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Rashid [163]
3 years ago
5

What is the farthest that unmanned space exploration has reached?

Physics
1 answer:
lapo4ka [179]3 years ago
7 0
I would say D. because the voyager has gone a little past the boundaries of the solar system

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PLEASE HELP!!!!
weqwewe [10]
The answer is A. <span>Some work input is used to overcome friction. </span>
6 0
3 years ago
A 25 kg mass is hanging from two cables, each with their own tension. Cable 1 is connected to the
Georgia [21]

Answer:

a. one line down one line to the right one live to the northwest from the object

b. t1=190 t2=310

Explanation:

5 0
2 years ago
What can change of matter from one state to another state?
Anna35 [415]

Answer:

answer is Heating

Explanation:

take a solid and heat it it will become a liquid

8 0
2 years ago
Which of the following is a false statement? Select one:
Art [367]

Answer:

The false statement is in option 'd': The center of mass of an object must lie within the object.

Explanation:

Center of mass is a theoretical point in a system of particles where the whole mass of the system is assumed to be concentrated.

Mathematically the position vector of center of mass is defined as

\overrightarrow{r}_{com}=\int \overrightarrow{r}_{i}dm

where,

\overrightarrow{r}_{i} is the position vector of the mass dm.

As we can see for homogenous symmetrical objects such as a sphere,cube,disc the center of mass is located at the centroid of the shapes itself but in many shapes it is located outside the body also.

Examples of shapes in which center of mass is located outside the body:

1) Horseshoe shaped body.

2) A thin ring.

In many cases we can make shapes of bodies whose center of mass lies outside the body.

6 0
3 years ago
g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn
krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

            T² = (\frac{4\pi }{G M_s} ) r³             (1)

           

in this case the period of the season is

            T₁ = 93 min (60 s / 1 min) = 5580 s

            r₁ = 410 + 6370 = 6780 km

            r₁ = 6.780 10⁶ m

for the satellite

           T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

            T² = K r³

            K = T₁²/r₁³

            K = \frac{ 5580^2}{ (6.780 10^6)^2}

            K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

            r³ = T² / K

            r³ = 86400² / 9.99 10⁻¹⁴

            r = ∛(7.4724 10²²)

            r = 4.21 10⁷ m

this distance is from the center of the earth

7 0
3 years ago
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