Ask question

Ask question

All categories

2 years ago

12

The position of a dragonfly that is flying parallel to the ground is given as a function of time by r⃗ =[2.90m+(0.0900m/s2)t2]i^

− (0.0150m/s3)t3j^.
At what value of t does the velocity vector of the insect make an angle of 36.0 ∘ clockwise from the x-axis? ... t=??

2 answers:

Arlecino [84]2 years ago

7 0

The solution for this problem is:

r = [(2.90 + 0.0900t²) i - 0.0150t³ j] m/s²
this is for t in seconds and r in meters

v = dr/dt = [0.180t i - 0.0450t² j] m/s²

tan(-36.0º) = -0.0450t² / 0.180t
0.7265 = 0.25t
t = 2.91 s is the velocity vector of the insect

White raven [17]2 years ago

7 0

The velocity vector of the insect make an angle of 36.0° clockwise from the x-axis at t = 2.91 s

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

$r = [2.90 +(0.0900t^2)]i - [(0.0150)t^3]j$

To find the velocity function, we will derive the position function above.

$v = dr/dt = [0 +(0.0900)(2t)]i - [(0.0150)(3t^2)]j$

$v = [0.18t]i - [0.045t^2]j$

Next to calculate the angle , we use this following formula:

$\tan \theta = \frac{v_y}{v_x}$

$\tan 36^o = \frac{0.045t^2}{0.18t}$

$\tan 36^o = \frac{0.045t}{0.18}$

$t = \frac{\tan 36^o \times 0.18}{0.045}$

$t \approx 2.91 ~ s$

<h2>Conclusion:</h2>

The velocity vector of the insect make an angle of 36.0° clockwise from the x-axis at t = 2.91 s

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

You might be interested in

If a freely suspended vertical spring is pulled in downward direction and then released, which type of wave is produced in the s

larisa [96]

Answer:

longitudinal wave

Explanation:

it is perpendicular to the direction of the wave

3 0

2 years ago

Read 2 more answers

The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo

irina1246 [14]

Answer:

$p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

Explanation:

first write the newtons second law:

F $_{s}$ =δma $_{s}$

Applying bernoulli,s equation as follows:

∑ $δp+\frac{1}{2} ρδV^{2} +δγz=0\\$

Where, $δp$ is the pressure change across the streamline and $V$ is the fluid particle velocity

substitute $ρg$ for {tex]γ[/tex] and $g_{0}-cz$ for $g$

$dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0$

integrating the above equation using limits 1 and 2.

$\int\limits^2_1 \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

there the bernoulli equation for this flow is $p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

note: $ρ$ =density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

4 0

2 years ago

How far could a rabbit run if it ran 36km/h for 5.0min?

Korolek [52]

3 kilometers, it is just 5/60 or 1/12 multiplied by 36.

4 0

2 years ago

Read 2 more answers

Write a hypothesis about the effect of the fan speed on the acceleration of the cart. Use the "if . . . then . . . because . . .

iragen [17]

As an object accelerates i.e., change it's velocity(either direction or speed), the position of the object depends on two factor; If the acceleration was direction based then it might have a zero displacement for eg: if it travels in circle. or it might have a net displacement if it travels in a straight line, quantitatively
$s = ut + \frac{1}{2} at {}^{2}$
where,
s = displacement
u = initial velocity
v = final velocity
a = acceleration
t = time

Now, for the hypothesis;
There is no direct relationship between fan speed and acceleration but anyways generally speaking if we do have a relationship that with more fan speed we have a larger displacement of air i.e., a more force i.e., greater acceleration
Thus, it can be said, well not exactly scientific, that with a greater fan speed there will be greater acceleration. if fan speed is increased then acceleration will be more.
:)

8 0

2 years ago

Read 2 more answers

A battery that produces a potential difference V is connected to a 10-W light bulb. Later, the 10-W bulb is replaced with a 20-W

uysha [10]

Answer:

The battery supply the greatest current with the 20-W bulb

Explanation:

The electrical power at a certain time by a two-terminal device is the product of the potential difference between the terminals and the current intensity that passes through the device. For this reason the power is proportional to the current and voltage. Mathematically can be written as:

$P=VI$

Let:

$P_1=Electric\hspace{3}power\hspace{3}dissipated\hspace{3}by\hspace{3}the\hspace{3}10-W\hspace{3}bulb\\P_2=Electric\hspace{3}power\hspace{3}dissipated\hspace{3}by\hspace{3}the\hspace{3}20-W\hspace{3}bulb$

The voltage remains constant for every case so:

$I_1=Current\hspace{3}supply\hspace{3}by\hspace{3}the\hspace{3}battery\hspace{3}with\hspace{3}the\hspace{3}\hspace{3}10-W\hspace{3}bulb\\I_2=Current\hspace{3}supply\hspace{3}by\hspace{3}the\hspace{3}battery\hspace{3}with\hspace{3}the\hspace{3}\hspace{3}20-W\hspace{3}bulb$

Then:

$P_1=V*I_1=10W\\P_2=V*I_2=20W$

Analyzing the equations it makes sense that the battery supplies the highest current with the 20 W bulb, because the power only depends on the voltage and the current, if the voltage remains constant, the only reason for the power to increase is due to current.

3 0

2 years ago