V² = U² + 2aS
27.4² = 13.5² + 2*241a
750.76 = 182.25 + 482a
750.76 - 182.25 = 482a
568.51/482
a = 1.18m/s^2
A car stereo pulls 2.89 A from a car's battery. If the battery has a voltage of 12 V, how much power does it use?
In order to calculate the power, we can multiply the voltage and the current:
[tex]P=I\cdot U[/tex]So we have:
[tex]\begin{gathered} P=2.89\cdot12\\ \\ P=34.68\text{ W} \end{gathered}[/tex]Therefore the power is 34.68 W.
SORRY IF ITS IN THE WRONG SUBJECT. I NEED SOMEONE ANSWER ASAP. text to answer the question. Most games that are played with a standard deck of playing cards are called trick games. In such a game, each player will take a turn playing a card, and whoever plays the winning card takes them all. These cards make up a trick, which the winner puts face-side down in a stack before playing the first card for the next round. Would the information in the text be an effective source to help answer the research question "What are the most popular magic tricks? (1 point) O No, because magic tricks that use playing cards are unpopular. O No, because the source is about card games instead of magic tricks. OYes, because the source is written by an expert on the subject. OYes, because the source explains a type of trick.
Answer: the answer should be "No, because the source is about card games instead of magic tricks."
Explanation: sorry for answering 2 weeks later
Answer:
B
Explanation:
Two charges are separated by 6.74 cm. Object A has a charge of 5.0 μ C , while object B has a charge of 7.0 μ C . What is the force on Object A?
ANSWER
69.34 N
EXPLANATION
Given:
• The charge of object A, qA = 5 x 10⁻⁶ C
,• The charge of object B, qB = 7 x 10⁻⁶ C
,• The distance between the objects, r = 6.74 cm = 0.0674 m
Find:
• The force on Object A, FA
The magnitude of the force between two charged particles, q₁ and q₂, separated by a distance r is given by Coulomb's Law,
[tex]F=k\frac{q_1q_2}{r^2}[/tex]Where k is Coulomb's constant and has a value of approximately 9x10⁹ Nm²/C²,
[tex]F=9\times10^9\frac{Nm^2}{C^2}\cdot\frac{5\times10^{-6}C\cdot7\times10^{-6}C}{0.0674^2m^2}\approx69.34N[/tex]Hence, the force on object A is 69.34 N.
4. A 45.8 kg block is placed on an inclined plane that is 44.2 degrees from the horizontal. What isthe acceleration of the block? Ignore friction.Challenge Question: 5. The same block in problem #4 is placed on the same inclined plane from problem #4. However, now there is friction. If the coefficient of kinetic friction is 0.25, what is the Friction force?(I already answered question 4, i just need help with question 5)
Explanation
Step 1
free body diagram
Step 2
sum of the forces on x-axis
a)x
[tex]\begin{gathered} F=\text{mg}\cos \text{ }\theta=ma \\ \text{replace} \\ F=\text{ 45.8}\cdot9.81\cdot\cos \text{ 44.2=ma} \\ 322.10=45.8a \\ a=\frac{322.10}{45.8} \\ a=703\text{ }\frac{\text{m}}{s^2} \end{gathered}[/tex]
1) The net external force on a golf cart is 390 N north. If the cart has a total mass of 270 kg, what arethe magnitude and directions of its acceleration?
Given data
*The net external force on a golf cart is F = 390 N
*The cart has a total mass is m = 270 kg
The formula for the magnitude of the acceleration of the cart is given as
[tex]a=\frac{F}{m}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} a=\frac{390}{270} \\ =1.44m/s^2 \end{gathered}[/tex]Hence, the magnitude of the acceleration of the cart is 1.44 m/s^2. The direction of the acceleration is towards the north direction.
An object with a mass of 5kg is moving with a force of 25N. What is the object's acceleration?
According to Newton's second law, F=ma
Where F is the force acting on the object, m is the mass of the object, and a is the acceleration of the body due to the applied force.
It is given in the question,
m=5 kg
F=25 N
On substituting the known values in the above equation,
[tex]25=5\times a[/tex]On rearranging the above equation,
[tex]a=\frac{25}{5}=5m/s^2[/tex]Therefore the object's acceleration is 5 m/s²
A 1000-watt kettle is connected to a 220-volt power source. Calculate the resistance of the kettle
The power of kettle is given as,
[tex]P=\frac{V^2}{R}[/tex]Plug in the known values,
[tex]\begin{gathered} 1000\text{ W=}\frac{(220V)^2}{R} \\ R=\frac{(220V)^2}{1000\text{ W}}(\frac{1\text{ ohm}}{1V^2W^{-1}})_{} \\ =48.4\text{ ohm} \end{gathered}[/tex]Thus, the resistance of the kettle is 48.4 ohm.
Calculate the power of a 2057 Ω night light plugged into a 120 V source.
Lets write the data down:
R = 2057 Ω
V = 120 V
The power of the night light is given by:
[tex]P=\frac{V^2}{R}[/tex]Plug in the data in the formula:
[tex]\begin{gathered} P=\frac{120^2}{2057} \\ P=\frac{14,400}{2057} \\ P\approx7.0W \end{gathered}[/tex]The power of the night light is 7.0 W.
A/An _____ is described as a device that detects current differences and then opens the circuit preventing electrocution.circuit breakerfuseground fault interruptershort circuit
ANSWER
ground fault interrupter
EXPLANATION
A ground fault interrupter is an automatic switch that measures the difference of current between the "hot" and neutral wires. If this difference is greater than a fixed values (only a few milliamperes) the switchs opens and stops the circulation or current.
If a person is in contact with a wire in the circuit the current is diverted through the person's body to the ground. Therefore the ground fault interrupter will detect a difference of current. Hence this device is used to prevent electrocution.
A force acting on an object does no work if1. a machine is used to move the object.2. the force is greater than the force of friction.3. the force acts perpendicular to the direction of the object’s motion.4. the object accelerates.
The work done of any object is defined by
[tex]W=Fd\cos \theta[/tex]Here, F is the force acting on the object.
d is the displacement of the object due to the force applied.
[tex]\theta[/tex]is the angle between force and displacement.
For work done to become zero, the applied force can be zero, or displacement can be zero, or cosine angle is zero.
Cosine angle can be zero if force and displacement are perpendicular to each other.
Thus, if the force acts perpendicular to the direction of the object's motion, no work is done..
Find the direction and magnitude of the vectors (a) A = (25m)x + (-12m)y(b) B = (2.0m)x + (15m)y and(c) A + B
ANSWER
[tex]\begin{gathered} a)\theta=-25.6\degree;|A|=27.73m \\ b)\theta=82.4\degree;|B|=15.13m \\ c)\theta=6.3\degree^{};|A+B|=27.17m \end{gathered}[/tex]EXPLANATION
To find the direction of a two-dimensional vector, we apply the formula:
[tex]\theta=\tan ^{-1}(\frac{a_y}{a_x})[/tex]To find the magnitude of a two-dimensional vector, we apply the formula:
[tex]|a|=\sqrt[]{(a^{}_y)^2+(a_x)^2}[/tex]a) The direction of vector A is:
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{-12m}{25m}) \\ \theta=\tan ^{-1}(-0.48) \\ \theta=-25.6\degree \end{gathered}[/tex]The magnitude of vector A is:
[tex]\begin{gathered} |A|=\sqrt[]{(25m)^2+(-12m)^2} \\ |A|=\sqrt[]{625m^2+144m^2}=\sqrt[]{769m^2} \\ |A|=27.73m \end{gathered}[/tex]b) The direction of vector B is:
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{15m}{2m}) \\ \theta=\tan ^{-1}(7.5) \\ \theta=82.4\degree \end{gathered}[/tex]The magnitude of vector B is:
[tex]\begin{gathered} |B|=\sqrt[]{(2.0m)^2+(15m)^2} \\ |B|=\sqrt[]{4m^2+225m^2}=\sqrt[]{229m^2} \\ |B|=15.13m \end{gathered}[/tex]c) First, we have to find the sum of A and B:
[tex]\begin{gathered} A+B=(25m)x+(-12m)y+(2.0m)x+(15m)y \\ A+B=(25m+2.0m)x+(-12m+15m)y \\ A+B=(27m)x+(3m)y \end{gathered}[/tex]The direction of the vector (A + B) is:
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{3m}{27m}) \\ \theta=\tan ^{-1}(0.1111) \\ \theta=6.3\degree^{} \end{gathered}[/tex]The magnitude of the vector (A + B) is:
[tex]\begin{gathered} |A+B|=\sqrt[]{(27m)^2+(3m)^2} \\ |A+B|=\sqrt[]{729m^2+9m^2}=\sqrt[]{738m^2} \\ |A+B|=27.17m \end{gathered}[/tex]Hi I do know the answer to this it’s be answered for me twice already just looking for a bit of a deeper explanation physical and why and how?
4.1
The free body diagram is shown below
Newton's first law of motion states that a body at rest or in motion would continue to be at rest or in motion unles an external force acts on it causing it to start or stop moving.
Sum of forces in the horizontal direction, Fx = - Fr + F
Sum of forces in the vertical direction, Fy = N + FSinθ - mg
Since the motion is along the horizontal axis, There would not be an acceleration in the y axis. Thus, Fy = 0
Force = mgCosθ
4.3) From the information given,
mass, m = 50kg
Force =
Forcce = 50 x 9.81Cos20
4.4)
Fy = 0
N + FSinθ - mg = 0
N = mg - FSinθ
N = 50 * 9.81 - FSin20
N = 490.5 - FSin20
N = 490.5 - 0.342F
4.5)
Frictional force = coefficient of friction x normal force
From the information given,
coefficient of friction = 0.4
Frictional force = 0.4(490.5 - 0.342F)
A 200-turn solenoid is 20.0 cm long and carries a current of 3.25 A.1. Find the magnetic field inside the solenoid in [mT].2. Find the force in [µN] exerted on a 15.0x10-6 C charged particle moving at 1050 m/s through the interior of the solenoid, at an angle of 11.5° relative to the solenoid’s axis.
Given,
Number of turns, N=200
Length of teh solenoid, l=20 cm=0.2 m
Current, I=3.25 A.
Charge is
[tex]q=15\times10^{-6}C[/tex]The velocity is v=1050 m/s
Angle is
[tex]\theta=11.5^o[/tex]To find
a. Magnetic field inside the solenoid
b. The force
Explanation
a. The magnetic field is
[tex]B=\mu_onI[/tex]n is the number of turns per unit length.
Thus,
[tex]\begin{gathered} n=\frac{N}{l} \\ \Rightarrow n=\frac{200}{0.2}=1000 \end{gathered}[/tex]So,
[tex]B=4\pi\times10^{-7}\times1000\times3.25=4.08\times10^{-3}=4.08\text{ mT}[/tex]b. The magnetic force is given by:
[tex]\begin{gathered} F=\text{qvBsin}\theta \\ \Rightarrow F=15\times10^{-6}\times1050\times4.08\times10^{-3}\sin 11.5=1.28\times10^{-5}N \end{gathered}[/tex]Conclusion
a.The magnetic field is 4.08 mT
b.The magnetic force is
[tex]1.28\times10^{-5}N[/tex]what is the greatest mass of groceries that can be lifted safely with this bag given that the bag is raised with an acceleration of 1.80 m/s^2
We will have the following:
According to the image:
First, we remember that:
[tex]F=m\cdot a[/tex]Now, we will determine he maximum mass will be:
[tex]52.0N=m\cdot(1.8m/s^2)\Rightarrow m=\frac{^{}52.0N}{1.80m/s^2}[/tex][tex]\Rightarrow m=\frac{260}{9}kg\Rightarrow m\approx28.9kg[/tex]So, the maximum mass will be approximately 28.9 kg.
Which choice is a valid way to construct a motion diagram?1) Add together the average speeds of the various objects in motion.2) Take a series of photographs at equal time intervals of a moving object, perpendicular to the direction of motion; overlay the images to see how the position changes with time.3) Draw vectors to represent the speeds involved.4) Add vectors in a head-to-tail manner to determine the resultant vector.
A motion diagram displays the location of an object at various equally spaced time intervals in the same diagram. This allows us to visualize the motion of the object.
From the described procedures, the one which would be useful to make a motion diagram, is:
Option 2) Take a series of photographs at equal time intervals of a moving object, perpendicular to the direction of motion; overlay the images to see how the position changes with time.
Imagine an asteroid at rest in space that cracks into two pieces. The first piece moves at 1.4 times the speed of the second piece. Calculate the ratio of the first piece’s mass to the second piece’s mass
Since the total linear momentum of the system was 0 before the asteroid cracking into two pieces, then the magntude of the linear momentum of each piece must be the same after the cracking.
Then:
[tex]m_1v_1=m_2v_2[/tex]If the speed of the first piece is 1.4 times the speed of the second, then:
[tex]\begin{gathered} v_1=1.4v_2 \\ \Rightarrow m_1\times1.4v_2=m_2v_2 \\ \Rightarrow1.4m_1=m_2 \\ \Rightarrow m_1=\frac{m_2}{1.4} \\ \Rightarrow\frac{m_1}{m_2}=\frac{1}{1.4} \end{gathered}[/tex]Then, the ratio of the first piece's mass to the second piece's mass is 1:1.4
It takes 10 J of energy to move a 2 C charge from point A to point B. What is the potential difference between points A and B?Group of answer choices20 V0.2 V5 VNone of the above
Take into account that potential difference is defined as the quotient between the energy requierd to move a charge a from a point A to a point B. Use the following formula:
[tex]\Delta V=\frac{\Delta U}{q}[/tex]where ΔV is the potential difference, ΔU is the energy and q is the charge.
In this case, ΔU = 10J and q = 2C. By replacing these values into the previous formula and by simplifying you obtain:
[tex]\Delta V=\frac{10J}{2C}=5V[/tex]where 1J/C = 1V (for the units of potential difference)
Hence, the potential difference is 5V
What is the power rating of a heating coil with a resis-tance of 12 12 that draws a current of 20 A?
According to the problem, we have the following
[tex]\begin{gathered} R=12 \\ I=20 \end{gathered}[/tex]We have to use Ohm's law
[tex]V=I\cdot R[/tex]Let's replace the given information
[tex]V=20\cdot12=240[/tex]Hence, the power is 20 Volts.Please helpppppppppp - the diagram shows the trajectory of a ball that is thrown horizontally from a top of a building. The ball’s vertical and horizontal velocity vectors along with the resultant factors are also indicated if the ball takes three seconds to reach the ground how fast is it moving by the time it reached the ground
What is the pH of a pure WATER solution if the concentration of H+ in water is 10-7 ?
Given data:
Comcentration of H+ ion in water,
[tex]\lbrack H^+\rbrack=10^{-7}\text{ }\frac{mol}{l}[/tex]The pH value is given as,
[tex]pH=-\log _{10}\lbrack H^+\rbrack_{}[/tex]Substituting all known values,
[tex]pH=-\log _{10}\lbrack10^{-7}\rbrack[/tex]Since,
[tex]\log _{10}(a^b)=b\times\log _{10}(a)[/tex]Therefore,
[tex]\begin{gathered} pH=-7\times(-\log _{10}\lbrack10\rbrack) \\ =7\log _{10}\lbrack10\rbrack \end{gathered}[/tex]Since,
[tex]\log _{10}\lbrack10\rbrack=1[/tex]Therefore,
[tex]\begin{gathered} pH=7\times1 \\ =7 \end{gathered}[/tex]Therefore, pH of a pure water solution is 7.
Let f(x) = 5* Let g(x) = 54 - 7 = - g Which statement describes the graph of g(x)with respect to the graph of f(x)? O g(x)is translated 7 units right from f(x). O g(x)is translated 7 units down fromf (x). O g(x)is translated 7 units left fromf (x). O g(x)is translated 7 units up fromf (x).
Answer:
[tex]\text{Second Option}[/tex]Explanation: The two functions are:
[tex]\begin{gathered} f(x)=5^x \\ g(x)=5^x-7 \end{gathered}[/tex]The graphs of these two functions are as follows:
From this we can infer that the answer is:
A pendulum has a mass of 3kg and is lifted to a height of .3m. What is the maximum speed of the pendulum
Given data
*The given mass of the pendulum is m = 3 kg
*The given height is h = 0.3 m
The formula for the maximum speed of the pendulum is given as
[tex]v_{\max }=\sqrt[]{2gh}[/tex]*Here g is the acceleration due to the gravity
Substitute the values in the above expression as
[tex]\begin{gathered} v_{\max }=\sqrt[]{2\times9.8\times0.3} \\ =2.42\text{ m/s} \end{gathered}[/tex]Hence, the maximum speed of the pendulum is 2.42 m/s
The vertices of two rectangles are A(−5,−1),B(−1,−1),C(−1,−4),D(−5,−4) and W(1,6),X(7,6),Y(7,−2),Z(1,−2). Compare the perimeters and the areas of the rectangles. Are the rectangles similar? Explain.Perimeter of ABCD: , Area of ABCD: Perimeter of WXYZ: , Area of WXYZ:
In order to compare the perimeters and areas, let's first find two adjacent sides of each rectangle.
From ABCD, let's calculate AB and BC:
A and B have the same y-coordinate, so the length is the difference in x-coordinate:
AB = -1 - (-5) = -1 + 5 = 4
B and C have the same x-coordinate, so the length is the difference in y-coordinate:
AB = -1 - (-4) = -1 + 4 = 3
Therefore the perimeter and area are:
[tex]\begin{gathered} P=4+3+4+3=14 \\ A=4\cdot3=12 \end{gathered}[/tex]Now, for rectangle WXYZ, let's use WX and XY:
W and X have the same y-coordinate, so the length is the difference in x-coordinate:
WX = 7 - 1 = 6
X and Y have the same x-coordinate, so the length is the difference in y-coordinate:
XY = 6 - (-2) = 6 + 2 = 8
So the perimeter and area are:
[tex]\begin{gathered} P=6+8+6+8=28 \\ A=6\cdot8=48 \end{gathered}[/tex]In order to check if the rectangles are similar, let's check the following relation:
[tex](\frac{P_1}{P_2})^2=\frac{A_1}{A_2}[/tex]So we have:
[tex]\begin{gathered} (\frac{14}{28})^2=\frac{12}{48} \\ (\frac{1}{2})^2=\frac{1}{4} \\ \frac{1}{4}=\frac{1}{4}\text{ (true)} \end{gathered}[/tex]Since the relation is true, so the rectangles are similar.
True or false. Iron fillings can indicate the presence of a magnetic field
Iron Filings are small shavings of ferromagnetic material.
Ferromagnetic material aligns with a magnetic field.
So, Iron fillings can indicate the presence of a magnetic field
True
a student throws a rock straight down words from a bridge into a river below if initial speed of the rock is 10.0 Ms and it takes 2.1 s to reach the river how high is the bridge
Since this is a free fall and all the motion is going down, let's make down the positive direction (that way we won't use minus signs). A free fall is an uniform accelerated motion, this means that we can use the formula:
[tex]y-y_0=v_0t+\frac{1}{2}at^2[/tex]In this case we know the initial velocity is 10 m/s, the time is 2.1 and the acceleration of gravity is 9.8 m/s^2. Plugging this values we have that:
[tex]\begin{gathered} y-y_0=10(2.1)+\frac{1}{2}(9.8)(2.1)^2 \\ y-y_0=42.609 \end{gathered}[/tex]This means that the rock fell 42.609 meters. Therefore, the bridge is 42.609 m tall.
Your friend is flying a hot air balloon and you are recording them from below. When you start filming, the hot air balloon is at an angle of 63 degrees above you while staying 100 feet away from where it launched. After 5 seconds, the angle has increased to 67.9 degrees. How fast is your friend rising in feet per seconds and miles per hour.
Answer:
10 ft/s and 6.818 mi/hr
Explanation:
We can represent the situation with the following diagram:
When the angle is 63 degrees, the value of h can be calculated using a trigonometric function as:
[tex]\tan 63=\frac{h_i}{100}[/tex]Because h is the opposite side and 100 ft is the adjacent side. Solving for h, we get:
[tex]\begin{gathered} h_i=100\times\tan 63 \\ h_i=196.26\text{ ft} \end{gathered}[/tex]In the same way, when the angle is 67.9 degrees, we can calculate the height as follows:
[tex]\begin{gathered} \tan 67.9=\frac{h_f}{100} \\ h_f=100\times\tan 67.9 \\ h_f=246.27ft \end{gathered}[/tex]Now, we can calculate the speed in ft per second as follows:
[tex]s=\frac{h_f-h_i}{t}=\frac{246.27ft-196.24ft}{5s}=10\text{ ft/s}[/tex]Finally, 1 mile = 5280 ft and 1 hour = 3600 seconds, so we can convert to miles per hour as:
[tex]10\text{ ft/s }\times\frac{1\text{ mile}}{5280\text{ ft}}\times\frac{3600}{1\text{ hour}}=6.818\text{ mi/hr}[/tex]Therefore, the answers are:
10 ft/s and 6.818 mi/hr
Two objects a distance apart are experiencing 40 N of force. How much force wouldthere be if you DOUBLED the distance between them?
We know that the masses experience a 40 N force between them, which is a gravitational force due to their mass. So, if we double their distance between them, the force will decrease due to Newton's Gravitational Law.
[tex]F=G\cdot\frac{m_1\cdot m_2}{d^2_{12}}[/tex]Let's use 2d.
[tex]F=G\cdot\frac{m_1\cdot m_2}{(2d)^2}=G\cdot\frac{m_1\cdot m_2}{4d^2}[/tex]As you can notice, the force would be divide by 4, so let's do that.
[tex]F=\frac{40N}{4}=10N[/tex]Therefore, if we double their distance, their force would be 10 N.A park ranger driving on a back country road suddenly sees a deer in his headlights 20m ahead. The ranger, who is driving at 11.4 m/s, immediately applies the brakes andslows down with an acceleration of 3.80 m/s2. How much distance is required for theranger's vehicle to come to rest? Only enter the number, not the units,
Given data
*The speed of the ranger who is driving at 11.4 m/s
*The given acceleration is a = -3.80 m/s^2
The formula for the distance covered by the ranger's vehicle is given by the kinematic equation of motion as
[tex]\Delta x=\frac{v^2-v^2_0}{2a}[/tex]*Here v = 0 m/s is the initial speed of the ranger's vehicle
Substitute the values in the above expression as
[tex]\begin{gathered} \Delta x=\frac{0^2-(11.4)^2}{2\times(3.80)} \\ =17.1\text{ m} \end{gathered}[/tex]The distance is required for the ranger's vehicle to come to rest is calculated as
[tex]\begin{gathered} D=20-17.1 \\ =2.9\text{ m} \end{gathered}[/tex]The stopping time is calculated as
[tex]v=v_0+at[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} 0=11.4+(-3.80)(t) \\ t=3.00\text{ s} \end{gathered}[/tex]Use g = 10 m/s/s for this problem. A force is applied to a block through tension in a rope, at an angle to the horizontal as shown. The block is being pulled across a rough surface. The mass of the block is 71 kg. The tension in the rope is 1197 N. The angle from horizontal is 15 degrees. The coefficient of friction μ is 0.3.
Answer:
14.6 m/s²
Explanation:
First, we will make the free body diagram
Since the net vertical force is equal to 0 because the block is not moving up, we can write the following equation
[tex]\begin{gathered} F_{nety}=F_n+F_{Ty}-mg=0 \\ F_n+F_T\sin15-mg=0 \end{gathered}[/tex]Then, we can solve for the normal force and replace Ft = 1197 N, m = 71 kg and g = 10 m/s²
[tex]\begin{gathered} F_n=mg-F_T\sin15 \\ F_n=(71\text{ kg\rparen\lparen10 m/s}^2)-(1197N)(0.26) \\ F_n=400.2\text{ N} \end{gathered}[/tex]Now, we can write the following equation for the net horizontal force
[tex]\begin{gathered} F_{net}=F_{Tx}-F_f=ma \\ F_T\cos15-\mu F_n=ma \end{gathered}[/tex]Where μ is the coefficient of friction and a is the acceleration of the block. Solving for a and replacing Ft = 1197 N, m = 71 kg, Fn = 400.2 N and μ = 0.3, we get
[tex]\begin{gathered} a=\frac{F_T\cos15-\mu F_n}{m} \\ \\ a=\frac{(1197\text{ N\rparen}\cos15-0.3(400.2N)}{71\text{ kg}} \\ \\ a=14.6\text{ m/s}^2 \end{gathered}[/tex]Therefore, the acceleration of the block is 14.6 m/s²
A falcon with a mass [tex]m_{1}= 1.21kg[/tex] is diving at a speed of [tex]v_{1}=25.8m/s[/tex] at an angle of[tex]tetha=39.7*[/tex] below horizontal. A pigeon whose mass is [tex]m_{2}=0.62kg[/tex] is flying the the positive x direction at a speed of [tex]v_{2}=8.6m/s[/tex]. The falcon catches the pigeon, and they move as one. Neglect gravity and air resistance.
(a) write an expression for the x component of the final velocity
(b) write an expression for the y component of the final velocity
(c) what is the magnitude, in meters per seconds, of the final velocity?
(e) what is the angel, in degrees below the horizontal, that the final velocity makes with the x axis?
a ) The expression for the x component of the final velocity = 16 m / s
b ) The expression for the y component of the final velocity = 10.92 m / s
c ) The magnitude of final velocity = 19.37 m / s
m1 = 1.21 kg
m2 = 0.62 kg
Along x-axis,
u1 = u cos θ
u1 = 25.8 cos 39.7°
u1 = 19.87 m / s
u2 = 8.6 m / s
According to law of conservation of momentum
m1u1 + m2u2 = ( m1 + m2 ) vx
( 1.21 * 19.87 ) + ( 0.62 * 8.6 ) = ( 1.21 + 0.62 ) vx
vx = ( 24 + 5.3 ) / 1.83
vx = 16 m / s
Along y-axis,
u1 = u sin θ
u1 = 25.8 sin 39.7°
u1 = 16.5 m / s
u2 = 0
m1u1 = ( m1 + m2 ) vy
( 1.21 * 16.5 ) = ( 1.21 + 0.62 ) vy
vy = 19.98 / 1.83
vy = 10.92 m / s
v = √ vx² + vy²
v = √ 16² + 10.92²
v = √ 256 + 119.25
v = √ 375.25
v = 19.37 m / s
tan θ = vy / vx
tan θ = 10.92 / 16
tan θ = 0.68
θ = 34.22°
Therefore,
a ) The expression for the x component of the final velocity = 16 m / s
b ) The expression for the y component of the final velocity = 10.92 m / s
c ) The magnitude of final velocity = 19.37 m / s
d ) The angle below the horizontal, that the final velocity makes with the x axis = 34.22°
To know more about law of conservation of momentum
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