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yulyashka [42]
3 years ago
14

On a safari, a team of naturalists sets out toward a research station located 4.63 km away in a direction 38.7 ° north of east.

After traveling in a straight line for 2.13 km, they stop and discover that they have been traveling 25.9 ° north of east, because their guide misread his compass.
What are (a) the magnitude and (b) the direction (as a positive angle relative to due east) of the displacement vector now required to bring the team to the research station?
Physics
1 answer:
aleksandrvk [35]3 years ago
5 0

Exactly, it is a vector subtraction problem.  Let t=theta,

t1=38.7, d1=4.63;  t2=25.9, d2=2.13

v1=d1 <cos(t1), sin(t1)>=<3,6134, 2.8949>

v2=d2 <cos(t2), sin(t2)>=<1.9161, 0.9304>

Final vector

v3 = v1-v2

=<v3x, v3y>

=<1.6973, 1.9645>

where

v3x=d1*cos(t1)-d2*cos(t2)

v3y=d1*sin(t1)-d2*sin(t2)

The final vector v3 has therefore a magnitude of

||v3||=sqrt(1.6973^2+1.9645^2)=2.5962, and a direction of

theta=atan(1.9645/1.6973)=49.17 degrees  north of east

Note: all three vectors are in the direction of the first quadrant.

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Many Amtrak trains can travel at a top speed of 42.0 m/s. Assuming a train maintains that speed for several hours, how many kilo
777dan777 [17]

Answer:

605 km

Explanation:

Hello

the same units of measure should be used, then

Step 1

convert  42 m/s ⇒   km/h

1 km =1000 m

1 h = 36000 sec

42 \frac{m}{s}*\frac{1\ km}{1000\ m}=0.042\ \frac{km}{s}\\ 0.042\ \frac{km}{s}\\

0.042\ \frac{km}{s}*\frac{3600\ s}{1\ h} =151.2 \frac{km}{h}\\ \\Velocity =151.2\ \frac{km}{h}

Step 2

find kilometers traveled after 4  hours

V=\frac{s}{t}\\ \\

V,velocity

s, distance traveled

t. time

now, isolating s

V=\frac{s}{t} \\s=V * t\\

and replacing

s=V * t\\s=151.2\frac{km}{h}*4 hours\\ s=604.8 km\\

S=604.8 Km

Have a great day

4 0
3 years ago
Give an example of a compound machine. Explain how at least two simple machines are part of this complex machine.
11Alexandr11 [23.1K]

Answer:

Bicycle

Explanation:

A compound machine is a machine which is a combination of simple machines.

Simple machines are like the pulley, inclined plane or a screw.

Suppose a bicycle is considered, it has more than one simple machine combined together, for it to work. Wheel and axle is one of them and the beam which is pivoted at a fixed hinge is another simple machine in it.

The pedals of the bicycle function as the lever.

6 0
3 years ago
Discuss the correlation or connection between stars with a higher mass and the amount of fuel they have to work with
dimulka [17.4K]
Larger stars have a higher amount of fuel in order to keep the process of nuclear fusion going.
3 0
3 years ago
a passenger elevator operates at an average of 8 m/s if the 60th floor is 219 m above the first floor how long does it take the
MAVERICK [17]

Answer:

The universal sign for choking is __________.

A.

two balled fists pressing the abdomen

B.

pointing at an open mouth

C.

two hands grasping the neck

D.

pretending to cough

Explanation:

4 0
3 years ago
A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a
Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

Z_{R} = R

R = resistance

Z_{L} = jωL

ω = voltage source angular frequency, L = inductance

Z_{C} = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

Z_{R} = 217Ω

Z_{L} = j(220)(0.875) = j192.5Ω

Z_{C} = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = Z_{R} + Z_{L} + Z_{C}

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

v_{R} = I×Z_{R}

v_{L} = I×Z_{L}

v_{C} = I×Z_{C}

Given values:

I = (0.0569∠1.147+220t)A

Z_{R} = 217Ω = (217∠0)Ω

Z_{L} = j192.5Ω = (192.5∠π/2)Ω

Z_{C} = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

v_{R} = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

v_{L} = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

v_{C} = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

v_{R} = (12.35∠1.147+220t)V

v_{L} = (10.95∠2.718+220t)V

v_{C} = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

v_{R} = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

v_{L} = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

v_{C} = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0
3 years ago
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