The diagram below shows the path of a planet around a star.

Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are

Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

$d_{12} = \sqrt{3^{2}+3^{2} } = \sqrt{18}$ m : distance from q₁ to q₂

(d₁₂)² = 18 m²

$d_{13} =\sqrt{1^{2}+3^{2} } = \sqrt{10}$ m : distance from q₁ to q₃

(d₁₃)² = 10 m²

$d_{14} =\sqrt{3^{2}+2^{2} } = \sqrt{13}$ m : distance from q₁ to q₄

(d₁₄)² = 13 m²

K= 8.98755 × 10⁹ N *m²/C²

q₁= 7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45° , β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

$F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }$

$F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }$

Fn₁= 23.95*10⁻⁹ N

8 0

3 years ago

Question 2.

boyakko [2]

The tensile stress of the wire supporting 2 kg mass is determined as 6.1 x 10⁷ N/m².

<h3>Tensile stress of the wire</h3>

The tensile stress of the wire is calculated as follows;

σ = F/A

where;

A is area of the wire

A = πr² = πD²/4

where;

D is diameter = 0.64 mm

A = π x (0.64 x 10⁻³)²/4

A = 3.22 x 10⁻⁷ m²

σ = F/A = (mg)/A = (2 x 9.8)/( 3.22 x 10⁻⁷)

σ = 6.1 x 10⁷ N/m²

Learn more about tensile stress here: brainly.com/question/25748369

#SPJ1

3 0

2 years ago

Me ajudem, Por favor!!!!!!

zzz [600]

Answer:

a) $a=4\,\frac{m}{s^2}$

b) $V(t)=4\,t\,+3$

c) $V(1)=7 \,\frac{m}{s} \\$

d) Displacement = 22 m

e) Average speed = 11 m/s

Explanation:

a)

Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:

$slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}$

Therefore, acceleration is $a=4\,\frac{m}{s^2}$

b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:

$y=m\,x+b\\V(t)=4\,t\,+3$

c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:

$V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}$

d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):

Displacement = $\frac{(7+15)\,2}{2} = 22\,\,m$

e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:

Average velocity = $\frac{22}{2} = 11\, \,\frac{m}{s}$

3 0

3 years ago

_______ from the mantle rises and flows out through a vent onto Earth's surface as _______. This molten material, sometimes alon

anygoal [31]

Magma from the mantle rises and flows out through a vent onto Earth's surface as lava. This molten material, sometimes along with ash, cinders, and rock, builds up a mountain around the vent. The vent and its mountain together are called a sediment.

hope this helps

5 0

3 years ago

Read 2 more answers

Can someone summarize the technique of gel electrophoresis in one or two sentences?

sertanlavr [38]

There is a gel (agarose) that is placed in abuffer-filled box and an electrical field is applied via a power supply. The negative terminal is at the far end (black wire), this causes DNA migrates toward the anode (red wire).

6 0

3 years ago