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Vikki [24]
3 years ago
9

If you were lost in space how would you get back to the space station explain scientifically

Physics
1 answer:
iris [78.8K]3 years ago
5 0

You would use a object called The big dipper to find the closest space station.

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Which statement did Ernest Rutherford make about atoms?
PtichkaEL [24]

Answer:

Option A

Explanation:

Ernest Rutherford concluded that the atom has a small, dense center which constitutes the mass of the whole atom. He called it a "Nucleus". He also said that most of the space in the atom is empty.

6 0
4 years ago
3. Which statement best describes the research projects that are funded by private
SOVA2 [1]

Answer:

b. Research projects for a specific cause

Explanation:

Research funded by private foundations are usually for a specific cause. These type of research are usually properly scrutinized, and must have a high probability of success. Most of these projects may not be to benefit the consumers, but are usually done with a special purpose in mind.

5 0
3 years ago
HELP I DONT HAVE MUCH TIME ILL time BRAINLIEST 10poINTSSSSSS HURYYYY
Soloha48 [4]
1.the sun must be lined up with the moon and earth all must be aligned for this to happen
6 0
3 years ago
Read 2 more answers
While driving north at 25 m/s during a rainstorm you notice that the rain makes an angle of 38 degrees with the vertical. while
ikadub [295]

While returning back in opposite direction driver see that rain drops are falling vertically down

so horizontal speed of rain with respect to driver must be Zero

while moving towards the north direction driver see that the rain drops makes an angle 38 degree with the vertical

tan\theta = \frac{v_y}{v_x}

tan38 = \frac{v_y}{25 + 25}

v_y = 39 m/s

so the speed of rain is

v = \sqrt{25^2 + 39^2}

v = 46.3 m/s

also the angle is given as

\theta = tan^{-1}\frac{39}{25}

\theta = 57.3 degree

8 0
4 years ago
Two charged concentric spherical shells have radii 8.83 cm and 15.4 cm. The charge on the inner shell is 5.03 × 10⁻⁸ C and that
REY [17]

Answer:

a. E(r) = 3.67 x 10⁴ N/C

b. E(r) = 5.38 x 10⁴ N/C

Explanation:

r1 = 8.83 cm

r2 = 15.4 cm

Charge on the inner shell = 5.03 x 10^-⁸ C

Charge on the outer shell = 2.77 x 10^-⁸ C

a)   To find the electric field at r = 11.1 cm,

r₁ = 8.83 cm < r = 11.1 cm < r₂ = 15.4 cm

E(r) = <u>1              </u> <u>q₁    </u>

          4\pi∈₀        r₂  

E(r) = <u>( 8.99 x 10⁹ N.m²/C² )  ( 5.03 x 10⁻⁸)</u>

                                 (0.111m)²

E(r) = 3.67 x 10⁴ N/C

b)   To find the electric field at r = 36.1 cm

since r₁  < r₂ < r = 36.1 cm

E(r) = <u>1              </u> <u>q₁  ₊  q₂   </u>

          4\pi∈₀           r₂  

E(r) = <u>( 8.99 x 10⁹ N.m²/C² )  ( 5.03 + 2.77 )(1 x 10⁻⁸)</u>

                                             (0.361m)²

E(r) = 5.38 x 10⁴ N/C

3 0
4 years ago
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