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zhuklara [117]
3 years ago
7

Dos masas están conectadas por una cuerda ligera que pasa por una polea sin rozamiento. Determine la aceleración de las masas y

la tensión de la cuerda si m Kg A  20 , m Kg B  50 y  0.20  K A Y B
Physics
1 answer:
Alexandra [31]3 years ago
5 0

Answer:A

Explanation:

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2. In the pendulum lab, the period (swing time) was O a. the independent variable O b.graphed on the vertical-axis O c.graphed o
Nadya [2.5K]
B is appropriate answer
7 0
3 years ago
It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, andthat the value of n in the expon
trapecia [35]

Answer:8.76\times 10^{-3} min^{-1}

Explanation:

Given

n=5

0.3 fraction recrystallize after 100 min

According to Avrami equation

y=1-e^{-kt^n}

where y=fraction Transformed

k=constant

t=time

0.3=1-e^{-k(100)^5}

e^{-k(100)^5} =0.7

Taking log both sides

-k\cdot (10^{10}=\ln 0.7

k=3.566\times 10^{-11}

At this Point we want to compute t_{0.5}\ i.e.\ y=0.5

0.5=1-e^{-kt^n}

0.5=e^{-kt^n}

0.5=e^{-3.566\times 10^{-11}\cdot (t)^5}

taking log both sides

\ln 0.5=-3.566\times 10^{-11}\cdot (t)^5

t^5=1.943\times 10^{10}

t=114.2 min

Rate of Re crystallization at this temperature

t^{-1}=8.76\times 10^{-3} min^{-1}

3 0
3 years ago
The electron gun in a television tube accelerates electrons (mass = 9.11x10^-31 kg, charge = 1.6 x 10^-19C) from rest to 3 x10^7
stich3 [128]

Answer:

Electric field acting on the electron is  127500 N/C.

Explanation:

It is given that,

Mass of an electron, m=9.11\times 10^{-31}\ kg

Charge on electron, q=1.6\times 10^{-19}\ C

Initial speed of electron, u = 0

Final speed of electron, v=3\times 10^7\ m/s

Distance covered, s = 2 cm = 0.02 m

We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}

a=2.25\times 10^{16}\ m/s^2

According to Newton's law, force acting on the electron is given by :

F = ma

F=9.1\times 10^{-31}\times 2.25\times 10^{16}

F=2.04\times 10^{-14}\ N

Electric force is given by :

F = q E, E = electric field

E=\dfrac{F}{q}

E=\dfrac{2.04\times 10^{-14}}{1.6\times 10^{-19}}

E = 127500 N/C

So, the electric field is 127500 N/C. Hence, this is the required solution.

8 0
3 years ago
How do you calculate change in position? A. initial position times two B. final position plus initial position C. final position
e-lub [12.9K]
The answer is C. Final position minus initial position.
5 0
3 years ago
Starting from rest, a 2.1x10^-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exert
Rainbow [258]

Answer:

a) The flea's speed when it leaves the ground is v=1.88m/s

b) The flea move 18cm upward while it is pushing off

Explanation:

Hi

<u>Knwons</u>

Mass m=2.1\times 10^{-4} kg, Work W=3.7\times 10^{-4} J and Force F=0.41N

a) Here we are going to use W=\frac{1}{2} mv^{2}, so v=\sqrt{\frac{2W}{m} }= \sqrt{\frac{2(3.7\times 10^{-4} J)}{2.1\times 10^{-4} kg} }=1.88m/s

a) Here we are going to use W=mgh, so h=\frac{W}{mg}= \frac{3.7\times 10^{-4} J}{(2.1\times 10^{-4} kg)(9.8m/s^{2})} =0,1797m or 18cm approx.

7 0
3 years ago
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