The dimensions and unit of C from the expression above is [ML^-3] and kgm^-3
What is dimension and units ?
Despite the fact that the solution to any engineering problem must incorporate units, dimensions and units are frequently misunderstood. While units are arbitrary names that are correlated to certain dimensions to make it relative, dimensions are tangible quantities that can be measured (e.g., a dimension is length, whereas a metre is a relative unit that describes length). Through a conversion factor, all units for the same dimension are related to one another.
given in question drag force depends upon the product of the cross sectional area of the object and the square of its velocity
and drag force can be given by
F = CAv^2 ----------- (1)
It is given that
Dimension of mass = [M]
Dimension of length = [L]
Dimension of time = [T]
So, by using above dimension we can write
the dimension of force,
F = [ MLT^-2]
dimension of cross-section area,
A = [L^2]
and dimension of velocity
v = [LT^-1]
now, by putting these values in equation (1), we will get
F = CAv^2
[ MLT^-2] = C[L^2][LT^-1]^2
C = [ML^-3T^0]
C = [ML^-3]
The dimensions and unit of C from the expression above is [ML^-3] and kgm^-3
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Hence, the dimension of constant C will be,
A marble was thrown vertically up the air and came back down to the samelevel in 1.6 s. (using g = 9.81 ms?).(a) What is it initial velocity?(b) What would be the highest height it can reach?(c) How long would it take to reach the highest point?
ANSWER:
(a) 15.7 m/s
(b) 12.56 m
(c) 0.8 sec
STEP-BY-STEP EXPLANATION:
We can calculate the initial velocity using the following formula:
(a)
[tex]\begin{gathered} v=u+a\cdot t \\ v=0 \\ \text{therefore:} \\ u=-a\cdot t \\ \text{ replacing} \\ u=-(-9.81\cdot1.6) \\ u=15.7\text{ m/s} \end{gathered}[/tex](b)
We calculate the height in this way:
[tex]\begin{gathered} h=\frac{(-u)^2}{2g} \\ \text{ replacing} \\ h=\frac{(-15.7)^2}{2\cdot9.81} \\ h=12.56\text{ m} \end{gathered}[/tex](c)The time to reach this height would be half the time it takes for the entire journey, that is, 0.8 seconds
A 27.6 kg block is pulled along a rough level surface at constant velocity by the force . The figure shows the free-body diagram for the block. FN represents the normal force on the block; and f represents the force of kinetic friction. The coefficient of kinetic friction is 0.30, what is the strength of P?
The strength of P, the pulling force is 352 N.
What is friction?Friction is the force that acts to oppose the motion of an object moving over another object at their surfaces of contact.
The force of friction depends on the following:
mass/weight of the objectnature of the surface of contactConsidering the free-body diagram of the given object, the forces acting on the object are as follows:
FN = normal force acting in the opposite direction and equal to the weight of the object, mgmg = the weight of the object acting downwardsP = the pulling force in the forward directionf = frictional force acting in an opposite direction to PSolving for P:
P = f + mg
μ = f/N
f = μ * N
N = mg
f = μ * m * g
f = 0.3 * 27.6 * 9.81
f = 81.2 N
P = 81.2 N + 27.6 * 9.81
P = 352 N
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What is the efficiency of a block and tackle if you pull 20 m of rope with a force of 600 N to raise 200kg piano 5 m?
W out = mass * distance * gravity
E in = Force * distance
Efficiency = W out / E in = (200 kg * 5m*9.8N) / (600N * 20m) = 0.81
0.81 x 100 = 81 %
The wiring in a house must be thick enough so it does not become so hot as to start a fire.part aWhat diameter must a copper wire be if it is to carry a maximum current of 34 A and produce no more than 1.6 W of heat per meter of length?
Given:
The maximum current in the circuit is,
[tex]i=34\text{ A}[/tex]The power per length is,
[tex]\frac{P}{l}=1.6\text{ W/m}[/tex]To find:
The diameter of the copper wire
Explanation:
The power (P) produced by current i, through a copper wire of resistance R and length l is given by,
[tex]\begin{gathered} Pl=i^2R \\ \frac{R}{l}=\frac{P}{i^2} \\ \frac{R}{l}=\frac{1.6}{34\times34} \end{gathered}[/tex]Now,
[tex]\begin{gathered} R=\frac{\rho l}{A} \\ R=\frac{\rho l}{\pi r^2} \end{gathered}[/tex]The resistivity of copper is,
[tex]\rho=1.72\times10^{-8}\text{ ohm.m}[/tex]So, we can write,
[tex]\begin{gathered} \frac{R}{l}=\frac{\rho}{\pi r^2} \\ \frac{1.6}{34\times34}=\frac{1.72\times10^{-8}}{\pi r^2} \\ r^2=\frac{1.72\times10^{-8}\times34\times34}{1.6} \\ r=3.5\times10^{-3}\text{ m} \\ diamer\text{ is,} \\ 2r=7.0\times10^{-3}\text{ m} \end{gathered}[/tex]Hence, the diameter is,
[tex]7.0\times10^{-3}\text{ m}[/tex]A star of mass 3.0e30 kg is moving in a circle of radius 1.0e12 metres, with a period of 100 years. This is due to the gravity of a second unseen object. what is the mass of the unseen object in kg
The mass of the unseen object is 1.21 x 10²⁴ kg.
What is the centripetal acceleration of the star?
The centripetal acceleration of the star is calculated as follows;
a = v²/r
where;
v is the linear speed of the starr is the radiusv = 2πr/T
where;
T is period of the motion = 100 years = 3.154 x 10⁹ sv = (2π x 1 x 10¹²) / (3.154 x 10⁹)
v = 1,992.1 m/s
a = (1992.1)² / ( 1 x 10¹²)
a = 3.97 x 10⁻⁶ m/s²
The centripetal force of the star is calculated as follows;
F = ma
F = (3 x 10³⁰) x ( 3.97 x 10⁻⁶)
F = 1.19 x 10²⁵ N
The mass of the unseen object is calculated as follows;
F = mg
m = F/g
where;
g is the acceleration due to gravitym = (1.19 x 10²⁵) / (9.8)
m = 1.21 x 10²⁴ kg
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Compute, in centimeters and in meters, the height of a basketball player who is 6 ft 1 in. tall into cm and m
We will have that:
6ft 1in is:
[tex]6ft(\frac{12in}{1ft})+1in=73in[/tex]Then:
[tex]73in(\frac{2.54cm}{1in})=185.42cm[/tex]So, the solution is both meters and cm are respectively.
1.8542 m and 185.42 cm.
If the electric intensity between two parallel plates, placed 1cm apart is 104 NC-1and the direction of the field of this intensity is vertically upward. Find the force on an electron in this field and compare the force on the electron with the electron’s weight
The force of a charge in an electric field is:
[tex]\vec{F}_e=q\vec{E}[/tex]In this case we know the electric field is:
[tex]\vec{E}=104\hat{j}[/tex]and that the charge is that of the electron, then we have:
[tex]\begin{gathered} \vec{F}_e=-1.6\times10^{-19}(104\hat{j}) \\ \vec{F}_e=-1.664\times10^{-17}\text{ N} \end{gathered}[/tex]Therefore, the magnitude of the force is
[tex]1.664\times10^{-17}\text{ N}[/tex]and in points down.
The weight of the electron is:
[tex]\begin{gathered} W=1.67\times10^{-27}(9.98) \\ W=1.6366\times10^{-26} \end{gathered}[/tex]Making the quotient between the force we have:
[tex]\frac{1.664\times10^{-17}}{1.6366\times10^{-26}}=1.02\times10^9[/tex]Therefore, the electric force is approximately 1e9 times the weight.
Draw In figure what the compass needles would show if that current is as shown in the figure. Make the arrow the “north” direction. (Please refer to picture)
The direction of the magnetic field around a current-carrying conductor is given by the right-hand thumb rule. In the figure, the direction of the flow of the current is into the plane of the paper.
Therefore the needles of the compass would align as,
In the above figure, the arrowheads indicate the north direction.
carts, bricks, and bands
10. Predict the acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks.
a. Approximately 0.16 m/s2
b. Approximately 0.50 m/s2
c. Approximately 0.64 m/s2
d. Approximately 1.00 m/s2
D. The acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks is approximately 1 m/s².
What is acceleration?The acceleration of an object is the rate of change of velocity of the object with time.
The acceleration that would occur when four rubber bands were used to pull a cart loaded with two bricks, is determined by applying Newton's second law of motion as follows.
a = F/m
where;
a is accelerationF is the applied forcem is the massLet the mass of a brick = mass of a band = m
the mass of a cart = 2 bricks = 2m
a = (force applied by 4 rubber) / (mass of 1 cart + mass of 2 brick)
a = (4m) / (2m + 2m)
a = (4m)/(4m)
a = 1 m/s²
Thus, the acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks is approximately 1 m/s².
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Ms. Terry travels to another planet where she drops a hamburger freely from rest. On this particular planet, g=5 m/s2. How far has the hamburger fallen after 4 s?
Answer:
40 m
Explanation:
d = 1/2 a t^2
= 1/2 * 5 m/s^2 * (4 s)^2 = 1/2 * 5 * 16 = 40 m
The Student Council conducted a vote to determine whether the homcoing dance should have live muscis ora DJ. The number of students voting for live music was 215. The number of students voting of a DJ was 645.What percent of the students voted for the DJ?(Hint: Find the total first)
Divide the number of students who voted for the DJ over the total number of students to find the percent of the students that voted for the DJ.
Since 215 students voted for live music and 645 students voted for the DJ, the total number of students was:
[tex]215+645=860[/tex]Then, the percent of the students who voted for the DJ was:
[tex]\frac{645}{860}=0.75=75\text{ \%}[/tex]Therefore, the answer is: 75%
A baseball is rolling along a tabletop with avelocity of 3.9 m/s to the right. The tabletopis 1.1 m above the ground. The ball rolls offthe edge of the table and falls to theground.A.) What is the ball's final vertical Velocity?B.) How long does the ball take to fall?C.) how far from the table does the ball land?
To answer this question we need to notice that once the ball starts falling we have a projectile motion; which means that horizontally we have a rectilinear motion and vertically we have an uniformly accelerated motion.
Then we can use the following equations for each direction:
[tex]\begin{gathered} \text{ Horizontal motion:} \\ x=x_0+v_{0x}t \\ \text{ Vertical motion:} \\ a=\frac{v_f-v_0}{t} \\ y=y_0+v_0t+\frac{1}{2}at^2 \\ v_f^2-v_0^2=2a(y-y_0) \end{gathered}[/tex]Since the ball is moving down in the vertical direction we will think that down is the positive direction vertically.
a)
We know that the ball is rolling to the right when it rolls off the edge of the table, this means that vertically the initial velocity is zero; we also know that the ball will fall for 1.1 m and that the acceleration is the gravitational acceleration. Then we can use the third vertical motion equation to find the final velocity, plugging the values we know we have that:
[tex]\begin{gathered} v_f^2-0^2=2(9.8)(1.1) \\ v_f=\sqrt{2(9.8)(1.1)} \\ v_f=4.64 \end{gathered}[/tex]Therefore, the final vertical velocity is 4.64 m/s.
b)
To determine the time we can use the second vertical equation with the values we know:
[tex]\begin{gathered} 1.1=0+0t+\frac{1}{2}(9.8)t^2 \\ 4.9t^2=1.1 \\ t^2=\frac{1.1}{4.9} \\ t=\sqrt{\frac{1.1}{4.9}} \\ t=0.474 \end{gathered}[/tex]Therefore, it takes 0.474 s for the ball to fall.
c)
While the ball is falling it is also moving horizontally, in this direction we know the initial velocity is 3.9 m/s; using the horizontal equations we have:
[tex]\begin{gathered} x=0+(3.9)(0.474) \\ x=1.85 \end{gathered}[/tex]Therefore, the ball lads 1.85 m from the table.
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
21. An object m is tied to one end of a string, moves in a circle with a constant speed v
on a horizontal frictionless table. The second end of the string is connected to a big
mass M and goes through a small hole in the table. What is the value of M if it stays
in equilibrium?
I
(B) v²/rmg
(A) mv²/rg
(C) rg/mv²
(D) mv²r/g
The value of M that goes through a small hole in the table if it stays in equilibrium is mv²/rg ( A )
The force acting on object m is the centripetal force.
[tex]F_{c}[/tex] = m v² / r
Mass = m
Velocity = v
Radius = r
The force acting on object M is the gravitational force,
[tex]F_{g}[/tex] = M g
g = Acceleration due to gravity
Since the system is at equilibrium,
[tex]F_{c}[/tex] - [tex]F_{g}[/tex] = 0
m v² / r = M g
M = m v² / r g
Therefore, the value of M if it stays in equilibrium is mv²/rg ( A )
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Which has more kinetic energy: a 0.0014-kg bullet traveling at 397 m/s or a 5.9 107-kg ocean liner traveling at 13 m/s (25 knots)?
the bullet has greater kinetic energy
the ocean liner has greater kinetic energy
Justify your answer.
Ek-bullet =
J
Ek-ocean liner =
J
Answer:
Ocean Liner because its mass is 5000 times more and the bullet's velocity squared is only 1600 times more. This means that the kinetic energy of the ocean liner will be roughly 3 times greater.
Explanation:
The kinetic energy of an object is dependent on mass and velocity.
E = ( 1 / 2 )mv²;
Energy is measured in Joules which is equal to kilogram metres squared per second squared.
1J = ( kg )( m² )( s⁻² ) or 1J = ( kg )( m² ) / ( s² )
We can substitute the mass and velocity directly into the equation because the question gives the values in metres and kilograms.
BULLET:
E = ( 1 / 2 )( 0.0014kg )( 397m / s )²;
E = ( 0.0007kg )( 157609m² )( s⁻² );
E = 110.3263kgm²s⁻²;
E = 110.3263J;
I will just skip some steps because you get the idea.
OCEAN LINER:
E = ( 1 / 2 )( 5.9107kg )( 13m / s )²;
E = 499.45415J;
johns mass is 92.0kg on the earth, what is his mass on mars where g= 3.72m/s^2
The mass is a fundamental property of all matter.
Mass is not the same as weight, the weight depends on the gravity while the mass does not. Therefore the John's mass is the same in mars.
Do you think we can see the planets without telescope? explain your answer
Answer:
No, how do you tell that?
Telescope is an example of technology that use on seeing the object on far place try to do that, you acn see a planet like small pieceof like an tungaw
Accomplishment of SPTA Officers and all parents this 1st Quarter.. Enjoy watching..Mabbalo
Explanation:
HOPE IT HELP TO YOUWhat is the index of refraction for a medium where light has a velocity of 2.75 x10³ m/s?0.9176.751.090.250
n = c/v, where:
n: index of refraction
c: speed of light in a vacuum (3*10^8 m/s)
v: speed of light in medium
n = (3*10^8)/(2.75*10^8)
n = 3/2.75
n = 1.09
A) A recipe calls for 5.0 qt of milk. What is this quantity in cubic centimeters?Express your answer in cubic centimeters.B) A gas can holds 2.0 gal of gasoline. What is this quantity in cubic centimeters?Express your answer in cubic centimeters.
A.
[tex]\begin{gathered} 1qt=946.353cm^3 \\ 5qt=\text{?} \\ \text{quantity}=946.353\times5 \\ \text{quantity}=4731.765cm^3 \end{gathered}[/tex]B.
[tex]\begin{gathered} 1\text{ gal=}3785.41\text{ }cm^3 \\ 2\text{ gal}=\text{?} \\ \text{quantity}=2\times3785.41 \\ \text{quantity}=7570.82cm^3 \end{gathered}[/tex]A bunny and a tortoise start a race from rest. The bunny accelerates at a rate "a" for a time "t" until it reaches itsmaximum speed (vb) after traveling a distance d_b. The tortoise accelerates at a rate one fourth as great as the bunny andtakes three times as long to reach its maximum speed of vt after traveling a distance d_t.a) What is the maximum speed of the tortoise? Answer in terms of vbb)How far did the tortoise travel? Answer in terms of db
We are given that a Bunny starts a race from rest, this means that its initial speed is zero:
[tex]v_{b0}=0,(1)[/tex]We are also given that it accelerates at a rate "a" for a time "t" and reaches its maximum speed. We can use the following equation of motion to relate the final speed with its acceleration:
[tex]v_b=at,(2)[/tex]We are also given that it travels a distance "d". We can use the following equation of motion to relate the distance travel with the velocity and the acceleration:
[tex]2ad_b=v^2_b,(3)[/tex]Now. We turn our attention to the turtle. We are told that its acceleration is one-fourth of the acceleration of the bunny, therefore, we have:
[tex]a_t=\frac{1}{4}a[/tex]And it also says that it takes 3 times as long to reach its maximum speed, therefore the time of the turtle is:
[tex]t_t=3t[/tex]Now we use the equation of motion that relates velocity with time and acceleration for turtle:
[tex]v_t=a_tt_t[/tex]Since we are required to express this velocity in terms of the final velocity of the bunny we replace the values of acceleration and time of the turtle that we determined previously:
[tex]v_t=(\frac{1}{4}a)(3t)[/tex]Rearranging terms:
[tex]v_t=\frac{3}{4}at[/tex]From the equation of motion from the bunny we know that the product of acceleration and time of the bunny equals its final velocity , therefore:
[tex]v_t=\frac{3}{4}v_b[/tex]And thus we have determined the final velocity of the turtle in terms of the final velocity of the bunny.
Now we are told to determine the final velocity in terms of the distance traveled by the bunny. First, we will use the equation of motion that relates the distance, the acceleration, and the time with the final velocity:
[tex]2a_td_t=v^2_t[/tex]Now we replace the value of the acceleration:
[tex]2(\frac{1}{4}a)d_t=v^2_t[/tex]Simplifying:
[tex]\frac{1}{2}ad_t=v^2_t[/tex]Now we use the relationship between velocities we determined in part A:
[tex]\frac{1}{2}ad_t=(\frac{3}{4}v_b)^2[/tex]Simplifying:
[tex]\frac{1}{2}ad_t=\frac{9}{16}v^2_b[/tex]From equation (3) we can replace the value of the velocity of the bunny:
[tex]\frac{1}{2}ad_t=\frac{9}{16}\times2ad_b[/tex]We can cancel out the acceleration:
[tex]\frac{1}{2}d_t=\frac{9}{16}\times2d_b[/tex]Now we multiply both sides by 2:
[tex]d_t=2\times\frac{9}{16}\times2d_b[/tex]Simplifying:
[tex]d_t=\frac{9}{4}d_b[/tex]And thus we have found the distance traveled by the turtle in terms of the distance traveld by the bunny.
Jacob, who has a mass of 74.8 kg, is sitting in a chair at rest. Calculate his weight in newtons
and round the answer to one decimal place.
Jacob's weight when sitting in a chair with a mass of 74.8 kg is 733.0 N
What is weight?Weight can be defined as the gravitational pull of a body.
To calculate Jacob's weight, we use the formula below.
Formula:
W = mg............. Equation 1Where:
W = Jacob's weightm = Jacob's massg = Acceleration due to gravity.From the question,
Given:
m = 74.8 kgg = 9.8 m/s²Substitute these values into equation 1
W = (74.8×9.8)W = 733.0 NHence, the weight of Jacob is 733.0 N.
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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
Mechanical energy is the form of energy associated with the ,or of an object.
Mechanical energy is the form of energy associated with the motion or position of an object.
Explanation:Mechanical energy can be defined as the energy possessed by an object as a result of its motion or position.
Mechanical energy is divided into two.
Kinetic energy and potential energy
Kinetic energy is the energy possessed by an object as a result of its motion
Potential energy is the energy possessed by an object as a result of its position.
Therefore, we can conclude that mechanical energy is the form of energy associated with the motion or position of an object.
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
A locomotive enters a station with an initial velocity of 19 m/s and slows down at a rate of as it goes through. If the station is 175 m long, how fast is it going when the nose leaves the station?
ANSWER
[tex]9m\/s[/tex]EXPLANATION
Parameters given:
Initial velocity, u = 19 m/s
Acceleration, a = -0.8m/s² (the train is slowing down)
Distance traveled, s = 175 m
To find the final velocity of the train, we have to apply one of Newton's equations of motion:
[tex]v^2=u^2+2as[/tex]where v = final velocity
Substituting the given values into the equation, the final velocity of the train as its nose leaves the station is:
[tex]\begin{gathered} v^2=19^2+(2\cdot-0.8\cdot175) \\ v^2=361-280 \\ v^2=81 \\ \Rightarrow v=\sqrt[]{81} \\ v=9m\/s \end{gathered}[/tex]That is the answer.
If the input force on the lever in problem 5 is 135 N, what is the net force on the rock?
The net force on the rock is determined as 76.2 N.
What is the net force on the rock?The net force on the rock is obtained by subtracting the weight of the rock from the input force on the lever.
F(net) = Input force - weight of the rock
The weight of the rock is calculated as follows;
W = mg
where;
m is mass of the rock = 6 kgg is acceleration due to gravity = 9.8 m/s²W = 6 x 9.8
W = 58.8 N
F(net) = 135 N - 58.8 N
F(net) = 76.2 N
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The complete question is below:
If the input force on the lever in problem 5 is 135 N, what is the net force on the rock? if the mass of the rock is 6kg.
Will mark as brainlist
Which of the following best represents R= A - B ?
Please help, it’s due soon!
Option C is the best representation for the resultant vector representing the resultant R=A+B because the result is represented by the closing side of the triangle, so it is the correct answer.
What exactly is a vector quantity?A measure with both magnitude and direction It is typically represented by an arrow with the same direction as the quantity and a length proportional to the magnitude of the quantity. A vector, while including magnitude and demand, does not have a position. Vector quantities are physical quantities that are distinguished by the presence of both magnitude and direction. Displacement, force, torque, momentum, acceleration, velocity, and so on. Vector quantities are physical quantities that have distinct magnitude and direction definitions. For example, suppose a boy is riding a bike at 30 km/hr in the north-east direction. In nature, vectors have velocity, acceleration, force, electromagnetic fields, and importance. (Weight is the force produced by gravity's acceleration acting on a mass.) A Scalar is a quantity or phenomenon that has only magnitude and no specific direction.To learn more about vector quantity, refer to:
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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]An AC generator produces an output voltage E = 200 sin96.084 t volts. What is the frequency of the voltage ?
We can write the output voltage E in the following form:
E = E0*sin(ωt), where ω is angular period.
ω = 96.084
We also know that ω = 2πf, where f is frequency.
96.084 = 2πf; f = 96.084/2π
f = 15.2922 Hz
I need help with this question There is 4 answersa)b)c)d)
Given,
The velocity of the joggers, v=-3.50 m/s
The mass of Jim, M=100 kg
The mass of Tom, m=59 kg
(a) The kinetic energy of the system is given by,
[tex]\begin{gathered} E_a=\frac{1}{2}mv^2+\frac{1}{2}Mv^2 \\ =\frac{1}{2}v^2(m+M) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} E_a=\frac{1}{2}\times3.50^2(59.0+100) \\ =973.88\text{ J} \end{gathered}[/tex]Thus the kinetic energy of the system is 973.88 J
(b)
The total momentum of the system is given by,
[tex]\begin{gathered} p_b=mv+Mv \\ =(m+M)v \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} p_b=(59+100)\times3.50 \\ =556.5\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Thus the total momentum of the system is 556.5 kg m/s
(c)
Given that the velocity of Tom is -v
The total kinetic energy of the system is given by,
[tex]\begin{gathered} E_c=\frac{1}{2}Mv^2+\frac{1}{2}m(-v)^2 \\ =\frac{1}{2}(Mv^2+m(-v)^2) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} E_c=\frac{1}{2}(100\times3.50^2+59\times(-3.50)^2) \\ =973.88\text{ J} \end{gathered}[/tex]Thus the total kinetic energy of the system, in this case, is 973.88 J
(d)
The total momentum of the system is given by,
[tex]\begin{gathered} p_d=Mv+m(-v) \\ =v(M-m) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} p_d=3.50(100-59) \\ =143.5\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Thus the total momentum of the system, in this case, is 143.5 kg m/s