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tangare [24]
3 years ago
15

Imagine two fixed charges on the x axis. Charge one is +q and is located to the left of charge two which is equal to -4q. Where

on the axis (other than at an infinite distance) is there a point at which their net electric field is zero? a) Between the chargesb) To the left of the chargesc) To the right of the charg
Physics
1 answer:
givi [52]3 years ago
7 0

Answer: B)To the left of the charges.

Explanation: between the charges the electric field will not cancel but will be added since electric field lines from both charges point in the same direction. To the right of the charge the -4q will take over as it’s strength overcomes the strength of the +q charge. At this point the magnitude of +q will never reach a magnitude strong enough to cancel the -4q. To the left, it is further away from -4q and is closer to +q and electric field lines point in different direction

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What is the ratio of escape speed from earth to circular orbital speed? ignore air resistance.
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About 40 000 km/h
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8 0
3 years ago
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You push a 15N cat in a box horizontally across the floor for 5m. What is the work done on the cat? W=Fx
scZoUnD [109]

Answer:

w=fx

w=(15)(5)

w=75

the angle between the deplacement and N is 0

3 0
2 years ago
A horizontal spring-mass system has low friction, spring stiffness 160 N/m, and mass 0.3 kg. The system is released with an init
anygoal [31]

Answer:

(a) 0.38 m

(b) 2.78 m/s

(c) 0.11 watt

Explanation:

mass, m = 0.3 kg

spring constant, K = 160 N/m

initial compression, d = 12 cm = 01.2 m

initial speed, u = 3 m/s

(a) Let the maximum stretch is y.

Use conservation of energy

Initial potential energy + initial kinetic energy = final potential energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x K x y²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 160 x y²

2.304 + 0.00432 = 160 y²

y = 0.38 m

y = 38 cm

(b) Let v is the maximum speed.

The speed is maximum when the stretch in the spring is zero, so by use of conservation of energy

Initial potential energy + initial kinetic energy = final kinetic energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x m x v²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 0.3 x v²

2.304 + 0.00432 = 0.3 v²

v = 2.78 m/s

(c) The time period of the spring mass system is given by

T=2\pi\sqrt{\frac{m}{K}}

T=2\pi\sqrt{\frac{0.3}{160}}

T = 0.272 second

Energy dissipated per cycle = 0.03 J

Power, P = 0.03 / 0.272 = 0.11 Watt

5 0
3 years ago
If 478 watts of power are used in 14 seconds,how much work was done
zepelin [54]

Answer:

6692J

Explanation:

Power is defined as the rate at which work is being done.

So,

    Power  = \frac{workdone}{time }  

  Work done  = Power x time

Given parameters:

Power  = 478watts

Time  = 14s

So;

 Work done  = 478 x 14  = 6692J

6 0
3 years ago
Wo siblings are arguing over what way to pull a 10 kg wagon. Boris wants to pull it to the right. Boris puts 220 N of force on t
Mrrafil [7]

Answer:

a) Then net force acting on the wagon  30 [N] in a negative direction to the left

b)Acceleration of the wagon 3 m/s² in a negative direction to the left

Explanation:

The coordinates x and y show the positive direction in x-axis ( to the right) and negative direction to the left. In north-south direction positive y values are to the north and negative values to the south

The free-body diagram shows 4 forces acting

In y-axis:

The weight ( P = m*g =  10 [kg]*9,8 [m/s²] = 98 [N] negative

Normal reaction force  Fn = P = 98 [N]  positive ( surface is frictionless)

∑Fy = 0     P - Fn = 0     P = Fn  = 98 [N]

In x-axis:

Boris pulling force to the right (positive ) 220 [N]

Natasha pulling force to the left ( negative) 250 [N]

∑Fx = m*a

220 [N] - 250 [N]  = 10 Kg * a                      [N] = Kg * m /s²

-30 [kg*m/s²]  = 10 * Kg * a

-30/10  =  - 3  m/s²  = a

Sign (-) means the direction of acceleration vector is to the left (the same direction of the movement )

Then net force acting on the wagon  30 [N] in negative direction to the left

Acceleration of the wagon 3 m/s² in negative direction to the left

5 0
3 years ago
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