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777dan777 [17]
3 years ago
15

Can somone pls help me??!! i’m very stuck

Physics
1 answer:
Alchen [17]3 years ago
3 0

Answer:

its the third one

Explanation:

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A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its loc
MArishka [77]

Answer:

New location at time 3.01 is given by: (7.49, 2.11)

Explanation:

Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:

V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11

With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:

distance=v\,*\, t

distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11

Therefore, adding these displacements in component form to the original particle's position, we get:

New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)

6 0
2 years ago
Which statement best describes an atom? (2 points)
alekssr [168]

Answer:

A group of protons and neutrons that are surrounded by electrons  I think that's the answer...

Explanation:

7 0
2 years ago
Read 2 more answers
Cars A and B are racing each other along the same straight road in the following manner:
zysi [14]

Answer:

\frac{x_{o}}{v_B-V_A} =t

Explanation:

Represent the car's position as a function

x_o= "head start"

x_{A}(t) = x_{o} + v_{A}t\\x_B(t)=v_Bt \\

Remember: v_B>v_A

"cathching up means" that x_A(t)=x_B(t)

x_{o} + v_{A}t =v_Bt\\x_{o} = v_Bt -v_{A}t\\x_{o} = t(v_B-V_A)\\\frac{x_{o}}{v_B-V_A} =t, where \ v_B>v_A

8 0
3 years ago
A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a
Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C =  50 - x cm

We know that ,electric field is given as

E=K\dfrac{q}{r^2}

K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

3 0
3 years ago
Why does damp soil help excess electrons move
Marina CMI [18]

Explanation:

  • Because where there is water, energy is conducted, and <u>moves </u>the electrons.
  • They also move excess electrons because pure water is a poor conductor

3 0
3 years ago
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