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Alla [95]
2 years ago
5

If this charged soot particle is now isolated (that is, removed from the electric field described in the previous part), what wi

ll be the magnitude E of the electric field due to the particle at distance 1.00 meter from the particle?
Physics
1 answer:
kicyunya [14]2 years ago
7 0

The magnitude of the electric field due to the particle at the given distance is 9\times 10^9 q.

<h3>Electric field strength</h3>

The electric field strength of a charged particle is the force per unit charge in the given field.

The electric field strength of a charge is given as;

E = \frac{F}{q} \\\\E = \frac{kq}{r^2}

where;

  • k is Coulomb's constant
  • q is the charge
  • r is the distance

E = \frac{9\times 10^9 \times q}{1^2} \\\\E = 9\times 10^9 q

Thus, the magnitude of the electric field due to the particle at the given distance is 9\times 10^9 q.

Learn more about electric field here: brainly.com/question/14372859

You might be interested in
How much heat is evolved when 1234 g of water condenses to a liquid at 100.°c?
lbvjy [14]
During the phase transition vapour --> liquid water, the temperature of the water does not change; the molecules of water release heat and the amounf of heat released is equal to
Q=mL_e
where
m is the mass of the water
L_e is the latent heat of evaporation.

For water, the latent heat of evaporation is L=2257 kJ/kg, while the mass of the water is
m=1234 g=1.234 kg

so, the amount of heat released in the process is
Q=mL_e = (1.234 kg)(2257 kJ/kg)=2785 kJ
6 0
3 years ago
A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?a.What is the speed of the mountain lio
Arisa [49]

Answer:

What is the speed of the mountain lion as it leaves the ground?

9.98m/s

At what angle does it leave the ground?

50.16°

Explanation:

This is going to be long, so if you want to see how it was solved refer to the attached solution. If you want to know the step by step process, read on.

To solve this, you will need use two kinematic equations and SOHCAHTOA:

d = v_it + \dfrac{1}{2}at^{2}\\\\vf = vi + at

With these formulas, we can derive formulas for everything you need:

Things you need to remember:

  • A projectile at an angle has a x-component (horizontal movement) and y-component (vertical movement), which is the reason why it creates an angle.
  • Treat them separately.
  • At maximum height, the vertical final velocity is always 0 m/s going up. And initial vertical velocity is 0 m/s going down.
  • Horizontal movement is not influenced by gravity.
  • acceleration due to gravity (a) on Earth is constant at 9.8m/s

First we need to take your given:

10.0 m long (horizontal) and maximum height of 3.0m (vertical).

d_x=10.0m\\d_y=3.0m

What your problem is looking for is the initial velocity and the angle it left the ground.

Vi = ?     Θ =?

Vi here is the diagonal movement and do solve this, we need both the horizontal velocity and the vertical velocity.

Let's deal with the vertical components first:

We can use the second kinematic equation given to solve for the vertical initial velocity but we are missing time. So we use the first kinematic equation to derive a formula for time.

d_y=V_i_yt+\dfrac{1}{2}at^{2}

Since it is at maximum height at this point, we can assume that the lion is already making its way down so the initial vertical velocity would be 0 m/s. So we can reduce the formula:

d_y=0+\dfrac{1}{2}at^{2}

d_y=\dfrac{1}{2}at^{2}

From here we can derive the formula of time:

t=\sqrt{\dfrac{2d_y}{a}}

Now we just plug in what we know:

t=\sqrt{\dfrac{(2)(3.0m}{9.8m/s^2}}\\t=0.782s

Now that we know the time it takes to get from the highest point to the ground. The time going up is equal to the time going down, so we can use this time to solve for the intial scenario of going up.

vf_y=vi_y+at

Remember that going up the vertical final velocity is 0m/s, and remember that gravity is always moving downwards so it is negative.

0m/s=vi_y+-9.8m/s^{2}(0.782s)\\-vi_y=-9.8m/s^{2}(0.782s)\\-vi_y=-7.66m/s\\vi_y=7.66m/s

So we have our first initial vertical velocity:

Viy = 7.66m/s

Next we solve for the horizontal velocity. We use the same kinematic formula but replace it with x components. Remember that gravity has no influence horizontally so a = 0:

d_x=V_i_xt+\dfrac{1}{2}0m/s^{2}(t^{2})\\d_x=V_i_xt

But horizontally, it considers the time of flight, from the time it was released and the time it hits the ground. Also, like mentioned earlier the time going up is the same as going down, so if we combine them the total time in flight will be twice the time.

T= 2t

T = 2 (0.782s)

<em>T = 1.564s</em>

<em>So we use this in our formula:</em>

<em>d_x=V_i_xT\\\\10.0m=Vi_x(1.564s)\\\\\dfrac{10.0m}{1.564s}=V_i_x\\\\6.39m/s=V_i_x</em>

Vix=6.39m/s

Now we have the horizontal and the vertical component, we can solve for the diagonal initial velocity, or the velocity the mountain lion leapt and the angle, by creating a right triangles, using vectors (see attached)

To get the diagonal, you just use the Pythagorean theorem:

c²=a²+b²

Using it in the context of our problem:

Vi^{2}=Viy^2+Vix^2\\Vi^2=(7.66m/s)^2+(6.39m/s)^2\\\sqrt{Vi}=\sqrt{(7.66m/s)^2+(6.39m/s)^2}\\\\Vi=9.98m/s

The lion leapt at 9.98m/s

Using SOHCAHTOA, we know that we can TOA to solve for the angle, because we have the opposite and adjacent side:

Tan\theta=\dfrac{O}{A}\\\\Tan\theta=\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{7.66m/s}{6.39m/s}\\\\\theta=50.17

The lion leapt at an angle of 50.16°.

6 0
3 years ago
DETERMINE THE BCD EQUIVALENT OF 1100111110101001
Solnce55 [7]

Answer: It states that the BCD equivalent would be 0001000100000000000100010001000100010000000100000001000000000001.

6 0
3 years ago
Drawing a shows a displacement vector (450.0 m along the y axis). In this x, y coordinate system the scalar components are Ax 0
Alisiya [41]

Answer:

x ’= 368.61 m,  y ’= 258.11 m

Explanation:

To solve this problem we must find the projections of the point on the new vectors of the rotated system  θ = 35º

            x’= R cos 35

            y’= R sin 35

           

The modulus vector can be found using the Pythagorean theorem

            R² = x² + y²

            R = 450 m

we calculate

            x ’= 450 cos 35

            x ’= 368.61 m

            y ’= 450 sin 35

            y ’= 258.11 m

4 0
3 years ago
You are investigating how objects move when they are dropped from different heights. To collect your data, you drop a 1 kg weigh
ASHA 777 [7]

The time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

The given parameters;

<em>Mass of the first object, m1 = 1 kg</em>

<em>Mass of the second object, m2 = 5 kg</em>

The final velocity of the objects during the downward motion is calculated as follows;

v_f = v_0 + gt\\\\v_f = 0 + gt\\\\\v_f = gt

The time of motion of the object from the given height is calculated as;

h = v_0 t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} }

The time of motion of each object is independent of mass of the object.

Thus, the time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

Learn more about time of motion here: brainly.com/question/2364404

3 0
3 years ago
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