The sequence that shows the energy transformations taking place in these lights are radiant energy → electric energy → radiant energy.
What is law of conservation of conservation of energy?
The principle or law of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another.
Based on this law, the energy of a substance can be converted from one form to another
For example, energy can be converted as follows;
potential energy to kinetic energychemical energy to electric energyelectrical energy to sound energyetc,The sequence of energy that takes place on street lights powered by solar panels is given as follows;
Radiant energy (light energy from sun) to electrical energy (converted by photo voltaic cell of the panels) to radiant energy (light given by the street lights).
Thus, the sequence that shows the energy transformations taking place in these lights are radiant energy → electric energy → radiant energy.
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Answer:
C
Explanation:
I did the test
Isaac Newton in his 1670's lectures on optics and later in his 1704 publicationOpticks, he described light as composed of in his corpuscular hypothesisarguing that the perfectly straight lines of reflection demonstrated this nature.linesparticlesphotonswaves
The perfectly straight lines of reflection demonstrated in the nature is due to particle nature of the light. Therefore, Newton described that the light consists of 'particles' which means second option is correct.
A student accidentally knocks a book off a table (it started at rest). If the book hits the ground in .5 seconds, how fast was it going when it hit the ground?
The velocity of a book that fell off a table and hit the ground after 5 seconds is 49 m/s.
What is velocity?Velocity can be defined as the ratio of displacement to time.
To calculate how fast the book will hit the ground, we use the formula below.
Formula:
v = u+gt............ Equation 1Where:
v = Velocity of the book before it hit the groundu = Initial velocity of the bookt = timeg = Acceleration due to gravityFrom the question,
Given:
u = 0 m/st = 5 secondsg = 9.8 m/sSubstitute these values into equation 1
v = 0+5×9.8v = 49 m/sHence, the velcoity of the book before it hit the ground is 49 m/s.
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If you have 100.g of a radioactive isotope with a half-life of 10. years, how much of the isotopewill you have left after 20. years?
Use the radioactive decay formula using t_0 as the half life:
[tex]A=A_0\cdot2^{-t/t_0}[/tex]Substitute the initial amount of radioactive element A_0=100g, the half life t_0=10y and the time period t=20y to find the remaining amount after 20 years:
[tex]\begin{gathered} A=100g\times2^{-20y/10y} \\ =100g\times2^{-2} \\ =100g\times\frac{1}{4} \\ =25g \end{gathered}[/tex]Therefore, after 20 years, there are 25 grams left.
In lacrosse, a ball is thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. If the angular velocity of the ball about the elbow joint is 31.3 rad/s and the ball is 1.45 m from the elbow joint, what is the velocity of the ball?
The linear velocity of the ball whose angular velocity is 31.3 rad/s about the elbow joint will be 45.385 m/s.
What is angular velocity?Angular velocity is the rate of change of angular displacement with respect to time. Mathematically -
ω = dθ/dt
from this we can write -
dθ = ω dt
∫dθ = ω ∫dt
θ₂ - θ₁ = ω(t₂ - t₁)
Δθ = ω Δt
ω = Δθ/Δt
Given is a ball thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. The angular velocity of the ball about the elbow joint is 31.3 rad/s and the ball is 1.45 m from the elbow joint.
The relation between linear velocity and the angular velocity of a body undergoing circular motion is given by -
v = rω
From this we can write -
ω = 31.3 rad/s
r = 1.45 m
Substituting the values -
v = rω
v = 1.45 x 31.3
v = 45.385 m/s
Therefore, the linear velocity of the ball whose angular velocity is 31.3 rad/s about the elbow joint will be 45.385 m/s.
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carts, bricks, and bands
Which one of the following changes would increase the amount of mass?
a. Increase the number of bricks resting upon the cart.
b. Increase the number of bands that are used to pull the cart.
c. Decrease the number of bands that are used to pull the cart.
d. Use a cart that is identical in every way, except for the color that it is painted.
The change that would increase the amount of mass is Increase in the number of bricks resting upon the cart. That is option A.
What is mass?Mass is defined as the amount or quantity of matter that makes up an object which can be measured in grams and Kilograms.
The mass of an object is dependent on the change of its weight, that is, an increase in the weight of an object increases its mass.
Since the bricks had the same mass as the cart, adding one brick to the cart would double the mass of the object. Adding two bricks to the cart would triple the mass of the object.
That is to say, when more bricks are being laid on the cart, it's weight increase leading to an increase in its mass also.
Therefore, it can be concluded that increase the amount of mass is Increase the number of bricks resting upon the cart.
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Pls quick will mark brainliest.
To BEST avoid any accidents when swimming, you should NEVER swim:
A. with a parent.
B. with a partner.
C. alone.
D. near a lifeguard.
Answer:
Explanation:
c. Alone
Answer:
C. alone.
Explanation:
Which one of the pulley systems below has the best force advantage?Select one:a. Pulley Ib. Pulley IIc. Pulley IIId. The force advantage is the same.
a. Pulley I
Explanation
A pulley system is a collection of one or more wheels which are used with a. rope or chain to make it easier to lift things, The main advantage in the use of pulleys is that the effort becomes less as compared to the normal lifting of the weights
the ideal mechanical advantage is equal to the number of rope segments pulling up on the object. The more rope segments that are helping to do the lifting work, the less force that is needed for the job.
so
Step 1
let's check the number of pulleys on each system
so, the firs system has the more rope segements pulling up (4)
therefore, the answer is
a. Pulley I
I hope this helps you
8) If the volume of the liquid in graduated cylinder B is 90 mL, then whatis the volume of the rock?AYour answer8060B100180
Answer:
20 mL
Explanation:
The volume of the rock is equal to the difference of volume of A and B. So, it is equal to
90 mL - 70 mL = 20 mL
Because 90 mL is the volue in cylinder B and 70 mL is the volume in cylinder A.
Therefore, the volume of the rock is 20 mL
Bart, mass 32.4 kilograms, and Milhouse, mass 27.6 kilograms, play on the schoolyard seesaw. If Bart and Milhouse want to sit 4.0 meters apart, how far from the center of the seesaw should Bart sit? Include units in your answer. Answer must be in 3 significant digits.
Given data:
* The mass of the Bart is m_1 = 32.4 kg.
* The mass of the Milhouse is m_2 = 27.6 kg.
* The distance between the Millhouse and Bart is d = 4 m.
Solution:
To balance the seesaw, the net moment about the center should be zero.
The diagrammatic representation of the given system is,
The distance between the Bart and Milhouse can be written as,
[tex]\begin{gathered} d=d_1+d_2 \\ 4=d_1+d_2 \\ d_2=4-d_1 \end{gathered}[/tex]where d_2 is the distance of Milhouse from the center and d_1 is the distance of Bart from the center,
Consider the moment as positive if it is in an anticlockwise direction and negative if it is a clockwise direction.
Thus, the net moment about the center is,
[tex]M=m_1d_1-m_2d_2[/tex]Substituting the known values,
[tex]\begin{gathered} 0=32.4\times d_1-27.6\times(4-d_1) \\ 0=32.4\times d_1-110.4+27.6d_1 \\ 0=60d_1-110.4 \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} 60d_1=110.4 \\ d_1=\frac{110.4}{60} \\ d_1=1.84\text{ m} \end{gathered}[/tex]Thus, the distance of the Bart from the center is 1.84 meters.
An object has a position function x(t) = 5t m. (a) What is the velocity as a function of time? (b) Graph the position function and the velocity function.
Considering the given position function, it is found that:
a) The velocity function is: v(t) = 5 m/s.
b) The functions are graphed at the end of the answer.
Position and velocity functionThe position function in this problem, after t seconds, is defined according to the following rule:
s(t) = 5t.
The velocity function is the derivative of the position function, hence it is calculated as follows:
v(t) = s'(t) = [5t]' = 5 m/s. (position in meters, hence velocity in meters per second).
The derivative rule applied was the power rule, [5t]' = 5[t'] = 5 x 1 x t^(1 - 1) = 5t^0 = 5.
These two functions are graphed at the end of the answer, considering a domain of t ≥ 0, as time cannot assume negative values.
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A cat chases a mouse across a 0.66 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor taylor (jdt3899) – Homework 3, 2d motion 22-23 – tejeda – (LermaHPHY1 1) 3 2.4 m from the edge of the table. The acceleration of gravity is 9.81 m/s 2 . What was the cat’s speed when it slid off the table?
The cat’s speed when it slid off the table will be 6.552 m/s
The branch of physics that defines motion with respect to space and time, ignoring the cause of that motion, is known as kinematics. Kinematics equations are a set of equations that can derive an unknown aspect of a body’s motion if the other aspects are provided.
a = -g = 9.8 m[tex]/s^{2}[/tex]
using equation of motion
x = u(horizontal )*t + 1/2 * a (horizontal) * [tex]t^{2}[/tex]
since , a (horizontal) = 0
x = u(horizontal )*t
u = x / t equation 1
similarly
y = u(vertical)*t + 1/2 * a (vertical) * [tex]t^{2}[/tex]
u(vertical) = 0
t = [tex]\sqrt{2y / a}[/tex] equation 2
substituting the value of equation 2 in equation 1
u = x / [tex]\sqrt{2y / a}[/tex]
= [tex]\sqrt{\frac{-9.81}{2*-0.66} } * 2.4[/tex]
= 6.552 m/s
The cat’s speed when it slid off the table will be 6.552 m/s
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Physics
Hi dears how could we solve this question basically how to plug it in calculator
The new length of the steel bar is 11.65 cm.
What is linear expansion?
Linear expansion can be defined as the increase in length of a material due to increase in the temperature of the material.
Mathematically, the linear expansion of a material is given as;
ΔL = L₀αΔT
where;
ΔL is the change in length or increase in lengthL₀ is the initial length of the steelα is the coefficient of linear expansion of steel = 11 x 10⁻⁶/⁰CΔT is change in temperatureThe change in the length of the steel is calculated as follows;
ΔL = (11.5 cm) x (11 x 10⁻⁶/⁰C) x (1221 ⁰C - 22 ⁰C)
ΔL = 0.152 cm
The new length of the steel bar is calculated as follows;
L = ΔL + L₀
L = (0.152 cm) + (11.5 cm)
L = 11.65 cm
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A football player runs from his own goal line to the opposing team's goal line, returning to his twenty-yard line, all in 27.0 s. Calculate his average speed and the magnitude of his average velocity. (Enter your answers in yards/s.)HINTApply the definitions of average speed and average velocity.Click the hint button again to remove this hint.(a) Calculate his average speed. ____yards/s(b) Calculate the magnitude of his average velocity. ____yards/s
time = 27 s
d = 20 yard
a) speed = distance / time
d1 = 100 yard ( own goal to opposing team's goal line)
d2 = 80 yard ( returning to 20 yard line)
d= d1+d2
d= 100 + 80 = 180 yards
Speed = 180 y / 27s = 6.667 y/s
b) velocity = displacement / time
d = 100 - 80 = 20 y
Velocity = 20 / 27 = 0.74 y/s
n a slap shot, a hockey player accelerates the puck from a velocity of 5 m/s to 30 m/s in the same direction. If the puck moves over a distance of 10 m during this process, what was the acceleration?
Given data
*The given initial velocity of the puck is u = 5 m/s
*The given final velocity of the puck is v = 30 m/s
*The given distance is s = 10 m
The formula for the acceleration is given by the kinematic equation of motion as
[tex]\begin{gathered} v^2=u^2+2as \\ a=\frac{v^2-u^2}{2s} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} a=\frac{(30)^2-(5)^2}{2\times10} \\ =43.75m/s^2 \end{gathered}[/tex]Hence, the acceleration of the puck is a = 43.75 m/s^2
A 75 kg criminal wants to escape from the 5th story window of the jail, 24 m above the ground. He has a rope, which can only support a tension force of 650 N.
a. What is the maximum acceleration he can slide down without breaking his "rope?"
Answer:
a = 1.1 m/s²
Explanation:
Given:
P = 650 N
m = 75 kg
g = 9.8 m/s²
____________
a - ?
Forse:
P = m*(g - a)
650 = 75*(10 - a)
10 - a = 650 /75
Acceleration:
a = 9.8 - 650 / 75 = 9.8 - 8.7 = 1.1 m/c²
calculate how much heat is released when 500g of platinum is cooled from 250.0K to 240.0K
Answer:
Q = 665 J
Explanation:
Given:
m = 500 g = 0.5 kg
T₁ = 250.0 K
T₂ = 240.0 K
c = 133 J / ( kg*⁰K) - Specific heat capacity of platinum
___________
Q - ?
Heat is released:
Q = c*m*( T₁ - T₂) = 133*0.5*(250.0 - 240.0) =
= 133*0.5*10 ≈ 665 J
Object a attracts objects be the gravitational force of 10 N from a given distance the distance between the two objects is doubled what is the new force of attraction between
We are given that two objects are being attracted by a gravitational force between each other of 10N. The gravitational force between two masses is given by the following equation:
[tex]F_g=G\frac{m_Am_B}{r^2}[/tex]Where:
[tex]\begin{gathered} F_g=\text{ Gravitational force} \\ m=\text{mass} \\ G=\text{ Gravitational constant} \\ r=\text{ distance between the masses} \end{gathered}[/tex]Replacing the given values for the 10N force:
[tex]10=G\frac{m_Am_B}{r^2_1}[/tex]Where:
[tex]r_1=\text{ initial distance}[/tex]Now we will solve for the product of the masses and the gravitational constant by multiplying both sides by the distance squared:
[tex]10r^2_1=Gm_Am_B[/tex]Now, the product of the masses and the gravitational constant won't change if we double the distance, therefore, if we apply the equation for the gravitational force for the new distance we get:
[tex]F_{g2}=G\frac{m_Am_B}{r^2_2}[/tex]We can replace the value we determined earlier:
[tex]F_{g2}=\frac{10r^2_1}{r^2_2}[/tex]Since the distance is double, we have:
[tex]r_2=2r_1[/tex]Replacing in the previous equation:
[tex]F_{g2}=\frac{10r^2_1}{(2r_1)^2}[/tex]Solving the square:
[tex]F_{g2}=\frac{10r^2_1}{4r^2_1}[/tex]Now we can cancel out the distances squared:
[tex]F_{g2}=\frac{10}{4}[/tex]Solving the operation:
[tex]F_{g2}=2.5[/tex]Therefore, doubling the distance the new gravitational force is 2.5N.
A snowmobiler travels 65km [37° E of S]. How far east does she travel?
ANSWER:
39.1 km
STEP-BY-STEP EXPLANATION:
To better understand the problem, we make a sketch, like this:
Therefore, we can determine the distance east with the cosine function, like this:
[tex]\begin{gathered} \cos\theta=\frac{\text{ adjacent}}{\text{ hypotenuse}} \\ \\ \theta=53\degree \\ \\ \text{ adjacent =}E \\ \\ \text{ hypotenuse = 65} \\ \\ \text{ We replacing:} \\ \\ \cos53\degree=\frac{E}{65} \\ \\ E=65\cdot\cos53\degree \\ \\ E=39.1\text{ km} \end{gathered}[/tex]The distance to the east is 39.1 km
A type of wave that is a combination of a transverse and a longitudinal wave is called aQuestion 17 options:Slinky waveSurface waveSound waveLight wave
The type of wave that is a combination of longitudinal and Transverse wave is called Surface wave
The mass of the Moon is about 1/80th of the mass of Earth. The force exerted by Earth on the Moon is about 80 times thatexerted by the Moon on Earth.Select one:O TrueO False
According to Newton's Third Law of Motion, the force that object A exerts to object B has the same magnitude as the force that object B to object A, but in the opposite direction:
[tex]\vec{F}_{AB}=-\vec{F}_{BA}[/tex]Then, the force exerted by Earth on the Moon has the same magnitude as the force exerted by the Moon on the Earth.
Therefore, the given statement is false.
A car initially traveling eastward turns north by traveling in a circular path at uniform speed as in the figure below. The length of the arc ABC is 219 m, and the car completes the turn in 32.0 s.
(a) What is the acceleration when the car is at B located at an angle of 35.0°? Express your answer in terms of the unit vectors î and ĵ.
(b) Determine the car's average speed.
(c) Determine its average acceleration during the 32.0-s interval.
a.) the acceleration when the car is at B located at an angle of 35.0° is 2.689i -0.42818j
b.) the car's average speed is v= 6.84375 m/s
c.) average acceleration during the 32.0-s interval is (−0.181 i+0.181 j)m/s²
What is acceleration?Acceleration is described as the rate of change of the velocity of an object with respect to time.
(a) The car’s speed around the curve is found from
v= 219/32.0
v= 6.84375 m/s
This is the answer to part (b) of this problem. We calculate the radius of the curve from
(1/4) X 2πr = 219 m
which gives r = 139.4 m
The car’s acceleration at point B is then
ar = (V²/r ) towards the center
= ( 6.84375)² / 139.4 at 35.0° north of west
= (2.9761 m/s²)(cos 35.0)(-i) + (sin 35.0j)
= -2.689i -0.42818j
(b) From part (a), v= 6.84375 m/s
(c) We find the average acceleration from
A avg = (Vf - Vi)/ change in t
A avg = ( 6.84375 j - 6.84375 i ) / 32.0 s
A avg = (−0.181 i+0.181 j)m/s²
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A 2000 kg car is stopped by applying a braking force of 5000 newtons.
Determine the acceleration caused by this braking force.
A 2000 kg car is stopped by applying a braking force of 5000 newtons. The acceleration caused by this braking force is (a)=2.5 m/s²
What is acceleration?When a object start with a velocity and ends with different velocity, so the change in velocity in a given time is called the acceleration of the object. It is a vector quantity. It can be measured in m/s².
How can we calculate the acceleration?To calculate the acceleration we are using the formula here is ,
F=ma
Here we are given by the question is,
F= The amount of force applied on the car. = 5000N.
m= The mass of the car. = 2000kg
We have to calculate the change in acceleration = a m/s².
Now we put the values in above equation we get,
F=ma
Or, a= F/m
Or, a=5000/2000
Or, a=2.5 m/s².
Now from the above calculation we can say that, The acceleration caused by this braking force is (a)=2.5 m/s².
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Which of the following is true for an isolated system? I. Matter is able to freely enter or exit the system.II. Heat is able to freely enter or exit the system.III. Work is able to freely enter or exit the system.II onlyNone of the aboveI or II onlyI only
Note that, an isolated system does not allow:
• the exchange of energy
,• the exchange of matter
Therefore, in an isolated system, neither
hea
There is no _________ movement in a longitudinal wave.A. HorizontalB. Back and forthC. VerticalD. Parallel
Explanation
A longitudinalwave is in which the particles of the medium vibrate in the direction of the line of advance of the wave.Longitudinal waves cause the medium to move parallel to the direction of the wave.
A longitudinal wave can be set up for example in a streched spring by compressing the coils in a small region, and releasing the compressed region,
the back and forth motions of the coils of the spring is in the same direction that the wave travels
so, in a longitudinal wave there is not Vertical movement, so the answer is
C. Vertical
Forces that are equal in magnitude
but opposite in direction will:
Answer:
The two forces equal in magnitude but acting opposite in direction on a body are called balanced forces
Explanation:
Convert each quantity to the indicated units.
a. 3.01 g to cg
b. 6200 m to km
c. 0.13 cal/g to kcal/g
Answer:
Explanation:
a) 301 cg
b) 6.2 km
c) 0.00013
Sean and Greg are on a job site standing on two beams 11.0 ft apart. they need to lift their crate of tools midway between them with ropes up 33.5 ft to where they are working. (a) What is the angle between the ropes when the crate is on the ground? (b) How much force do Sean and Greg need to exert on the ropes when lifting the 115-lb crate off the ground? (c) How much force do both Sean and Greg need to exert when the crate is 5.75 ft below them? (d) Explain why the force to lift the crate changes as it moves closer to them crate is 5.75 ft below them? (d) Explain why the force to lift the
crate changes as it moves closer to them.
a ) The angle between the ropes when the crate is on the ground = 18.4°
b ) Force exerted by Sean and Greg = 260.8 N
c ) Force exerted when the crate is 5.75 ft below them = 354.9 N
a ) The angle between the ropes,
Distance between crate and a person = 33.5 ft
Distance between Sean and Greg = 11 ft
Consider only one person side of the rope and the midway point between Sean and Greg. This forms a right angled triangle.
sin θ = 5.5 / 33.5
θ = [tex]sin^{-1}[/tex] ( 0.16 )
θ = 9.2°
Angle between the ropes = 2 θ
Angle between the ropes = 2 * 9.2
Angle between the ropes = 18.4°
b ) Force exerted,
Since both ropes pull the same amount of weight for the same amount of distance, their tensions will be equal,
T = T1 = T2
Resolving the tension into its horizontal and vertical component.
[tex]T_{y}[/tex] = T cos θ
[tex]T_{x}[/tex] = T sin θ
Since there is not time component mentioned assuming the crate is pulled at a constant velocity. Therefore acceleration will be zero and hence net force in vertical direction will be zero.
m = 115 lb = 52.16 kg
∑ [tex]F_{y}[/tex] = 0
T cos θ + T cos θ - m g = 0
2 T cos 9.2° = 52.16 * 9.8
T = 511.17 / 1.96
T = 260.8 N
c ) When the crate is 5.75 ft below them,
tan θ = 5.5 / 5.75
θ = [tex]tan^{-1}[/tex] ( 0.96 )
θ = 43.8°
∑ [tex]F_{y}[/tex] = 0
T cos θ + T cos θ - m g = 0
2 T cos 43.8° = 52.16 * 9.8
T = 511.17 / 1.44
T = 354.9 N
d ) The force to lift the crate changes as it moves closer to them is because in a shorter cable the horizontal force increases and the vertical force decreases. So it becomes harder to pull as it gets closer to the destination.
Therefore,
a ) The angle between the ropes when the crate is on the ground = 18.4°
b ) Force exerted by Sean and Greg on the ropes when lifting the 115-lb crate off the ground = 260.8 N
c ) Force exerted by Sean and Greg when the crate is 5.75 ft below them = 354.9 N
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How many moles of a gas sample are in a 5.0 L container at 251 K and 370 kPa?(The gas constant is8.31L kPamol K)Round your answer to one decimal place and enter the number only with no units.
Given:
The volume of the gas, V=5.0 L
The temperature of the gas, T=251 K
The pressure of the gas, P=370 kPa
The gas constant, R=8.31 L kPa/(mol K)
To find:
The moles of the gas sample.
Explanation:
From the ideal gas equation,
[tex]PV=\text{nRT}[/tex]Where n is the moles of the gas present.
On substituting the known values,
[tex]\begin{gathered} 370\times5.0=n8.31\times251 \\ \Rightarrow n=\frac{370\times5.0}{8.31\times251} \\ =0.9\text{ mol} \end{gathered}[/tex]Final answer:
The moles of the gas present in the sample is 0.9 mol
Hunter pushed a couch across the room. He did 800 J of work in 20 seconds.The couch weighed 500 N. How much power did he have?A. 16,000 WB. 1.6WC. 800 WD. 40 W
Power = 40 W
Option D
Explanation:The workdone by the Hunter = 800 Joules
The weight of the couch = 500 N
Time, t = 20 seconds
Power = Workdone/Time
Power = 800/20
Power = 40 W
Sheena can row a boat at 2.90 mi/h in still water. She needs to cross a river that is 1.20 mi wide with a current flowing at 2.00 mi/h. Not having her calculator ready, she guesses that to go straight across, she should head upstream at an angle of 25.0° from the direction straight across the river.What is her speed with respect to the starting point on the bank?How long does it take her to cross the river?How far upstream or downstream from her starting point will she reach the opposite bank? If upstream, enter a positive value and if downstream, enter a negative value.In order to go straight across, what angle upstream should she have headed?
Given
v = 2.90 mi/h
x = 1.20 mi
vs = 2.0 mi/h
θ = 25°
Procedure
Part a)
Velocity of the boat with respect to water stream is given as
[tex]\begin{gathered} v_b=v-v_s \\ v_b=(2.90\cdot\sin 25-2.00)\hat{j}+2.90\cdot\cos 25\hat{i} \\ v_b=-0.77\hat{j}+2.62\hat{i} \end{gathered}[/tex]magnitude of the speed is given as
[tex]\begin{gathered} v_b=\sqrt[]{0.77^2+2.62^2} \\ v_b=2.73mi/h \end{gathered}[/tex]Part b)
Time to cross the river is given as:
[tex]\begin{gathered} t=\frac{x}{v_x} \\ t=\frac{1.2mi}{2.90\cdot\cos25} \\ t=0.46h \end{gathered}[/tex]Part c)
Distance moved by boat in downstream is given as
[tex]\begin{gathered} x=v_yt \\ x=-0.77\cdot0.46 \\ x=-0.354mi \end{gathered}[/tex]Part d)
In order to go straight, we must net speed along the stream must be zero
so we will have
[tex]\begin{gathered} v\sin \theta=v_s \\ 2.90\sin \theta=2.00_{} \\ \sin \theta=\frac{2.00}{2.90} \\ \theta=43.60^{\circ} \end{gathered}[/tex]