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Ymorist [56]
3 years ago
13

A long uniformly charged thread (linear charge density λ= 2.5 C/m) lies along the x axis in the figure.(Figure 1) A small charge

d sphere (Q= -2.3 C) is at the point x=0cm, y=−5.0cm
What is the direction of the electric field at the point x=7.0cm, y=7.0cm?
E
⃗ thread and E Q represent fields due to the long thread and the charge Q, respectively.

What is the magnitude of the electric field at the point x=7.0cm, y=7.0cm

Physics
1 answer:
Kamila [148]3 years ago
5 0

Answer 1) The electric field at distance r from the thread is radial and has magnitude  

E = λ / (2 π ε° r)  

The electric field from the point charge usually is observed to follow coulomb's law:  

E = Q / (4  π ε° r^{2})  

Now, adding the two field vectors:  

E_{thread}  =  {2.5 / (22 π ε° X 0.07 ) ; 0}  

Answer 2) E_{q}  = {2.3 / (4 2 π ε°) ( - 7/ (√(84); -12 / (√84))

Adding these two vectors will give the length which is magnitude of the combined field.  

The y-component / x-component gives the tangent of the angle with the positive x-axes.

Please refer the graph and the attachment for better understanding.

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Y_Kistochka [10]

Answer:

F = 17.3 kN

Explanation:

The normal force must support the weight of the car plus provide for the needed centripetal acceleration.

F = m(g + v²/R ) = 1000(9.8 + 15²/30) = 17,300

6 0
3 years ago
An object accelerates from rest to 85m/s over a distance of 36m. What acceleration did it experience?
Marizza181 [45]
Suvat
we have s, u, v and we want a
the suvat equation with these values in is: v^2 = u^2 - 2as
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a = (-85^2 + 0^2)/-2*36 = 7225/72 = 100.3... ms^-2
6 0
3 years ago
Which statement is true about a planet’s orbital motion?
lana66690 [7]

Answer:

Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

Explanation:

The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.

When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.

The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.

The velocity of the orbit is given by the relation,

                                    V = \sqrt{\frac{GM}{R + h} }

Where

                   V - velocity of the orbit at a height h from the surface

                    R - Radius of the second object

                    G - Gravitational constant

                    h - height from the surface

The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

                         1/2 mV^{2} = GMm/R

Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

3 0
3 years ago
Which law states that that the direction of the induced current is such that the magnetic field resulting from the induced curre
Vikentia [17]

Lenz's law or option B for plato users

6 0
3 years ago
Read 2 more answers
A ball is droped from a height of 16m how much time will pass before the ball hits the ground​
sergey [27]

Answer:

The time is 1.8s

Explanation:

The ball droped, will freely fall under gravity.

Hence we use free fall formula to calculate the time by the ball to hit the ground

h= \frac{1}{2}g{t}^{2}

Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.

From the question,

h=16m

Also, let take

g = 9.8m{s}^{-2}

By substitution we obtain,

16= \frac{1}{2}\times 9.8{t}^{2}

\implies32=9.8{t}^{2}

Diving through by 9.8

\frac{32}{9.8}= \frac{ 9.8{t}^{2} }{9.8}

\implies{t}^{2} =3.265

square root both sides, we obtain

\implies t= \sqrt{3.265}

t=1.8s

4 0
4 years ago
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