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Firlakuza [10]
3 years ago
8

What is the difference between contact and action-at-a-distance forces?

Physics
1 answer:
Amiraneli [1.4K]3 years ago
4 0

Answer:

A

Contact forces are forces that require the actual contact (touching) of two pieces of matter. ... A field force is a force that works at a distance. No touching is required. Gravity is a good example of a field force, because it works whether or not an object is touching something or touching nothing at all.

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An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft
sergeinik [125]

Answer:

a) 80 V

b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

Explanation:

<u>Given :</u>

We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( K_{A}= 0) to the point B at distance l = 0.500m where its kinetic energy is (  K_{B}= 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.

<u>Required :</u>

<em>(a) We are asked to find the electric potential VB </em>

<em>(b) We want to determine the magnitude and the direction of the electric field E. </em>

<u> Solution </u>

(a) We are given the values for VA,K_{B} and q so we want to find a relationship between these three parameters and VB to get the value of VB.

As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:

                                   K_{A} +U_{A} =K_{B} +U_{B} .........................................(1)                                          

Where K_{A}= 0 and the potential energy U of the charge is given by U = q V

where V is the electric potential.  So, equation (1) will be in the form :

                                  0+qVA=K_{B} +qVB                      (Divide by q)

                                         VA=K_{B} /q + VB                  (solve for VB)

                                         VB=VA- K_{B}/q .......................................(2)

We get the relation between VB, VA and K_{B}, now we can plug our values for VA, K_{B} and q into equation (2) to get VB

                                         VB=VA- K_{B}/q

                                              =30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)

                                              =80 V

(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points

                                   VA-VB =\int\limits^1_0 {E} \, dl...................................(a)

                                               =E\int\limits^1_0 {} \, dl

                                   VA-VB=El                      (solve for E)

                                            E= VA-VB/l..................................(3)

Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E

                                            E= VA-VB/l

                                              =-100 N/C

The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

5 0
3 years ago
What is latent heat of fusion?
Naddik [55]

If you have a lump of solid at its melting point ... like ice at 32°F ...
you have to put a certain amount of heat into it just to change it
to water at 32°F.  That amount of heat, that's used just to change
a solid lump into liquid without changing its temperature, is called
the heat of fusion for that substance.

The number is different for every substance.

For water, it takes 336 joules of heat to melt 1 gram of ice
into 1 gram of water, all at 32°F (0°C).
That's an enormous latent heat of fusion ... more than almost any
other known substance.  That's why ice is such a good choice
when you need something to put in your drink to cool it down.
Ice absorbs a huge amount of heat before it melts and the drink
gets watered down.
5 0
3 years ago
Read 2 more answers
A body of mass 25kg, moving at 3 ms per second on a rough horizontal floor brought to rest after sliding through a distance of 2
erastova [34]
You have to solve this by using the equations of motion:
u=3
v=0
s=2.5
a=?
v^2=u^2+2as
0=9+5s
Giving a=-1.8m/s^2

Then using the equation:
F=ma
F is the frictional force as there is no other force acting and its negative as its in the opposite direction to the direction of motion.

-F=25(-1.8)
F=45N

Then use the formula:
F=uR
Where u is the coefficient of friction, R is the normal force and F is the frictional force.

45=u(25g)
45=u(25*10)

Therefore, the coefficient of friction is 0.18

Hope that helps




5 0
3 years ago
When serving a tennis ball, a player hits the ball when its velocity is zero (at the highest point of a vertical toss). The racq
Lilit [14]

Answer:60 gm

Explanation:

Given

initial velocity of ball u=0

Force exerted by racquet F=540 N

time period of force t=5\ ms

final velocity of ball v=45\ m/s

Racquet imparts an impulse to the ball which is given by

J=F\Delta t=\Delta P

J=540\times \Delta t=m(45-0)

m=\frac{540\times 5\times 10^{-3}}{45}

m=60\ gm

8 0
3 years ago
Can anyone help me with this??? Anyone familiar with this sorta project??
SIZIF [17.4K]

yah set up an experiment do u have the rocks with u?


7 0
2 years ago
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