

thus 5.75 eV=9.2\times 10^{-19} J

\lambda =\frac{6.626\times 10^{-34}\time 3\times 10^8}{9.2\times 10^{-19}}

\lambda =216.06 nm\approx 220 nm

6 0
3 years ago
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