What is primary succession? Give an example.

A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc

katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

$cos \theta = \frac {V_{river}}{V_{boat}}$

When solving for theta, we get that:

$\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})$

so now we can substitute the corresponding values:

$\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})$

Which yields:

$\theta = 60^{o}$

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

$V_{y}=5.60km/hr*cos(210^{o})$

$V_{y}=-4.85km/hr$

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

$t=\frac{y}{V_{y}}$

$t=\frac{5.60km}{4.85km/hr}=1.155hr$

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

$V_{ds}=V_{river}+V{boat}$

$V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr$

and now up stream:

$V_{us}=V_{boat}-V{river}$

$V_{us}=5.60km/hr-2.80km/hr=2.80km/hr$

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

$t_{ds}=\frac{x}{v_{ds}}$

$t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr$

and the time up stream:

$t_{us}=\frac{x}{v_{us}}$

$t_{us}=\frac{2.80km}{2,80km/hr}=1hr$

so the total time will be:

$t_{ds}+t_{us}=0.333hr+1hr=1.333hr$

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

$t=\frac{d}{v_{boat}}$

$t=\frac{5.60km}{5.60km/hr}$

so

t= 1hr

4 0

4 years ago

Read 2 more answers

A meter stick whose mass is 290 grams lies on ice. You pull at one end of the meter stick, at right angles to the stick, with a

katen-ka-za [31]

Answer

given,

mass of the stick = 290 grams = 0.29 Kg

Force on the stick on one side = F = 9 N

force acting perpendicular to stick.

magnitude of acceleration

rate of change of angular momentum is equal to Force

rate of change of angular momentum = 9 N

F = m a

$a = \dfrac{F}{m}$

$a = \dfrac{9}{0.29}$

a = 31.034 m/s²

Direction of motion will in the direction of force application or in the direction of change of velocity

5 0

4 years ago

A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the bod

11Alexandr11 [23.1K]

Answer:

X(t) = 13/13 cos(12t+α)

C =13/13

π/6 s

Explanation:

(A) A body with mass 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time t = 0 the body is pulled 1 m in to the right, stretching the spring, 3 set in motion with an initial velocity of 5 m/s to the left.

(a) Find X(t) in the form c • cos(w_o*t— α)

(b) Find the amplitude 3 Period of motion of the body 1

mass: m = 200g = 0.200 kg

displacement: ΔX = 20 cm = 0.20 m

Spring Constant: K = 9/0.20 = 45 N/m

IV: X(0) = 1m V(0) = -5 m/s

Simple Harmonic Motion: c•cos(cosw_t— α) = X(t)

Circular Frequency: w_o = √k/m= √36/(0.20) = 13 rad/s

X(0) = 1m =c_1

X'(0) = V(0) = c_2*w_o/w_o

= -5/12 = c_2

"radians Technically Unitless"

Amplitude: c = √ci^2 + c^2 ==> √1^2 + (-5/12)^2 = 1 m =13/13 = c

X(t) = 13/13 cos(12t+α)

since, C>0 : damped forced vibration c_1>0, c_2>0

phase angle 2π+tan^-1(c_2/c_1)

=2π+tan^-1(-5/12/1)= 5.884

period: T =2π/w_o

=π/6 s

6 0

4 years ago

A long wire carrying a 5.0 A current perpendicular to the (xy)-plane intersects the x-axis at x = -2.00 cm. A second parallel wi

zimovet [89]

Answer:

a. 05cm from x axis

b. 8cm from x axis

Explanation:

If the net magnetic field is zero and the currents are in the same direction then the thanks point is between the currents i1 and i2 as show in the attachment below

a. Given that i1= 5A and i2=3A

Let assume the null point is xcm from current i1, then the null point will be (4-x)cm from current i2 since the total length is 4cm.

Now the magnetic field of the current i1 from the null point= to magnetic field of current i2 from the null point

B1=B2

μi1/2πx=μi2/2π(4-x)

i1/x=i2/(4-x)

5/x=3/(4-x)

20-5x=3x

8x=20

8x=2.5cm

since from the left of x axis is 2cm, then the null point is 2.5-2 which 0.5cm from the origin x axis.

The null point is 0.5cm from the origin x axis

b. If both current are flowing in opposite direction, the null point lies outside of the current.

Then with same analysis let assume the first current i1 is xcm from the null point and since the total length is 4cm the second current i2 will be (x-4)cm from the null point.

Also the magnetic field of the current i1 from the null point = to magnetic field of current i2 from the null point

B1=B2

μi1/2πx=μi2/2π(x-4)

i1/x=i2/(x-4)

5/x=3/(x-4)

5x-20=3x

2x=20

x=10cm.

This shows that the distance of the null point from current i1 is 10cm and the current i1 is 2cm from the x axis, then the null point is 10-2=8cm from the origin x axis.

The null point is 8cm from the x axis.

Check the attachment to see the diagram of the current and the null points

6 0

4 years ago

Suppose a vector v makes an angle theta with respect to the y axis, what could be the x and y components of the vector v

gtnhenbr [62]

If the angle between the vector and the y-axis is Θ ,
then the y-component of the vector is

(length of the vector) times cosine( Θ ),

and the x-component of the vector is

(length of the vector) times sine( Θ ) .

8 0

3 years ago