Answer:
Fn₃= -28.3*10⁻⁶N (in direction -x) :net force exerted by two charges q₁,q₂ on a third charge q₃.
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces .
Equivalences
1nC= 10⁻⁹ C
Known data
k= 9 *10⁹ N*m² /C²
q₁=-15.0 nC=-15*10⁻⁹ C
q₂=+38 nC =+38*10⁻⁹ C
q₃= +46 nC= =+46*10⁻⁹ C
d₁₃= 0.65 m
d₂₃= 1.075 m
d₁₃: distance from q₁ to q3
d₂₃: distance from q₂ to q3
Graphic attached
The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.
The force F₂₃ of q₂ on q₃ is repulsive because the charges have equal signs and the forces : Force F₂₃ is directed to the left (-x).
The force F₁₃ of q₁ on q₃ is attractive because the charges have opposite signs. Force F₁₃ is directed to the left (-x)
Calculation of the net force exerted for q₁ and q₂ on the charge q₃
The net force exerted for q₁ and q₂ on the charge q₃ is the algebraic sum of the forces F₁₁₃ and F₂₃ because both acts on the x-axis.
Fn₃= F₁₃+F₂₃
To calculate the magnitudes of the forces exerted by the charges q₁, and q₂ on q₃ we apply Coulomb's law:
F₁₃=(k*q₁*q₃)/d₁₃²
F₁₃=(9*10⁹*15*10⁻⁹*46*10⁻⁹)/(0.65)² = 14698.2*10⁻⁹ N = 14.7*10⁻⁶N
F₁₃= 14.7*10⁻⁶N, in the direction of the x-axis negative (-x)
F₃=(k*q₂*q₃)/d₂₃²
F₂₃=(9*10⁹*38*10⁻⁹*46*10⁻⁹)/(1.075)² =13623.4*10⁻⁹ N= 13.6*10⁻⁶N
F₂₃=13.6*10⁻⁶N (in direction -x)
Net force on q₃
Fn₃= -14.7*10⁻⁶N - 13.6*10⁻⁶N = -28.3*10⁻⁶N
Fn₃= -28.3*10⁻⁶N (in direction -x)
Angular momentum is given by the length of the arm to the object, multiplied by the momentum of the object, times the cosine of the angle that the momentum vector makes with the arm. From your illustration, that will be:
<span>L = R * m * vi * cos(90 - theta) </span>
<span>cos(90 - theta) is just sin(theta) </span>
<span>and R is the distance the projectile traveled, which is vi^2 * sin(2*theta) / g </span>
<span>so, we have: L = vi^2 * sin(2*theta) * m * vi * sin(theta) / g </span>
<span>We can combine the two vi terms and get: </span>
<span>L = vi^3 * m * sin(theta) * sin(2*theta) / g </span>
<span>What's interesting is that angular momentum varies with the *cube* of the initial velocity. This is because, not only does increased velocity increase the translational momentum of the projectile, but it increase the *moment arm*, too. Also note that there might be a trig identity which lets you combine the two sin() terms, but nothing jumps out at me right at the moment. </span>
<span>Now, for the first part... </span>
<span>There are a few ways to attack this. Basically, you have to find the angle from the origin to the apogee (highest point) in the arc. Once we have that, we'll know what angle the momentum vector makes with the moment-arm because, at the apogee, we know that all of the motion is *horizontal*. </span>
<span>Okay, so let's get back to what we know: </span>
<span>L = d * m * v * cos(phi) </span>
<span>where d is the distance (length to the arm), m is mass, v is velocity, and phi is the angle the velocity vector makes with the arm. Let's take these one by one... </span>
<span>m is still m. </span>
<span>v is going to be the *hoizontal* component of the initial velocity (all the vertical component got eliminated by the acceleration of gravity). So, v = vi * cos(theta) </span>
<span>d is going to be half of our distance R in part two (because, ignoring friction, the path of the projectile is a perfect parabola). So, d = vi^2 * sin(2*theta) / 2g </span>
<span>That leaves us with phi, the angle the horizontal velocity vector makes with the moment arm. To find *that*, we need to know what the angle from the origin to the apogee is. We can find *that* by taking the arc-tangent of the slope, if we know that. Well, we know the "run" part of the slope (it's our "d" term), but not the rise. </span>
<span>The easy way to get the rise is by using conservation of energy. At the apogee, all of the *vertical* kinetic energy at the time of launch (1/2 * m * (vi * sin(theta))^2 ) has been turned into gravitational potential energy ( m * g * h ). Setting these equal, diving out the "m" and dividing "g" to the other side, we get: </span>
<span>h = 1/2 * (vi * sin(theta))^2 / g </span>
<span>So, there's the rise. So, our *slope* is rise/run, so </span>
<span>slope = [ 1/2 * (vi * sin(theta))^2 / g ] / [ vi^2 * sin(2*theta) / g ] </span>
<span>The "g"s cancel. Astoundingly the "vi"s cancel, too. So, we get: </span>
<span>slope = [ 1/2 * sin(theta)^2 ] / [ sin(2*theta) ] </span>
<span>(It's not too alarming that slope-at-apogee doesn't depend upon vi, since that only determines the "magnitude" of the arc, but not it's shape. Whether the overall flight of this thing is an inch or a mile, the arc "looks" the same). </span>
<span>Okay, so... using our double-angle trig identities, we know that sin(2*theta) = 2*sin(theta)*cos(theta), so... </span>
<span>slope = [ 1/2 * sin(theta)^2 ] / [ 2*sin(theta)*cos(theta) ] = tan(theta)/4 </span>
<span>Okay, so the *angle* (which I'll call "alpha") that this slope makes with the x-axis is just: arctan(slope), so... </span>
<span>alpha = arctan( tan(theta) / 4 ) </span>
<span>Alright... last bit. We need "phi", the angle the (now-horizontal) momentum vector makes with that slope. Draw it on paper and you'll see that phi = 180 - alpha </span>
<span>so, phi = 180 - arctan( tan(theta) / 4 ) </span>
<span>Now, we go back to our original formula and plug it ALL in... </span>
<span>L = d * m * v * cos(phi) </span>
<span>becomes... </span>
<span>L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( 180 - arctan( tan(theta) / 4 ) ) ] </span>
<span>Now, cos(180 - something) = cos(something), so we can simplify a little bit... </span>
<span>L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( arctan( tan(theta) / 4 ) ) ] </span>
I'm going to say a Human.
In one simple statement, Humans have survived all of earth's devastation because of evolution. At one point it is believed that the earth had changing conditions including that some of the food that we hunted in waters became extinct. Then, as lack of food grew, We gained the ability to walk on land. As this happened, we began to adapt to our surroundings, including gaining the ability to walk on two legs and use tools
The average velocity of the Batman is 0.722 m/s North-East.
<h3>What is average velocity?</h3>
Average velocity can be defined as the ratio of the total displacement to the total time of a body.
To calculate the average velocity of Batman, we use the formula below, first, we need to calculate the resultant displacement of the Batman.
- D = √[(d₁-d₃)²+ (d₂)²]............... Equation 1
Where:
- d₁ = first displacement of the Batman = 6 m North
- d₂ = second displacement of the Batman = 3 m East
- d₃ = third displacement of the Batman. = 4 m south
Substitute these values into equation 1
- D = √[(6-4)²+3²]
- D = √(2²+3²)
- D = √(4+9)
- D = √13
- D = 3.61 m North-East.
To calculate the average velocity, we use the formula below.
- V = D/T............ Equation 2
Where:
- D = 3.61 m North-East
- T = total time = 15 s
Substitute these values into equation 2
- V = 3.61/5
- V = 0.722 m/s North- East.
Hence, the average velocity of the Batman is 0.722 m/s North-East.
Learn more about average velocity here: brainly.com/question/13665920