Using the right amount of significant figures, calculate the answer to the following problem, 215.5+101.02555

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

Vector vector a has a magnitude of 29 units and points in the positive y-direction. when vector vector b is added to vector a ,

Nutka1998 [239]

Good morning.

We have:

$\mathsf{\overset{\to}{a} = 29\overset{\to}{j}}$

Where j is the unitary vector in the direction of the y-axis.

We have that

$\mathsf{\overset{\to}{a}+\overset{\to}{b} = -18\overset{\to}{j}}$

We add the vector -a to both sides:

$\mathsf{\overset{\to}{b} = -18\overset{\to}{j} -\overset{\to}{a} = -18\overset{\to}{j} -29\overset{\to}{j}}\\ \\ \mathsf{\overset{\to}{b}=-47\overset{\to}{j}}$

Therefore, the magnitude of b is 47 units.

5 0

3 years ago

Read 2 more answers

The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold

Liula [17]

Answer:

$L_{o}=0.1224m$

Explanation:

Given data

Force F=2 N

Length L=17 cm = 0.17 m

Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

$F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m$

6 0

4 years ago

A student travels 4.0m East, 8.0m South, 4.0m west, and finally 8.0m north. What is her displacement

Vedmedyk [2.9K]

Answer:

travel travelled 24 meters

3 0

3 years ago

Describe the different processes that lead to substantial internal heat sources for Jupiter and Saturn. Since these two objects

Softa [21]

Explanation:

The internal heat sources for Jupiter and Saturn derive from primordial heat resulting from the initial gravitational contraction of each planet. Jupiter also generates heat by slow contraction, which liberates substantial gravitational energy. A significant part of Saturn’s heat comes from the release of gravitational energy from helium separating from the lighter hydrogen and sinking to its core. What one considers to be a star is a matter of definition, as we discuss in more detail in the chapter on The Birth of Stars and the Discovery of Planets outside the Solar System. While both Jupiter and Saturn generate much of their energy internally, they are not large enough (by a significant factor) to support nuclear reactions in their interiors, and so are not considered to be stars.

6 0

4 years ago