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aalyn [17]
3 years ago
10

Dos masas de 8kg es tan unidas en el extremo de una varilla de aluminio de 400mm de longitud. La varilla está sostenida en su pa

rte media hora en círculos y solo puede soportar una tensión máxima de 800 N ¿cuál es la frecuencia máxima de revolución
Physics
1 answer:
enyata [817]3 years ago
7 0

Answer:

The maximum frequency of revolution is 3.6 Hz.

Explanation:

Given that,

Mass = 8 kg

Distance = 400 mm

Tension = 800 N

We need to calculate the velocity

Using centripetal force

F=\dfrac{mv^2}{r}

Where, F= tension

m = mass

v= velocity

r = radius of circle

Put the value into the formula

800=\dfrac{8\times v^2}{200\times10^{-3}}

v^2=\sqrt{\dfrac{800\times200\times10^{-3}}{8}}

v=4.47\ m/s

We need to calculate the maximum frequency of revolution

Using formula of frequency

f=\dfrac{v}{2\pi r}

Put the value into the formula

f=\dfrac{4.47}{2\pi\times200\times10^{-3}}

f=3.6\ Hz

Hence, The maximum frequency of revolution is 3.6 Hz.

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A plant collects sunlight to form glucose, and your friend proposes an idea for a fan. Conserved = saving
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A car initially traveling 7 m/s speeds up uniformly at a rate of 3 m/s2 until it reaches a velocity of 22 m/s. How much time did
dezoksy [38]

Answer:

t = 5 s

Explanation:

Data:

  • Initial Velocity (Vo) = 7 m/s
  • Acceleration (a) = 3 m/s²
  • Final Velocity (Vf) = 22 m/s
  • Time (t) = ?

Use formula:

  • \boxed{t=\frac{Vf - Vo}{a}}

Replace:

  • \boxed{t=\frac{22\frac{m}{s} -7\frac{m}{s}}{3\frac{m}{s^{2}}}}

Solve the subtraction of the numerator:

  • \boxed{t=\frac{15\frac{m}{s}}{3\frac{m}{s^{2}}}}

It divides:

  • \boxed{t=5\ s}

How much time did it take the car to reach this final velocity?

It took a time of <u>5 seconds.</u>

8 0
3 years ago
The position of an electron is given by , with t in seconds and in meters. At t = 3.99 s, what are (a) the x-component, (b) the
egoroff_w [7]

Answer:

A. Vx = 3.63 m/s

B. Vy = -45.73 m/s

C. |V| = 45.87 m/s

D. θ = -85.46°

Explanation:

Given that position, r, is given as:

r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk

Velocity is the derivative of position, r:

V = dr/dt = 3.63 - 11.46t^j

A. x component of velocity, Vx = 3.63 m/s

B. y component of velocity, Vy = -11.46t

t = 3.99 secs,

Vy = - 11.46 * 3.99 = -45.73 m/s

C. Magnitude of velocity, |V| = √[(-45.73)² + 3.63²]

|V| = √(2091.2329 + 13.1769)

|V| = √(2104.4098)

|V| = 45.87 m/s

D. Angle of the velocity relative to the x axis, θ is given as:

tanθ = Vy/Vx

tanθ = -45.73/3.63

tanθ = -12.6

θ = -85.46°

7 0
3 years ago
Two metal spheres are suspended from strings. The
nikdorinn [45]

Answer:

B. A repulsive force of 8.0*10^3 N.

Explanation:

As we know by Coulomb's law that the electrostatic force between two charges is given as

F = \frac{kq_1q_2}{r^2}

here we know that

q_1 = -2 \times 10^2 C

q_2 = -4 \times 10^{-8} C

r = 3.0 m

now we have

F = \frac{(9 \times 10^9)(2 \times 10^2)(4\times 10^{-8})}{3^2}

F = 8000 N

since both charges are similar charges so they will repel each other by the force we calculated above so correct answer will be

B. A repulsive force of 8.0*10^3 N.

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3 years ago
How much equal charge should be placed on the earth and the moon so that the electrical repulsion balances the gravitational for
kumpel [21]

As we know that electrostatic force between two charges is given as

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here we know that electrostatic repulsion force is balanced by the gravitational force between them

so here force of attraction due to gravitation is given as

F_g = 1.98 \times 10^{20} N

here we can assume that both will have equal charge of magnitude "q"

now we have

1.98 \times 10^{20} = \frac{kq^2}{r^2}

1.98 \times 10^{20} = \frac{(9\times 10^9)(q^2)}{(3.84 \times 10^8)^2}

1.98 \times 10^{20} = (6.10 \times 10^{-8}) q^2

now we have

q = 5.7 \times 10^{13} C

6 0
3 years ago
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