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3 years ago

Buco novia de 15n de edad

Physics

1 answer:

Zigmanuir [339]3 years ago

4 0

Yo tengo quince y soy de cali y estoy viviendo en mexico

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A horizontal spring with spring constant 210 Nm is compressed by 20 cm and then used to launch a 250 g box across the floor. The

Damm [24]

Answer:

ugmd = 1/2 kx²

d = (1/2 kx²) / (ugm)

= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)

= 7.4 m

ugmd = 1/2 mv²

v = √2ugd

= √(2(0.23)(9.81 m/s²)(7.4 m)

= 5.8 m/s

Explanation:

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4 years ago

Find the change in kinetic energy of a 1.0 kg fish leaping to the right at 12.0 m/s.

sp2606 [1]

Answer:

Explanation:

Given parameters:

Mass of fish = 1kg

Velocity = 12m/s

Unknown:

Change in kinetic energy = ?

Solution:

Kinetic energy is the energy due to the motion of a body. It is mathematically given as:

K.E = $\frac{1}{2}$ m v²

Now, insert the parameters and solve;

K.E = $\frac{1}{2}$ x 1 x 12 = 6J

The change in kinetic energy is 6J

4 0

3 years ago

Read 2 more answers

What is the motion of the particles in this kind of wave?

Musya8 [376]

The answer to this question is B I think

7 0

3 years ago

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What is the mass, in kilograms, of an avogadro's number of people, if the average mass of a person is 170 lb ?

andre [41]

About 4.7e25 kilograms

8 0

3 years ago

Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.4 m from the slits. The center of the 3

VARVARA [1.3K]

Answer:

Part a)

$f = 4.76 \times 10^{14} Hz$

Part b)

$d = 3.48 \times 10^{-4} m$

Part c)

$\theta = 0.311 degree$

Explanation:

Part a)

As we know that the speed of light is given as

$c = 3 \times 10^8 m/s$

$\lambda = 630 nm$

now the frequency of the light is given as

$f = \frac{c}{\lambda}$

so we have

$f = \frac{3 \times 10^8}{630 \times 10^{-9}}$

$f = 4.76 \times 10^{14} Hz$

Part b)

Position of Nth maximum intensity on the screen is given as

$y_n = \frac{n\lambda L}{d}$

so here we know for 3rd order maximum intensity

$y_3 = 0.76 cm$

n = 3

L = 1.4 m

$0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}$

$d = 3.48 \times 10^{-4} m$

Part c)

angle of third order maximum is given as

$d sin\theta = 3 \lambda$

$3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})$

$\theta = 0.311 degree$

8 0

3 years ago