In two or more complete sentences, explain the effect of gases being trapped in the atmosphere.

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Sav [3153]

Answer:

The third charged particle needs to be positioned at x = 0.458 m = 45.8 cm

Explanation:

To address this inquiry, we refer to Coulomb's law:

Two point charges (q₁, q₂) spaced apart by a distance (d) impart a mutual force (F) with a magnitude defined by the equation:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K: Coulomb's constant in N*m²/C²

q₁, q₂: Charges quantified in Coulombs (C)

d: distance in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Information

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem analysis

Refer to the accompanying graphic.

We assume a positive charge q₃, thus F₁₃ and F₂₃ exhibit repulsive forces that need to be equivalent so the resultant force is zero:

Utilizing equation (1), we will determine the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ from both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We discard 10⁻⁶ from both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve for the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

Choosing x₂ results in F₁₃ and F₂₃ acting in the same direction and unable to cancel each other, therefore we select x₁ as the accurate choice since at this stage the forces counteract each other.

x = 0.458m = 45.8cm

8 0

2 months ago

Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a

Yuliya22 [3333]

Answer:

Explanation:

a )

Each blade resembles a rod with its axis positioned near one end.

The moment of inertia for one blade is:

= 1/3 x m l²

where m stands for the mass of the blade

l represents the length of each blade.

Total moment of inertia for 3 blades is:

= 3 x $\frac{1}{3}$ x m l²

ml²

2 )

Details provided include:

m = 5500 kg

l = 45 m

Substituting these values produces:

moment of inertia of one blade:

= 1/3 x 5500 x 45 x 45

= 37.125 x 10⁵ kg.m²

Moment of inertia for 3 blades:

= 3 x 37.125 x 10⁵ kg.m²

= 111.375 x 10⁵ kg.m²

c )

Angular momentum

= I x ω

I denotes the moment of inertia of the turbine

ω symbolizes angular velocity

ω = 2π f

f indicates the rotational frequency of the blades

d )

We have I = 111.375 x 10⁵ kg.m² (Calculated)

f = 11 rpm (revolutions per minute)

= 11 / 60 revolutions per second

ω = 2π f

= 2π x 11 / 60 rad / s

Calculating angular momentum yields

= I x ω

111.375 x 10⁵ kg.m² x 2π x 11 / 60 rad / s

= 128.23 x 10⁵ kgm² s⁻¹.

4 0

2 months ago

For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo

ValentinkaMS [3465]

The ideal launch angle of 45° for achieving the greatest horizontal distance is only applicable when the starting height matches the final height.

In this scenario, you can demonstrate it as follows: 

the initial velocity is Vo 
the launch angle is α 

the initial vertical velocity is 
Vv = Vo×sin(α) 

horizontal velocity becomes 
Vh = Vo×cos(α) 

the total flight duration is the period required to return to a height of 0 m, thus 
d = v×t + a×t²/2 
where 
d = distance = 0 m 
v = initial vertical velocity = Vv = Vo×sin(α) 
t = time =? 
a = gravitational acceleration = g (= -9.8 m/s²) 
therefore 
0 = Vo×sin(α)×t + g×t²/2 
0 = (Vo×sin(α) + g×t/2)×t 
t = 0 (obviously, the projectile is at height 0 m at time = 0s) 
or 
Vo×sin(α) + g×t/2 = 0 
t = -2×Vo×sin(α)/g 

Now let's examine the horizontal distance. 
r = v × t 
where 
r = horizontal range =? 
v = horizontal velocity = Vh = Vo×cos(α) 
t = time = -2×Vo×sin(α)/g 
therefore 
r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) 
r = -(Vo)²×sin(2α)/g 

To find the extreme points of r (max or min) with respect to α, the first derivative of r with regards to α must be determined and set to 0. 

dr/dα = d[-(Vo)²×sin(2α)/g] / dα 
dr/dα = -(Vo)²/g × d[sin(2α)] / dα 
dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα 
dr/dα = -2 × (Vo)² × cos(2α) / g 

As Vo and g are constants that are not equal to 0, the only solution for dr/dα to equal 0 is when 
cos(2α) = 0 
2α = 90° 
α = 45°

4 0

1 month ago

Suppose that an owner of the same dog breed has also taken some measurements. They notice that the surface area of the dog has i

inna [3103]

Answer:

The surface area of the dog changes from A to 3A

Explanation:

It is stated that the dog's surface area has increased by a factor of 3 over four years.

We need to calculate the change in the relative surface area of the dog over this timeframe.

Let’s assume the initial surface area is A.

Since the surface area has been multiplied by 3,

it follows that the surface area after four years is equal to 3×A = 3A.

Thus, the dog's surface area transitions from A to 3A.

7 0

2 months ago

You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce

serg [3582]

The duration required for the seventh car to pass amounts to 13.2 seconds. The train's movement is characterized by uniform acceleration, enabling the application of suvat equations. Initially, we analyze the movement of the first car, utilizing the equation for distance s covered in time t, which corresponds to the length of one car, with u = 0 as the initial velocity and a representing acceleration, over t = 5.0 s. We can rearrange the equation reflecting L as the length of one car. This is similarly applicable for the initial seven cars, accounting for the distance of 7L and the required time t'. With constant acceleration retained, we can derive t' through substitution in the equation, leading to fundamental conclusions regarding the relationship exhibited in the graph of distance against time in uniformly accelerated motion.

8 0

2 months ago