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Misha Larkins [42]
3 years ago
6

Write the electric field at point p in component form. assume that x-axis is horizontal and points to the right, and y-axis poin

ts upward.
Physics
1 answer:
salantis [7]3 years ago
5 0
E = q/r^2

where r is perpendicular distance between the point and source charge

Ex = E cos theta

Ey = E sin theta

take theta angle between horizontal and line joining the point charge and the point

 
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A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. If g-9.
seraphim [82]

Answer:

0.31

Explanation:

horizontal force, F = 750 N

mass of crate, m = 250 kg

g = 9.8 m/s^2

The friction force becomes applied force = 750 N

According to the laws of friction,

Friction force = μ x Normal reaction of the surface

here, μ be the coefficient of friction

750 = μ x m g

750 = μ x 250 x 9.8

μ = 0.31

Thus, the coefficient of static friction is 0.31.

7 0
3 years ago
Today's topic
babunello [35]

Answer:

While a body is said to be in motion if it changes its position with respect to immediate surroundings.

A body is said to be in uniform motion if it covers equal distances in equal interval of time.

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4 0
2 years ago
What is kirchoff s law???
kolbaska11 [484]
There are two laws named for Kirchhoff.  The both concern electrical circuits.
Here they are in my own words:

1).  The sum of the voltage drops around any closed loop in a circuit is zero.

2).  The sum of the currents at any single point in a circuit is zero.
8 0
3 years ago
Read 2 more answers
Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t
earnstyle [38]
1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
B= \frac{\mu_0I}{2 \pi r}
where
\mu_0 is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance 
r=40.0 cm=0.40 m, 
which is the distance at which the other wire is located:
B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N

2) direction of the force: 
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.
6 0
3 years ago
I need help with this question ASAP I am on a timer
Tom [10]
First box and third box !
4 0
3 years ago
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