A"boat"is"moving"to"the"right"at"5"m/s"with"respect"to"the"water."A"wave"moving"to"the"left,"opposite"the"motion"of"the"boat."The"waves"have"2.0"m"between"the"top"of"the"crests"and"the"bottom"of"the"troughs."The"period"of"the"wave"is"8.3"s"and"their"wavelength"is"110"m."At"one"instant"the"boat"sits"on"a"crest"of"the"wave,"20"seconds"later,"what"is"the"vertical"displacement"of"the"boat

Answers

Answer 1
Answer:

Answer:

0.99m

Explanation:

Firs you calculate the relative velocity between the boat and the wave. The velocity of the boat is 5m/s and the velocity of the wave is given by:

v=\lambda f=\lambda(1)/(T)=(110m)(1)/(8.3s)=13.25(m)/(s)

the relative velocity is:

v'=13.25m/s-5m/s=8.25(m)/(s)

This velocity is used to know which is the distance traveled by the boat after 20 seconds:

x'=v't=(8.25m/s)(20s)=165m

Next, you use the general for of a wave:

f(x,t)=Acos(kx-\omega t)=Acos((2\pi)/(\lambda)x-\omega t)

you take the amplitude as 2.0/2 = 1.0m.

\omega=(2\pi)/(T)=(2\pi)/(8.3s)=0.75(rad)/(s)

by replacing the values of the parameters in f(x,t) you obtain the vertical displacement of the boat:

f(165,20)=1.0m\ cos((2\pi)/(110m)(165)-(0.75(rad)/(s))(20s))\n\nf(165,20)=0.99m


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How fast can the car take this curve this curve without skidding to the outside of the curve?

Answers

Lets write the data down. That will help us solve the problem later:

R = 36 m

θ = 18º

m = 1492 kg

μ = 0.67

g = 9.8 m/s²

Lets draw all the forces that act on the car:

In order to the car won't skidding to the outside of the curve we must have the centripetal force equals the friction force:

F_(cp)=f_a

(m.v^(2))/(R)=\mu.F_N

A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of gravity of his arm is 25 cm from the joint axis of shoulder abduction at point O • The weight of the arm W is 0.06 of the pitcher’s weight of 100 N • Deltoids muscles are at an angle θ of 15° with respect to the humerus and insert 15 cm from the joint axis at point A • Determine the force applied by the Deltoid muscles and the joint reaction force at the shoulder joint and its orientation β

Answers

I attached a Diagram for this problem.

We star considering the system is in equlibrium, so

Fm makes 90-(\theta+5) with vertical

Fm makes 70 with vertical

Applying summatory in X we have,

\sum F_x = 0

W+1.4-Fm cos(70)

We know that W is equal to

W= 0.06*100N = 6N

Substituting,

Fm cos (70) = W+1.4N

Fm cos (70) = 6N + 1.4N

Fm = (7.4)/(cos(70))

Fm = 21.636N

For the second part we know that the reaction force Fj on deltoid Muscle is equal to Fm, We can assume also that\beta  = \theta

A certain lightning bolt moves 50.0 c of charge. how many fundamental units of charge (qe) is this?

Answers

The lightning bolt that moves 50.0 C of charge is converted to qe units. The conversion factor is 1 coulumb is equivalent to 6.24150975·10^18 e. Thus, converting 50 C to e is:

50.0 C * (6.24150975x10^18 e)/1C
equals 
3.120754875x10^20 e 

Answer: There are 3.125\tims 10^(20) number of electrons.

Explanation:

We are given 50 Coulombs of charge and we need to find the number of electrons that can hold this much amount of charge. So, to calculate that we will use the equation:

nq_e=Q

where,

n = number of electrons

q_e Charge of one electron = 1.6* 10^(-19)C

Q = Total charge = 50 C.

Putting values in above equation, we get:

n* 1.6* 10^(-19)C=50C\n\nn=(50C)/(1.6* 10^(-19)C)\n\nn=3.125* 10^(20)C

Hence, there are 3.125\tims 10^(20) number of electrons.

A 5 kg bowling ball with a velocity of +10 m/s collides with a stationary 2 kg bowling pin. If the ball's final velocity is +8 m/s, what is the pin's final velocity?a 5 m/s
b 2.5 m/s
c 10 m/s
d 5.2 m/s

Answers

Answer:

The pin's final velocity is 5m/s

Explanation:

Step one:

given data

mass of ball m1=5kg

initial velocity of ball u1=10m/s

mass of pin m2=2kg

initial velocity of pin u2= 0m/s

final velocity of ball v2=8m/s

final velocity of pin v2=?

Step two:

The expression for elastic collision is given as

m1u1+m2u2=m1v1+m2v2

substituting we have

5*10+2*0=5*8+2*v2

50+0=40+2v2

50-40=2v2

10=2v2

divide both sides by 2

v2=10/2

v2=5m/s

The pin's final velocity is 5m/s

Nichrome wire, often used for heating elements, has resistivity of 1.0 × 10-6 Ω ∙ m at room temperature. What length of No. 30 wire (of diameter 0.250 mm) is needed to wind a resistor that has 50 ohms at room temperature?

Answers

Answer:

Length = 2.453 m

Explanation:

Given:

Resistivity of the wire (ρ) = 1 × 10⁻⁶ Ω-m

Diameter of the wire (d) = 0.250 mm = 0.250 × 10⁻³ m

Resistance of the wire (R) = 50 Ω

Length of the wire (L) = ?

The area of cross section is given as:

A=(1)/(4)\pi d^2\n\nA=(1)/(4)*\ 3.14* (0.250* 10^(-3))^2\n\nA=0.785* 6.25* 10^(-8)\n\nA=4.906* 10^(-8)\ m^2

We know that, for a constant temperature, the resistance of a wire is directly proportional to its length and inversely proportional to its area of cross section. The constant of proportionality is called the resistivity of the wire. Therefore,

R=\rho (L)/(A)

Expressing the above in terms of length 'L', we get:

L=(RA)/(\rho)

Plug in the given values and solve for 'L'. This gives,

L=(50* 4.906* 10^(-8))/(1* 10^(-6))\ m\n\nL=(2.453)/(1)=2.453\ m

Therefore, length of No. 30 wire (of diameter 0.250 mm) is 2.453 m.

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536 J of energy.
Calculate the power output of the cake mixer.
A
I DONT KNOW​

Answers

And electric light bulb mix it is used for nine. Two seconds in that temperature transfers one 536 J of energy calculate shin is 2+2 is 4-1 that’s three quick maths exclamation is calculate the power work output of the kitchen