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devlian [24]
3 years ago
10

Ralphie runs north 0.5 km, then turns east and runs 2.0 km, turns south and runs 1.5 km, turns west and runs another 1km. What i

s is distance he ran?
2√ km square root of 2 km
-2 km
2 km
5 km
Physics
1 answer:
Step2247 [10]3 years ago
7 0

Answer:

I think its 2km

Explanation:

bc if u put it out on a graph, it would essentially take about 4 jumps to get to your ending point, and assuming each jumb was 0.5 km. But if u dont trust me look at another answer bc I dont know if thats right or not.

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A star has a spectrum with lines indicating the presence of ionized helium. This star should be classified as a(n) _____ type st
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At a pressure of one billionth (10–9) of atmospheric pressure, there are about 2.7 × 1010 molecules in one cubic centimeter of a
Sergeeva-Olga [200]

Answer:

Required answer =  2.7 x 10^16 MOLECULES

Explanation:

We know that

1 m = 100 cm

so 1 m^3 = (100 cm)^3 = 1000000 cm^3 = 1 x 10^6 cm3

as per question, 2.7*10^{10} molecules in 1 cm^3

so number of molecules in 1 m^3

Number of molecules = ( 2.7 x 10^10 molecules /1 cm^3 ) x ( 1 x 10^6 cm3 / 1 m3)

Number of molecules = 2.7 x 10^10 x 1 x 10^6 = 2.7 x 10^16 per m3

Required answer =  2.7 x 10^16 MOLECULES

3 0
3 years ago
The glowing of a neon light is caused by electrons emitting energy as they (1 point)
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4 years ago
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where would the spaceprobe experience the strongest net (or total) gravitional force exerted on it by Earth and Mars
Arte-miy333 [17]

Answer:

r = 41.1 10⁹ m

Explanation:

For this exercise we use the equilibrium condition, that is, we look for the point where the forces are equal  

                  ∑ F = 0

                  F (Earth- probe) - F (Mars- probe) = 0

                  F (Earth- probe) = F (Mars- probe)

Let's use the equation of universal grace, let's measure the distance from the earth, to have a reference system

the distance from Earth to the probe is        R (Earth-probe) = r

the distance from Mars to the probe is        R (Mars -probe) = D - r

where D is the distance between Earth and Mars

                   

                 G  \ \frac{m \ M_{Earth}}{r^2} = G  \ \frac{m \ M_{Mars}}{(D-r)^2}

                 M_earth (D-r)² = M_Mars r²

                 (D-r) = \sqrt{ \frac{M_{Mars}}{ M_{Earth}} }    r

                  r ( 1 + \sqrt{ \frac{M_{Mars}}{M_{Earth}} }) = D

                  r = \frac{D}{ 1+ \sqrt{\frac{M_{Mars}}{ M_{Earth}} } }

We look for the values ​​in tables

                  D = 54.6 10⁹ m (minimum)

                  M_earth = 5.98 10²⁴ kg

                  M_Marte = 6.42 10²³ kg = 0.642 10²⁴ kg

                   

let's calculate

                  r = 54.6 10⁹ / (1 + √(0.642/5.98)  )

                  r = 41.1 10⁹ m

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3 years ago
I need to fill the gaps
andriy [413]

Answer:

1. combustion

2. oxygen

3. CO²

5 0
3 years ago
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