A boat that travels 3.00 m/s relative to the water is crossing a river that is 1.00 km wide. The destination on the far side of the river is 0.500 km downstream from the starting point. (a) If the river current is 2.00 m/s, in what direction should the boat be pointed in order to reach the destination? (b) How much time will the trip take?


Answer 1


a) 10.29° upstream

b) t=338.7s


If the river is 1km wide and the destination point is 0.5km away downstream, then the angle and distance the the boat has to travel is:

\alpha =atan((0.5)/(1))=26.56°


The realitve velocity of the boat respect to the water is:

V_(B/W)=[3*cos\beta ,3*sin\beta ]  where β is the angle it has to be pointed at.

From the relative mvement equations:

V_(B/W)=V_B-V_W  where V_B=[V*cos\alpha ,-V*sin\alpha ]

From this equation we get one equation per the x-axis and another for the y-axis. If we square each of them and add them together, we will get 2 equations:

(3*cos\beta )^2+(3*sin\beta )^2=(V*cos\alpha )^2+(-V*sin\alpha +2)^2

V^2-4*V*sin\alpha -5=0  Solving for V:

V = 3.3m/s   and   V=-1.514m/s   Replacing this value into one of our previous x or y-axis equations:

\beta =acos((V*cos\alpha )/(3) ) = 10.29°

The amount of time:

t = D/V = (1118m)/(3.3m/s) =338.7s

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Every few years, winds in Boulder, Colorado, attain sustained speeds of 45.0 m/s (about 100 mi/h) when the jet stream descends during early spring. show answer No Attempt Approximately what is the force due to the Bernoulli effect on a roof having an area of 205 m2? Typical air density in Boulder is 1.14 kg/m3 , and the corresponding atmospheric pressure is 8.89 × 104 N/m2 . (Bernoulli’s principle assumes a laminar flow. Using the principle here produces only an approximate result, because there is significant turbulence.)



The force exerted on the roof is F =2.37*10^(5)N


From the question we are told that

      The speed of the wind is v = 45.0 m/s

       The area of the roof is A = 205 m^2

       The air density of Boulder is \rho = 1.14 kg / m^3

        The atmospheric pressure is P_(atm) = 8.89 * 10^(4) N/ m^2

For a laminar flow the Bernoulli’s principle is  mathematically represented as

            P_1 + (1)/(2) \rho v_a ^2 + \rho g h_a = P_2 + (1)/(2)  \rho v_b ^2 + \rho h_b

Where  v_1 is the  speed of air in  the building

             v_b is the speed of air outside the building

             P_1 \ and \ P_2 are the pressure of inside and outside the house

             h_a \ and \ h_b are the height above and  below the roof

Now for  h_a = h_b

            The above equation becomes

                 P_1 + (1)/(2) \rho v_a ^2 = P_2 + (1)/(2) \rho v_b ^2

                 P_1 - P_2 = (1)/(2) \rho (v_b^2 - v_a^2)

Since pressure is mathematically represented as

           P = (F)/(A )

The above equation can be written as

             F  = (1)/(2) \rho ( v_b^2 - v_a ^2 ) A

The initial velocity is 0

    Substituting value  

                F = (1)/(2)  (1.14) [(45^2 - 0^2 ) ](205)

                F =2.37*10^(5)N


An implanted pacemaker supplies the heart with 72 pulses per minute, each pulse providing 6.0 V for 0.65 ms. The resistance of the heart muscle between the pacemaker’s electrodes is 550 Ω. Find (a) the current that flows during a pulse, (b) the energy delivered in one pulse, and (c) the average power supplied by the pacemaker.



a) Current = 11 mA

b) Energy = 66 mJ

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a) Voltage, V = IR

   Voltage, V = 6 V, Resistance, R = 550 Ω

   Current, I =(6)/(550)=0.011A=11mA

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Wavelength = (speed) / (frequency)

The speeds of the two possible signals are equal, just like
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Wavelength of 895 MHz = (3 x 10⁸ m/s) / (8.95 x 10⁸/s) = 0.335 m

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Water on Earth was (a) transported here by comets; (b) accreted from the solar nebula; (c) produced by volcanoes in the form of steam; (d) created by chemical reactions involving hydrogen and oxygen shortly after Earth formed.


Answer: Water on Earth was transported here by comets. The correct option is A.


Comets are made up of water with ice, rock and minerals.

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A swimmer heads directly across a river, swimming at 1.00 m/s relative to still water. He arrives at a point 41.0 m downstream from the point directly across the river, which is 73.0 m wide. What is the speed of the river current?



velocity of the river is equal to 0.56 m/s



velocity of swimmer w.r.t still water = 1 m/s

width of river = 73 m

he arrives to the point = 41 m

times = (distance)/(speed)      

times = (73)/(1)          

 t = 73 s                        

velocity = (distance)/(time)                  

                    = (41)/(73)                      

                    = 0.56 m/s                        

velocity of the river is equal to 0.56 m/s

A particle (q = 5.0 nC, m = 3.0 μg) moves in a region where the magnetic field has components Bx = 2.0 mT, By = 3.0 mT, and Bz = −4.0 mT. At an instant when the speed of the particle is 5.0 km/s and the direction of its velocity is 120° relative to the magnetic field, what is the magnitude of the acceleration of the particle in m/s2?


The acceleration of the particle is 38.87 kg.

Net magnetic field

The net magnetic field is calculated as follows;

B_(net) = √(B_x^2 + B_y^2 + B_z^2) \n\nB_(net) = √(2^2 + 3^2 + 4^2) = 5.385 \ mT

Magnetic force on the charge

The magnetic force on the charge is calculated as follows;

F = qvB * sin(\theta)\n\nF = 5* 10^(-9) * 5* 10^3 * 5.385 * 10^(-3) * sin(120)\n\nF = 1.166 * 10^(-7) \ N

Acceleration of the particle

The acceleration of the particle is calculated as follows;

a = (F)/(m) \n\na = (1.166 * 10^(-7))/(3 * 10^(-9)) \n\na = 38.87 \ kg

Learn more about magnetic force here: brainly.com/question/13277365


It is given that,

Charge on the particle, q=5\ nC=5* 10^(-9)\ C

Mass of the particle, m=3\ \mu g=3* 10^(-6)\ g=3* 10^(-9)\ kg

Magnetic field component, B_x=2\ mT,B_y=3\ mT,B_z=-4\ mT

Net magnetic field, B=√(2^2+3^2+4^2)=5.38\ mT=0.00538\ T

Speed of the particle, v = 5 km/s = 5000 m/s

Angle between velocity and magnetic field, \theta=120

Magnetic force is given by :

F=qvB\ sin\theta

F=5* 10^(-9)* 5000\ m/s* 0.00538* sin(120)

F=1.16* 10^(-7)\ N

Acceleration of the particle is given by, a=(F)/(m)

a=(1.16* 10^(-7)\ N)/(3* 10^(-9)\ kg)

a=38.6\ m/s^2

So, the acceleration of the particle is 38.6 m/s². Hence, this is the required solution.