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Alexandra [31]
3 years ago
14

As a positively charged object moves toward a negatively charged object, their potential energy increases. As a positively charg

ed object moves toward another positively charged object, their potential energy increases. As a negatively charged object moves toward another negatively charged object, their potential energy increases. Increasing the distance between two electrically charged objects may raise or lower their potential energy. All of the above are true.
Physics
1 answer:
Zielflug [23.3K]3 years ago
4 0

Answer:

As a positively charged object moves toward another positively charged object, their potential energy increases. True

As a negatively charged object moves toward another negatively charged object, their potential energy increases True

Explanation:

When two like charges move towards each other, there is increase in potential energy because the motion is in opposition to the direction if the field. Hence when two negative or two positive charges move towards each other, there is an increase in potential energy.

However, potential energy is inversely related to the distance of separation of the charges.

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A police car with its 300-Hz siren is moving toward a warehouse at 30 m/s, intending to crash through the door. The sound bounce
Lady bird [3.3K]

Answer: The frequency heard will be f = 275.675Hz

Explanation: When an object emitting sound is moving, it occurs a phenomenon called Doppler shift or Doppler effect. What happens is that the sound gets higher when the moving object comes closer the observer and becomes lower after it passes, This change is due to the quantity of waves that passes through an area in an unit of time.

The formula to calculate the Doppler effect is as follows

f = (\frac{c}{c+Vs}) · f₀

f is the observed frequency;

c is the speed of sound;

Vs is velocity of the source;

f₀ is the emitted frequency of source;

Substituting and calculating,

f = \frac{340}{340+30} · 300

f = 275.675 Hz

Thus, the frequency heard by the police officer is 275.675Hz.

8 0
3 years ago
An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl
svet-max [94.6K]

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 0 m/s

         Displacement, s = 192 m

         Acceleration, a = 9.81 m/s²

         Substituting

                      s = ut + 0.5 at²

                      192 = 0 x t + 0.5 x 9.81 x t²

                         t = 6.26 seconds

         Now we need to find horizontal distance traveled in 6.26 seconds by the package.

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 41 m/s

        Time, t = 6.26 s

         Acceleration, a = 0 m/s²

         Substituting

                      s = ut + 0.5 at²

                      s = 41 x 6.26 + 0.5 x 0 x 6.26²

                         s = 256.52 m

     The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

          Initial velocity, u = 0 m/s

         Time, t = 6.26 s

         Acceleration, a = 9.81 m/s²  

         Substituting

                      v = u+ at

                       v = 0 + 9.81 x 6.26 = 61.41 m/s

   Vertical component of velocity = 61.41 m/s      

4 0
3 years ago
Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f
stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2

When s = Db, t = 2t

s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2

Dividing the two equations

\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db

Hence, Da=(1/4)Db

3 0
3 years ago
Flying against the wind, an airplane travels 2670 km in 3 hours. Flying with the wind, the same plane travels 11,070 km in 9 hou
xxMikexx [17]

Answer:

speed of plane in still air = 1060 km/h

speed of wind = 170 km/h

Explanation:

Let teh speed of plane in still air is vp and the speed of air is va.

Irt travels 2670 km in 3 hours against the wind

So,

vp - va = 2670 / 3 = 890 km/h ..... (1)

It travels 11070 km in 9 hours along the wind.

vp + va = 11070 / 9 = 1230 km/h .... (2)

Adding both the equations

2 vp = 2120

vp = 1060 km/h

and va = 1230 - vp = 1230 - 1060 = 170 km/h

5 0
3 years ago
A jetliner has a velocity of 95 m/s. What is the displacement of the jetliner at t=3.0 seconds?
charle [14.2K]
Distance = speed / time

speed = 95 m/s
time = 3 s

distance = 95 / 3 m

displacement = 95/3 m or 32 m (2 s.f.)
5 0
3 years ago
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