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dsp73
3 years ago
8

Goslo and Speedo are twins. They have just celebrated their 20th birthday (Yay!). Speedo travels to the planet Zorth which is 40

light years away, he immediately returns. Neglect any time Speedo spent accelerating. If Speedo travels at 0.997 relative to the Earth, who will measure the proper time?
a. Goslo
b. Speedo
c. Nobody
d. Everybody
Physics
1 answer:
Wittaler [7]3 years ago
4 0

Answer:

a. Goslo

Explanation:

given data

Speedo travels = 40 light year

Speedo travels relative to the Earth = 0.997

solution

Goslo will measure here proper time and this time is expres as

time T =  Speedo travels ÷ Speedo travels relative to the Earth   ..........................1

put here value

time T = \frac{40}{0.997}

time T = 40.1 year

so Goslo will measure the proper time

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El dormitorio de Pablo es rectangular, y sus lados miden 3 y 4 metros. Ha decidido dividirlo en dos partes triangulares con una
Paraphin [41]

Answer:

 c = 5 m

Explanation:

this exercise you want to divide the rectangular room into two triangular rooms

                 

the area of ​​triangles is

           A = ½ base height

           A = ½ 4 3

          A = 6 m²

the length of the curtain can be found using the Pythagorean theorem

           c² = b² + a²

           c = √ (4² + 3²)

           c = 5 m

this is the length of the curtain

5 0
3 years ago
In the simplified version of Kepler's third law, P 2 = a3, the units of the orbital period P and the semimajor axis a of the ell
Orlov [11]

Answer:

The units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

Explanation:

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Therefore, if you square the year of each planet and divide it by the distance that it is from the star, you will get the same number for all the other planets.

Thus, the units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

8 0
3 years ago
The weather conditions
bixtya [17]

Answer:

ok confusion but we could figure it out right

Explanation:

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4 0
3 years ago
A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}

v_2=0.301\ m/s

So, the speed of the large cart after collision is 0.301 m/s.

4 0
3 years ago
A hot-air balloon is rising upward with a constant speed of 2.85 m/s. When the balloon is 2.50 m above the ground, the balloonis
NISA [10]

Answer:

Explanation:

You have to declare which way is plus -- up or down. I will use down.

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a = 9.81

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d = vi*t + 1/2 a t^2

-2.50 = -2.85*t + 1/2 * 9.81 * t^2

-2.50 = -2.85*t + 4.905 * t^2           transfer the left to the right.

-4.905 t^2 + 2.85*t + 2.50 = 0

Use the quadratic formula to solve for t.

It turns out that t = 1.06

5 0
3 years ago
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