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nadezda [96]
3 years ago
14

Please, help!!

Physics
1 answer:
pashok25 [27]3 years ago
6 0

Answer:

The net force in the middle particle is zero, and it has no direction.

Explanation:

First, the force that one charge does in other charge is:

F = K*q1*q2/r^2

where k is a constant, q1 and q2 are the charges, and r is the distance between the charges.

If the force is positive, the force is repulsive (pushes away the charge) if the force is negative, is attractive.

Now, we have that the middle charge has a charge of 4.00nC, and in each side at a distance of 4m, has a charge of 4.00nC.

i will write qL as the charge in the left, qR as the charge in the right, and qM as the charge in the middle.

The force that the charge in the right does to the charge in the middle is:

Fr = (k*(4.00nC)^2)/4m^2 = k*(nC/m)^2

And is positive, so this is a repulsive force, this means that if the charge is at the right of the middle charge, then this force pushes the middle charge to the left.

For the left charge we have the same:

Fl = (k*(4.00nC)^2)/4m^2 = k*(nC/m)^2

But in this case, the force pushes the particle to the right, then this force, that is equal in magnitude to the previous force, pushes it in the opposite direction, then the total force in the middle particle is anulated. This is because the two external particles are "pushing" the middle particle with the same force and in the opposite direction.

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How come we can see orange? In simple words.
IgorLugansk [536]

Answer:

<u><em>When sunlight shines through an orange solution, the violet, blue and green wavelengths are absorbed.</em></u> The other colors pass through.

8 0
3 years ago
You set out to design a car that uses the energy stored in a flywheel consisting of a uniform 101-kg cylinder of radius r that h
Ket [755]
Ok, assuming "mj" in the question is Megajoules MJ) you need a total amount of rotational kinetic energy in the fly wheel at the beginning of the trip that equals
(2.4e6 J/km)x(300 km)=7.2e8 J
The expression for rotational kinetic energy is

E = (1/2)Iω²  

where I is the moment of inertia of the fly wheel and ω is the angular velocity.  
So this comes down to finding the value of I that gives the required energy.  We know the mass is 101kg.  The formula for a solid cylinder's moment of inertia is

 I = (1/2)mR²

We want (1/2)Iω² = 7.2e8 J and we know ω is limited to 470 revs/sec.  However, ω must be in radians per second so multiply it by 2π to get 
ω = 2953.1 rad/s
Now let's use this to solve the energy equation, E = (1/2)Iω²,  for I:
I = 2(7.2e8 J)/(2953.1 rad/s)² = 165.12 kg·m²

Now find the radius R,

 165.12 kg·m² = (1/2)(101)R²,
√(2·165/101) = 1.807m

R = 1.807m
8 0
3 years ago
Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the
nydimaria [60]

Answer:F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

F_1=\frac{kq(-q)}{(L\sqrt{2})^2}

where  L\sqrt{2}=Distance between the two charges

F_1=-\frac{kq^2}{2L^2}

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

F_2=\frac{kq(-q)}{(L)^2}

The magnitude of force by both the  charge is same but at an angle of 90^{\circ}

thus combination of two forces at 2 and 3 will be

F'=\sqrt{2}\frac{kq^2}{2L^2}

Now it will add with force due to 1 charge

Thus net force will be

F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

6 0
3 years ago
A tow truck drags a stalled car along a road. The chain makes an angle of 30???? with the road and the tension in the chain is 1
My name is Ann [436]

Answer: work = 1,305kJ

Explanation:

angle= 30°

force= 1,500N

distance= 1,000m

The formula for work is : Work= force x distance, however there is an angle of 30° between the direction of force applied and the direction of motion, therefore force must be decomposed to its value on the horizontal axis which is the direction of motion by using the cosine of the very angle.

W= F×cos(α)×D

W= 1,500×cos (30)×1,000

W= 1,305kJ ( kilojoules)

3 0
3 years ago
The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh
Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

v^2=v^2_o-2ax

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}

Next, consider that the formula for the force is:

F=ma

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

F=(820kg)(0.66\frac{m}{s^2})=542.20N

Hence, the required force to stop the car is 542.20N

4 0
1 year ago
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