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OlgaM077 [116]
3 years ago
14

Erica throws a tennis ball against a wall, and it bounces back. Which force is responsible for sending the ball back to Erica?

Physics
1 answer:
Tresset [83]3 years ago
3 0

Answer:

d

Explanation:

becuase i am smart :)

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Answer: D.

Explanation:

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Marine magnetic anomalies result from seafloor spreading in conjunction with
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Earths magnetic field
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At a 1500 m race, Ken ran at an average speed of 200 m/min. How long did it take for Ken to finish the race? (meter = m, minute
Anettt [7]

speed =  \frac{distance}{time}  \\   =  > \frac{200m}{min}  =  \frac{1500m}{t}  \\  =  >  \frac{200m}{60s}  = \frac{1500m}{t} \\  =  >  \frac{10m}{3s}  = \frac{1500m}{t} \\  =  > t = 1500m \times  \frac{3s}{10m}  \\  =  > t = 150 \times 3s \\  =  > t = 450s \:

This is the answer.

Hope it helps!!

3 0
3 years ago
Particle A of charge 2.79 10-4 C is at the origin, particle B of charge -5.64 10-4 C is at (4.00 m, 0), and particle C of charge
kirill [66]

Answer:

a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N  g) 16.9 N

h) 24.3 N θ = 44.2º

Explanation:

a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.

So, Fcax = 0

b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.

Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²

Fyca = 29. 9 N

c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:

rbc² = (3.00 m)² + (4.00m)² = 25.0 m²

⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²

⇒ Fbc = 21.7 N

d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:

Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:

cos θ = x/r = 4.00 / 5.00 m =

Fcbx = 21.7*(-0.8) = -17.4 N

e) The  y component can be calculated in the same way, projecting the force over the y-axis, as follows:

Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N

f) The sum of both x components gives :

Fcx = 0 + (-17.4 N) = -17.4 N

g) The sum of both y components gives :

Fcy = 29.9 N + (-13.0 N) = 16.9 N

h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:

Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)²\sqrt{((17.4)^{2} +(16.9)^{2}} = 24.3 N

The angle from the horizontal can be found as follows:

Ф = arc tg (16.9 / 17.4) = 44.2º

4 0
3 years ago
A 1 uF parallel-plate capacitor is charged to 12 V and then disconnected from the voltage supply. The plate separation distance
nlexa [21]

Answer:

22

Explanation:

6 0
3 years ago
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