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olga2289 [7]
2 years ago
5

There are 15 water molecules in the air, and the relative

Physics
1 answer:
AlexFokin [52]2 years ago
6 0

30 30 30 30 30 30 30 30 30 30

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According to the diagram, which choice has the shortest wavelength?
makvit [3.9K]

Answer:

Blue visible light

Explanation:

the distance between the crests of the waves is the shortest

8 0
2 years ago
1. Take the speed at which the body moves as 10 m/s 2. Calculate the distance travelled by the body every second 3. Take the tim
exis [7]

Answer:

a)this graph is also a line     b) in both cases we have a uniform movement

Explanation:

In this exercise we have a uniform movement

     v = d / t

     d = v t

in the table we give some values ​​to make the graph

       t (s)    d (m)

        1         10

        2        20

        3         30

In the attached we can see the graph that is a straight line

we have another vehicle at v = 50 me / S

t (s)     d (m)

1         50

2        100

3         150

this graph is also a line

b) in both cases we have a uniform movement

3 0
3 years ago
Who is famous for having introduced the three laws of motion?
Digiron [165]

Isaac Newton is famous for the three laws

3 0
3 years ago
Read 2 more answers
¿Qué resistencia debe ser conectada en paralelo con una de 20 Ω para hacer una
ValentinkaMS [17]

Answer:

60 Ω

Explanation:

R(com) = 15 Ω

1/R(com) = 1/R1 + 1/R2 + 1/R3 ..... + 1/Rn

1/15 = 1/20 + 1/R2

1/R2 = 1/15 - 1/20

1/R2 = (4 - 3) / 60

1/R2 = 1/60

R2 = 60 Ω

así, la combinada de resistencia necesaria es 60 Ω

5 0
3 years ago
A particle of mass 10 g and charge 80 mC moves through a uniform magnetic field,in a region where the free-fall acceleration is .
Alex777 [14]

Answer:

B=1.223\frac{T\cdot m}{s}\frac{1}{v}

Explanation:

According to the first Newton's law, when a particle moves with constant velocity, the sum of forces on it is zero. So, we have:

F_m=F_g

Here F_m=qvBsin\theta is the magnetic force and F_g=mg is the gravitational force. The velocity of the particle is perpendicular to the magnetic field, so \theta=90^\circ. Replacing and solving for B:

qvBsin(90^\circ)=mg\\B=\frac{mg}{qv}\\B=\frac{mg}{q}\frac{1}{v}\\B=\frac{(10*10^{-3}kg)(9.8\frac{m}{s^2})}{80*10^{-3}C}\frac{1}{v}\\B=1.223\frac{T\cdot m}{s}\frac{1}{v}

4 0
3 years ago
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