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GuDViN [60]
3 years ago
7

En un m.A.S. La amplitud tiene un valor de 10 centimetros y el periodo es de 2 segundos calcular el valor de la velocidad de 0.8

y 1.4 segundos de haberse iniciado el movimiento
Physics
1 answer:
Ivanshal [37]3 years ago
6 0

Answer:

v1=18.46m/s

v2=29.8cm/s

Explanation:

We know that

A=10cm\\T=2s

the equation of the motion is

x=Acos(\omega t)\\

we can calculate w by using

\omega=\frac{2\pi}{T}=\frac{2\pi}{2}=\pi

Hence, we have that

x=10cm*cos(\pi t)\\

the speed will be

v=-\omega*Asin(\omega t)\\|v(0.8)|=|\pi*10cm*sin(\pi *0.8)|=18.46\frac{cm}{s}\\|v(1.4)|=|\pi*10cm*sin(\pi *1.4)|=29.8\frac{cm}{s}

hope this helps!

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A hammer has a mass of 1 kg. What is its weight (i) on Earth (ii) on the
solong [7]

Given mass= 1kg

Weight on earth = mg(gravity of earth) = 9.8N

weight on moon = mg(gravity of moon)= 1.62N

weight on outer space mg(gravity outer space = 0) = 0N

4 0
3 years ago
What profession is most likely to make use of amphoras and candelas
Monica [59]
Biologists probably
5 0
3 years ago
A block of amber is placed in water and a laser beam travels from the water through the amber. The angle of incidence is 35 degr
7nadin3 [17]
Answer: 1.88

Explanation

Applying Snell’s Law, sin(1)/sin(2) = n(2)/n(1), where n is the index of refraction and sin 1 and 2 being of incidence and refracted respectively.

1) sin35/sin24 = n(2)/1.33
2) 1.41 = n(2)/1.33
3) n(2) = 1.41 x 1.33
4) n(2) = 1.88

Hope this helps :)
7 0
3 years ago
How much work must be done to bring three electrons from a great distance apart to 5.0×10^−10 m from one another (at the corners
Inessa05 [86]

Answer:

1.38 x 10^-18 J

Explanation:

q = - 1.6 x 10^-19 C

d = 5 x 10^-10 m

the potential energy of the system gives the value of work done

The formula for the potential energy is given by

U =\frac{Kq_{1}q_{2}}{d}

So, the total potential energy of teh system is

U =\frac{Kq_{1}q_{2}}{d}+\frac{Kq_{2}q_{3}}{d}+\frac{Kq_{1}q_{3}}{d}

As all the charges are same and the distance between the two charges is same so the total potential energy becomes

U =3\times \frac{Kq^{2}}{d}

K = 9 x 10^9 Nm^2/C^2

By substituting the values

U =3\times \frac{9\times 10^{9}\times \ 1.6 \times 1.6 \times 10^{-38}}{5\times 10^{-10}}

U = 1.38 x 10^-18 J

6 0
3 years ago
Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive
Burka [1]

Answer:

1.91773\times 10^{37}\ kg

Explanation:

v = Orbital speed = 130 km/s

d = Diameter = 16 ly

r = Radius = \dfrac{d}{2}=\dfrac{16}{2}=8\ ly

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

1\ ly=9.461\times 10^{15}\ m

As the centripetal force balances the gravitational energy we have the following relation

\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow M=\dfrac{v^2r}{G}\\\Rightarrow M=\dfrac{130000^2\times 8\times 9.461\times 10^{15}}{6.67\times 10^{-11}}\\\Rightarrow M=1.91773\times 10^{37}\ kg

Mass of the the massive object at the center of the Milky Way galaxy is 1.91773\times 10^{37}\ kg

4 0
3 years ago
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