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Viefleur [7K]
2 years ago
14

How has technology impacted our society? Describe both positive and negative effects.​

Physics
1 answer:
KIM [24]2 years ago
8 0
Positive : The development and adoption of technology have helped societies raise productivity, inclusivity of services and improve overall well-being.

Negative: The negative impact of tech on society involves mass-made products, with most often blamed: social media.
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If a football player collides with a goal post, what forces are at work?
AlekseyPX
When a footballer collides with the goal post, the forces at work are the action and reaction forces. The player will exert an action force on the goal post, and then a reaction force from the goal post will stop the player. The reaction force call will cause pain and even injury to the player.
7 0
3 years ago
a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its
lilavasa [31]

Answer:

D/H =15

Explanation:

  • We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
  • At this point, we can find the value of H, applying the following kinematic equation:

       v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)

  • If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

       H = \frac{v_{0} ^{2} }{2*g} (2)

  • We can use the same equation, to find the value of D, as follows:

        v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)

  • In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
  • When we replace these values in (1), we find that  v₁ = -v₀.
  • Replacing in (3), we have:

        (4*v_{0})^{2} - (-v_{0}) ^{2}  = 2* g* D\\ \\ 15*v_{0}^{2}  = 2*g*D

  • Solving for  D:

       D = \frac{15*v_{0} ^{2} }{2*g}

  • From (2) we know that H can be expressed as follows:

       H = \frac{v_{0} ^{2} }{2*g}

  • ⇒ D = 15 * H

        \frac{D}{H} = 15

3 0
3 years ago
A 20 g sparrow flying toward a bird feeder mistakes the pane of glass in a window for an opening and slams into it with a force
Rama09 [41]
I found this using the app Socratic. When I took physics in high school it helped me so much.

5 0
3 years ago
Please follow me guys <br>​
Lilit [14]

Answer:

if i do give me brainliest ok ok

Explanation:

7 0
3 years ago
A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.
bija089 [108]

a)

i) Potential for r < a: V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

ii) Potential for a < r < b:  V(r)=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

iii) Potential for r > b: V(r)=0

b) Potential difference between the two cylinders: V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c) Electric field between the two cylinders: E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

Explanation:

a)

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

E=\frac{\lambda}{2\pi \epsilon_0 r}

where

\lambda is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

E=0

So the potential where the electric field is zero is constant:

V=const.

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density +\lambda and an equal negative charge density -\lambda. Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)

However, we know that the potential at b is zero, so

V(r)=V(b)=0

ii) The electric field in the region a < r < b instead it is given only by the positive charge +\lambda distributed over the surface of the inner cylinder of radius a, therefore it is

E=\frac{\lambda}{2\pi r \epsilon_0}

And so the potential in this region is given by:

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r} (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

And so, for r<a,

V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

b)

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

- Potential at the surface of the outer cylinder:

V(b)=0

Therefore, the potential difference is simply equal to

V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c)

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

The electric field is just the derivative of the electric potential:

E=-\frac{dV}{dr}

so we can find it by integrating the expression for the electric potential. We find:

E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
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