What are three basic conditions for a hurricane to form

During the middle of a family picnic, Barry Allen received a message that his friends Bruce and Hal

weeeeeb [17]

The kinematics of the uniform motion and the addition of vectors allow finding the results are:

The Barry's initial trajectory is 94.30 10³ m with n angles of θ = 138.8º

The return trajectory and speed are v = 785.9 m / s, with an angle of 41.2º to the South of the East

Vectors are quantities that have modulus and direction, so they must be added using vector algebra.

A simple method to perform this addition in the algebraic method which has several parts:

Vectors are decomposed into a coordinate system

The components are added

The resulting vector is constructed

Indicate that Barry's velocity is constant, let's find using the uniform motion thatthe distance traveled in ad case

v = $\frac{\Delta d}{t}$

Δd = v t

Where v is the average velocity, Δd the displacement and t the time

We look for the first distance traveled at speed v₁ = 600 m / s for a time

t₁ = 2 min = 120 s

Δd₁ = v₁ t₁

Δd₁ = 600 120

Δd₁ = 72 10³ m

Now we look for the second distance traveled for the velocity v₂ = 400 m/s

time t₂ = 1 min = 60 s

Δd₂ = v₂ t₂

Δd₂ = 400 60

Δd₂ = 24 103 m

In the attached we can see a diagram of the different Barry trajectories and the coordinate system for the decomposition,

We must be careful all the angles must be measured counterclockwise from the positive side of the axis ax (East)

Let's use trigonometry for each distance

Route 1

cos (180 -35) = $\frac{x_1}{\Delta d_1}$

sin 145 = $\frac{y_1}{\Delta d1}$

x₁ = Δd₁ cos 125

y₁ = Δd₁ sin 125

x₁ = 72 103 are 145 = -58.98 103 m

y₁ = 72 103 sin 155 = 41.30 10³ m

Route 2

cos (90+ 30) = $\frac{x_2}{\Delta d_2}$

sin (120) = $\frac{y_2}{\Delta d_2}$

x₂ = Δd₂ cos 120

y₂ = Δd₂ sin 120

x₂ = 24 103 cos 120 = -12 10³ m

y₂ = 24 103 sin 120 = 20,78 10³ m

The component of the resultant vector are

Rₓ = x₁ + x₂

R_y = y₁ + y₂

Rx = - (58.98 + 12) 10³ = -70.98 10³ m

Ry = (41.30 + 20.78) 10³ m = 62.08 10³ m

We construct the resulting vector

Let's use the Pythagoras' Theorem for the module

R = $\sqrt{R_x^2 +R_y^2}$

R = $\sqrt{70.98^2 + 62.08^2}$ 10³

R = 94.30 10³ m

We use trigonometry for the angle

tan θ ’= $\frac{R_y}{R_x}$

θ '= tan⁻¹ $\frac{R_y}{R_x}$

θ '= tan⁻¹ $\frac{62.08}{70.98}$

θ ’= 41.2º

Since the offset in the x axis is negative and the displacement in the y axis is positive, this vector is in the second quadrant, to be written with respect to the positive side of the x axis in a counterclockwise direction

θ = 180 - θ'

θ = 180 -41.2

θ = 138.8º

Finally, let's calculate the speed for the way back, since the total of the trajectory must be 5 min and on the outward trip I spend 3 min, for the return there is a time of t₃ = 2 min = 120 s.

The average speed of the trip should be

v = $\frac{\Delta R}{t_3}$

v = $\frac{94.30}{120} \ 10^3$

v = 785.9 m / s

in the opposite direction, that is, the angle must be

41.2º to the South of the East

In conclusion, using the kinematics of the uniform motion and the addition of vectors, results are:

To find the initial Barry trajectory is 94.30 10³ m with n angles of 138.8º

The return trajectory and speed is v = 785.9 m / s, with an angle of 41.2º to the South of the East

Learn more here: brainly.com/question/15074838

4 0

3 years ago

the other end of the pipe. With the aide of the pipe, does the applied force produce a smaller torque, a greater torque, or the

pav-90 [236]

Answer:

With the addition of the pipe we have a greater torque.

Explanation:

We need to complete the description of the problem, searchin in internet we have:

"Sometimes, even with a wrench, one cannot loosen a nut that is frozen tightly to a bolt. It is often possible to loosen the nut by slipping one end of a long pipe over the wrench handle and pushing at the other end of the pipe. With the aid of the pipe, does the applied force produce a smaller torque, a greater torque, or the same torque on the nut?"

With the addition of the pipe we have a greater torque, as it increases the distance or radius of torque.

We know that torque is defined, as the product of force by distance, in this way we have:

T = F * d

where:

T = torque [N*m]

F = force [N]

d = distance [m]

We can see in the above equation, that increasing the distance increases torque proportionally.

3 0

3 years ago

Lightning produces a maximum air temperature on the order of 104K, whereas a nuclear explosion produces a temperature on the ord

gtnhenbr [62]

Answer:

tex]2.898\times 10^{-7}\ \text{m}[/tex] ultraviolet region

$2.898\times 10^{-10}\ \text{m}$ x-ray region

Explanation:

T = Temperature

b = Constant of proportionality = $2.898\times 10^{-3}\ \text{m K}$

$\lambda$ = Wavelength

$T=10^4\ \text{K}$

From Wein's law we have

$\lambda=\dfrac{b}{T}\\\Rightarrow \lambda=\dfrac{2.898\times 10^{-3}}{10^4}\\\Rightarrow \lambda=2.898\times 10^{-7}\ \text{m}$

The wavelength of the radiation will be $2.898\times 10^{-7}\ \text{m}$ and it is in the ultraviolet region.

$T=10^7\ \text{K}$

$\lambda=\dfrac{2.898\times 10^{-3}}{10^7}\\\Rightarrow \lambda=2.898\times 10^{-10}\ \text{m}$

The wavelength of the radiation will be $2.898\times 10^{-10}\ \text{m}$ and it is in the x-ray region.

5 0

3 years ago

If the plotted points on a speed-time graph do not form a straight line, what do you know about the object's acceleration?

slega [8]

Remember, that while sped is constant, acceleration is not. Acceleration is when velicity changes. So the graph which shows the slop of a velocity vs time describes acceleration.
If we have the straight line on the graph it means that the slope is always the same whereas the non-linear graphs has a variable slope that changes depending on your point in the graph.
To conclude - if your graph is not a straight line it has variable acc at many points.

6 0

3 years ago

Read 2 more answers

Why does light refract when it encounters the glass in the lens

Kobotan [32]

Hey there!

because it slows down, which causes it to bend is the reason why light refracts when it encounters the glass in the lens

Hope this helps!

5 0

3 years ago

Read 2 more answers