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3 years ago

Por favor. Escribe en Notación científica las siguientes medidas:

Physics

1 answer:

Zolol [24]3 years ago

6 0

I don’t understand Spanish could you please translate

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A swimmer standing near the edge of a lake notices a cork bobbing in the water. While watching for one minute, she notices the c

timurjin [86]

Answer:

4Hz (240 cycles/60 seconds = 4 cycles/second)

Explanation:

hope this helped!

5 0

3 years ago

Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.

stiv31 [10]

2cos2(θ)−1 = 0
2cos2(θ)=1
cos2(θ)=1/2
2θ=arc cos1/2
2θ=60 degrees
θ=30 degrees

4 0

3 years ago

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

enyata [817]

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.

7 0

3 years ago

Sound waves require a medium to travel through, such as a solid, liquid or gas because

Natasha2012 [34]

A

because waves need air or something else to travel through

8 0

3 years ago

Read 2 more answers

A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air

Bess [88]

Answer:

Explanation:

given,

initial velocity of the ball = 20 m/s

angle of ramp = 22°

ball travel at a distance = 5 m

a) for friction less

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh$

$v^2 = u^2 - 2gh$

$v = \sqrt{u^2- 2 g h cos 22^0}$

$v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0 }$

v = 17.58 m/s

b) considering the friction

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh-\mu_kmgl$

$v^2 = u^2 - 2gh-2\mu_kmgl$

$v = \sqrt{u^2- 2 g h cos 22^0-2\mu_kgl}$

$v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0-2\times 0.15\times 9.8 \times 5 }$

v = 17.16 m/s

7 0

3 years ago