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?

Answers

Answer 1
Answer:

Answer:

a) 10.29° upstream

b) t=338.7s

Explanation:

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°

D=√(1^2+0.5^2)=1.118km

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.)

Answers

Answer:

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

Explanation:

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.

Answers

Answer:

a) Current = 11 mA

b) Energy = 66 mJ

c) Power = 101.54 W

Explanation:

a) Voltage, V = IR

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

   Current, I =(6)/(550)=0.011A=11mA

b) Energy = Current x Voltage = 6 x 0.011 = 0.066 J = 66 mJ

c) \texttt{Power=}(Energy)/(Time)=(0.066)/(0.65* 10^(-3))=101.54W    

Two microwave frequencies are authorized for use in microwave ovens: 895 and 2560 mhz. calculate the wavelength of each.

Answers

Wavelength = (speed) / (frequency)

The speeds of the two possible signals are equal, just like
all other forms of electromagnetic radiation.

Wavelength of 895 MHz = (3 x 10⁸ m/s) / (8.95 x 10⁸/s) = 0.335 m

Wavelength of 2560 MHz = (3 x 10⁸ m/s) / (2.56 x 10⁹/s) =  0.117 m 

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.

Answers

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

Explanation:

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

Alot of research and hypotheses has been made to prove the origin of water on planet earth. Extraplanetary source such as comets, trans-Neptunian objects, and water-rich meteoroids (protoplanets) are believed to have delivered water to Earth.

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?

Answers

Answer:

velocity of the river is equal to 0.56 m/s

Explanation:

given,

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?

Answers

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

Explanation:

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.