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Nadusha1986 [10]
2 years ago
14

Two particles, one with charge −7.97×10−6 C and the other with charge 6.91×10−6 C, are 0.0359 m apart. What is the magnitude of

the force that one particle exerts on the other?
Physics
1 answer:
kolbaska11 [484]2 years ago
7 0

Answer:

-384.22N

Explanation:

From Coulomb's law;

F= Kq1q2/r^2

Where;

K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2

q1 and q2 = magnitudes of the both charges

r= distance of separation

F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2

F= -495.65 × 10^-3/ 1.29 × 10^-3

F= -384.22N

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The change in momentum that occurs when a 1. 0 kg ball traveling at 4. 0 m/s strikes a wall and bounces back at 2. 0 m/s is.
Doss [256]

Answer:

The change is momentum is given by ∆p=p(inital) - p(final) =4-2=2 kg.m/s

Explanation:

momentum is the product of mass and velocity (speed)

So it's initial momentum would be:

p=mv=(1)(4)=4 kg.m/s

It's final momentum is given by:

p=mv=(1)(2)=2 kg.m/s

7 0
2 years ago
At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 fi
Amanda [17]

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

(5+0.18, 2-0.1)

= ( 5.18, 1.9)

3 0
3 years ago
Most chemical reactions can be divided into how many main groups?
tatyana61 [14]
There is synthesis
decomposition
double displacement
single displacement
combustion
metathesis
so i guess you could say 6
5 0
3 years ago
Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker poi
Nataly [62]

Answer

given,

mass of ball, m = 57.5 g = 0.0575 kg

velocity of ball northward,v = 26.7 m/s

mass of racket, M = 331 g = 0.331 Kg

velocity of the ball after collision,v' = 29.5 m/s

a) momentum of ball before collision

   P₁ = m v

   P₁ = 0.0575 x 26.7

   P₁ = 1.535 kg.m/s

b) momentum of ball after collision

   P₂ = m v'

   P₂ = 0.0575 x (-29.5)

   P₂ = -1.696 kg.m/s

c) change in momentum

    Δ P = P₂ - P₁

    Δ P = -1.696 -1.535

    Δ P = -3.231 kg.m/s

d) using conservation of momentum

  initial speed of racket = 0 m/s

  M u + m v = Mu' + m v

  M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)

  0.331 u' = 3.232

     u' = 9.76 m/s

change in velocity of the racket is equal to 9.76 m/s

5 0
3 years ago
The amount of time required for 2 successive wave crests to pass a fixed point is called wave ________.
mihalych1998 [28]
I believe this is known as wave period.
hope this helps!
7 0
2 years ago
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