Answer:
The arrangement with the greatest resistance is the light bulb of option C. 4 W, 4.5 V
Explanation:
The equation for electric power is
power P = IV
also, I = V/R,
substituting into the equation, we have
[tex]P = \frac{V^{2} }{R}[/tex]
[tex]R = \frac{V^{2} }{P}[/tex]
a) [tex]R = \frac{1.5^{2} }{0.8}[/tex] = 2.8 Ω
b) [tex]R = \frac{3^{2} }{6}[/tex] = 1.5 Ω
c) [tex]R = \frac{4.5^{2} }{4}[/tex] 5.06 Ω
d) [tex]R = \frac{6^{2} }{8}[/tex] = 4.5 Ω
from the calculations, one can see that the lightbulb with te greates resistance is
C. 4 W, 4.5 V
Three solid, uniform, cylindrical flywheels, each of mass 65.0 kg and radius 1.47 m, rotate independently around a common axis through their centers. Two of the flywheels rotate in one direction at 8.94 rad/s, but the other one rotates in the opposite direction at 3.42 rad/s.
Required:
Calculate the magnitude of the net angular momentum of the system.
Answer:
the angular momentum is 1015.52 kg m²/s
Explanation:
given data
mass of each flywheel, m = 65 kg
radius of flywheel, r = 1.47 m
ω1 = 8.94 rad/s
ω2 = - 3.42 rad/s
to find out
magnitude of the net angular momentum
solution
we get here Moment of inertia that is express as
I = 0.5 m r² .................1
put here value and we get
I = 0.5 × 65 × 1.47 × 1.47
I = 70.23 kg m²
and
now we get here Angular momentum that is express as
L = I × ω ...........................2
and Net angular momentum will be
L = 2 × I x ω1 - I × ω2
put here value and we get
L = 2 × 70.23 × 8.94 - 70.23 × 3.42
L = 1015.52 kg m²/s
so
the angular momentum is 1015.52 kg m²/s
The magnitude of the net angular momentum of the system will be "1015.52 kg.m²/s".
MomentumAccording to the question,
Flywheel's mass, m = 65 kg
Flywheel's radius, r = 1.47 m
ω₁ = 8.94 rad/s
ω₂ = 3.42 rad/s
We know,
The moment of inertia (I),
= 0.5 m r²
By substituting the values,
= 0.5 × 65 × 1.47 × 1.47
= 70.23 kg.m²
hence, The angular momentum be:
→ L = I × ω or,
= 2 × I × ω₁ - l × ω₂
= 2 × 70.23 × 8.94 - 70.23 × 3.42
= 1015.52 kg.m²/s
Thus the above answer is correct.
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Two people play tug of war. The 100-kg person on the left pulls with 1,000 N, and the 70-kg person on the right pulls with 830 N. Assume that neither person releases their grip on the rope with either hand at any time, assume that the rope is always taut, and assume that the rope does not stretch. What is the magnitude of the tension in the rope in Newtons
Answer:
The tension on the rope is T = 900 N
Explanation:
From the question we are told that
The mass of the person on the left is [tex]m_l = 100 \ kg[/tex]
The force of the person on the left is [tex]F_l = 1000 \ N[/tex]
The mass of the person on the right is [tex]m_r = 70 \ kg[/tex]
The force of the person on the right is [tex]F_r = 830 \ N[/tex]
Generally the net force is mathematically represented as
[tex]F_{Net} = F_l - F_r[/tex]
substituting values
[tex]F_{Net} = 1000-830[/tex]
[tex]F_{Net} = 170 \ N[/tex]
Now the acceleration net acceleration of the rope is mathematically evaluated as
[tex]a = \frac{F_{net}}{m_I + m_r }[/tex]
substituting values
[tex]a = \frac{170}{100 + 70 }[/tex]
[tex]a = 1 \ m/s ^2[/tex]
The force [tex]m_i * a[/tex]) of the person on the left that caused the rope to accelerate by a is mathematically represented as
[tex]m_l * a = F_r -T[/tex]
Where T is the tension on the rope
substituting values
[tex]100 * 1 = 1000 - T[/tex]
=> T = 900 N
Find the pressure difference (in kPa) on an airplane wing if air flows over the upper surface with a speed of 125 m/s, and along the bottom surface with a speed of 109 m/s. [Express answer in TWO decimal places]
Answer:
P= 2414.9 Pa
Explanation:
given
density of air , p = 1.29 kg/m³
speed of air over the upper surface , v₁ = 125 m/s
speed of air over the lower surface , v₂ = 109 m/s
the pressure difference on an airplane wing , P = 0.5 × p × ( v₁² - v₂²)
P = 0.5 × 1.29 × ( 125² - 109²)
P= 0.645(3744)
P = 2414.9 Pa
the pressure difference on an airplane wing is 2414.9 Pa
If 2 balls had the same volume but ball a has twice as much mass as babil which one will have the greater density
What is Ohm's Law, and how does it work in real life.
Explanation:
Ohms law states that the electrical current present in a metallic conductor is directly proportional to the potential difference between the metallic conductor and inversely proportional to the resistance therefore if the voltage is increased resistance also increases provided that temperature and other physical properties remains constant V=IR
Water is boiled at 1 atm pressure in a 20-cm-internal diameter polished copper pan on an electric range. If it is observed that the water level in the pan drops by 8.00 cm in 15 minutes, determine the inner surface temperature of the pan.
Answer:
11.3298W
Explanation:
The rate of heat transfer is determined from the enthalpy of vaporization at the give pressure obtained and the mass flow rate. The mass flow rate is determined from the volume of the boiled water, the given time interval and the specific volume of the saturated liquid.
Given that
1atm as the atmospheric pressure
Internal diameter = 20cm = 0.2m
Time = 15mins = (15×60)secs
Latent heat of vaporization (hevap) = 2256.6
Q = mh(evap)
= m/∆t . hevap
= V/αliq∆t ×h(evap)
D^2π∆h/4αliq ∆t × hevap
= 0.2^2 ×π×0.8×2256.5/4×0.001043×15×60
=0.04×3.142×0.08×2256.6/2.00256
= 22.68876/2.00256
Q = 11.3298W
An experiment is set up to test the angular resolution of an optical device when red light (wavelength ????r ) shines on an aperture of diameter D . Which aperture diameter gives the best resolution? D=(1/2)????r D=????r D=2????r
Explanation:
As per Rayleigh criterion, the angular resolution is given as follows:
[tex]\theta=\frac{1.22 \lambda}{D}[/tex]
From this expression larger the size of aperture, smaller will be the value of angular resolution and hence, better will be the device i.e. precision for distinguishing two points at very high angular difference is higher.
which of the following has the lowest density? A. Water B. Air C. Mineral Water D. Salt Water
Air has lower density than water, mineral Water, or salt water. (B)
Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of 38.0° with the normal to the surface, while in the glass it makes an angle of 26.0° with the normal. What is the index of refraction of the glass?
Answer:
n_glass = 1.404
Explanation:
In order to calculate the index of refraction of the light you use the Snell's law, which is given by the following formula:
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex] (1)
n1: index of refraction of vacuum = 1.00
θ1: angle of the incident light respect to normal of the surface = 38.0°
n2: index of refraction of glass = ?
θ2: angle of the refracted light in the glass respect to normal = 26.0°
You solve the equation (1) for n2 and replace the values of all parameters:
[tex]n_2=n_1\frac{sin\theta_1}{sin\theta_2}=(1.00)\frac{sin(38.0\°)}{sin(26.0\°)}\\\\n_2=1.404[/tex]
The index of refraction of the glass is 1.404
A current carrying wire is oriented along the y axis It passes through a region 0.45 m long in which there is a magnetic field of 6.1 T in the z direction The wire experiences a force of 15.1 N in the x direction.1. What is the magnitude of the conventional current inthe wire?I = A2. What is the direction of the conventional current in thewire?-y+y
Answer:
The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.
Explanation:
- To find the direction of the conventional current in the wire you use the following formula:
[tex]\vec{F}=i\vec{l}\ X\ \vec{B}[/tex] (1)
i: current in the wire = ?
F: magnitude of the magnetic force on the wire = 15.1N
B: magnitude of the magnetic field = 6.1T
l: length of the wire that is affected by the magnetic field = 0.45m
The direction of the magnetic force is in the x direction (+^i) and the direction of the magnetic field is in the +z direction (+^k).
The direction of the current must be in the +y direction (+^j). In fact, you have:
^j X ^k = ^i
The current and the magnetic field are perpendicular between them, then, you solve for i in the equation (1):
[tex]F=ilBsin90\°\\\\i=\frac{F}{lB}=\frac{15.1N}{(0.45m)(6.1T)}=5.5A[/tex]
The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.
a certain volume of dry air at NTP is allowed to expand five times of it original volume under adiabatic condition.calculate the final pressure.(air=1.4)
Answer:
Final pressure 0.105atm
Explanation:
Let V1 represent the initial volume of dry air at NTP.
under adiabatic condition: no heat is lost or gained by the system. This does not implies that the constant temperature throughout the system , but rather that no heat gained or loss by the system.
Adiabatic expansion:
[tex]\frac{T_1}{T_2} =(\frac{V_1}{V_2} )^{\gamma -1}[/tex]
273/T2=(5V1/V1)^(1.4−1)
273/T2=5^0.4
Final temperature T2=143.41 K
Also
P1/P2=(V2/V1)^γ
1/P2=(5V1/V1)^1.4
Final pressure P2=0.105atm
What is surface tension??
Answer:
Surface tension is the tendency of liquid surfaces to shrink into the minimum surface area possible. Surface tension allows insects (e.g. water striders), usually denser than water, to float and slide on a water surface.
Explanation:
Answer:
It is the tension of the surface film of a liquid caused by the attraction of the particles in the surface layer by the bulk of the liquid, which tends to minimize surface area.
The ball tends to come back to the centerline of the flow when it is pushed by an external disturbance. Explain this phenomenon using the curvature of streamlines.
Answer is given below
Explanation:
given data
we will consider here
Ping-Pong ball weighs = 3.1 g
diameter = 4.2 cm
solution
Whenever the ball is pushed, the length of the airflow along the outer edge increases and it accelerates. According to Bernoulli's equation. As the speed increases, the pressure decreases, so the pressure at the outer end is reduced. As the pressure at the outer edge is low, the extra air jet pushes it back to the center line.
A particle with charge q is to be brought from far away to a point near an electric dipole. Net nonzero work is done if the final position of the particle is on:__________
A) any point on the line through the charges of the dipole, excluding the midpoint between the two charges.
B) any point on a line that is a perpendicular bisector to the line that separates the two charges.
C) a line that makes an angle of 30 ∘ with the dipole moment.
D) a line that makes an angle of 45 ∘with the dipole moment.
Answer:
Net nonzero work is done if the final position of the particle is on options A, C and D
Explanation:
non zero work is done if following will be the final position of the charges :
A) Any point on the line through the charges of the dipole , excluding the midpoint between the two charges.
C) A line that makes an angle 30° with the dipole moment.
D) A line that makes an angle 45° with the dipole moment.
If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will Group of answer choices
Answer:
P' = 4 P
Therefore, the power dissipated by the circuit will becomes four times of its initial value.
Explanation:
The power dissipation by an electrical circuit is given by the following formula:
Power Dissipation = (Voltage)(Current)
P = VI
but, from Ohm's Law, we know that:
Voltage = (Current)(Resistance)
V = IR
Substituting this in formula of power:
P = (IR)(I)
P = I²R ---------------- equation 1
Now, if we double the current , then the power dissipated by that circuit will be:
P' = I'²R
where,
I' = 2 I
Therefore,
P' = (2 I)²R
P' = 4 I²R
using equation 1
P' = 4 P
Therefore, the power dissipated by the circuit will becomes four times of its initial value.
When one person was talking in a small room, the sound intensity level was 60 dB everywhere within the room. Then, there were 14 people talking in similar manner simultaneously in the room, what was the resulting sound intensity level?
A. 64 dB
B. 60 dB
C. 69 dB
D. 79 dB
E. 71 dB
Answer:
E= 71dB
Explanation:
See attached file for step by step calculation
According to the model in which active galactic nuclei are powered by supermassive black holes, the high luminosity of an active galactic nucleus primarily consists of
Answer:
the high luminosity of an active galactic nucleus primarily consists of light emitted by hot gas in an accretion disk that swirls around the black hole
Find the potential energy associated with a 79-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level. Take the zero of potential energy at sea level.
Answer:
P = 1470980 J
Explanation:
We have,
Mass of the hiker is 79 kg
It is required to find the potential energy associated with a 79-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level.
It is given by :
[tex]P=mgh\\\\P=79\times 9.8\times 1900\\\\P=1470980\ J[/tex]
So, the potential energy of 1470980 J is associated with a hiker.
Four point charges have the same magnitude of 2.4×10^−12C and are fixed to the corners of a square that is 4.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.
Answer:
7.2N/C
Explanation:
Pls see attached file
Rope BCA passes through a pulley at point C and supports a crate at point A. Rope segment CD supports the pulley and is attached to an eye anchor embedded in a wall. Rope segment BC creates an angle of ϕ = 51.0 ∘ with the floor and rope segment CD creates an angle θ with the horizontal. If both ropes BCA and CD can support a maximum tensile force Tmax = 120 lb , what is the maximum weight Wmax of the crate that the system can support? What is the
Answer:
Wmax = 63.65 ≈ 64 lb
Explanation:
1) A net force of 75.5 N is applied horizontally to slide a 225 kg crate across the floor.
a. Compute the acceleration of the crate?
Answer:
The acceleration of the crate is [tex]0.3356\,\frac{m}{s^2}[/tex]
Explanation:
Recall the formula that relates force,mass and acceleration from newton's second law;
[tex]F=m\,a[/tex]
Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:
[tex]F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}[/tex]
A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation between the plates is increased, what will happen to the charge on the capacitor and the electric potential across it
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. [tex]Q[/tex], the amount of charge stored in this capacitor, will stay the same.
The formula [tex]\displaystyle Q = C\, V[/tex] relates the electric potential across a capacitor to:
[tex]Q[/tex], the charge stored in the capacitor, and[tex]C[/tex], the capacitance of this capacitor.While [tex]Q[/tex] stays the same, moving the two plates apart could affect the potential [tex]V[/tex] by changing the capacitance [tex]C[/tex] of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
[tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex],
where
[tex]\epsilon[/tex] is the permittivity of the material between the two plates.[tex]A[/tex] is the area of each of the two plates.[tex]d[/tex] is the distance between the two plates.Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of [tex]\epsilon[/tex]. Neither will that change the area of the two plates.
However, as [tex]d[/tex] (the distance between the two plates) increases, the value of [tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex] will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula [tex]\displaystyle Q = C\, V[/tex] can be rewritten as:
[tex]V = \displaystyle \frac{Q}{C}[/tex].
The value of [tex]Q[/tex] (charge stored in this capacitor) stays the same. As the value of [tex]C[/tex] becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come back down to the level of the crossbar. Assume the crossbar of a soccer goal is 2.44 m above the ground. (a) How fast was the ball originally moving when it was kicked. (b) How much longer would it take the ball to reach the ground?
Answer:
(a) vo = 24.98m/s
(b) t = 5.09 s
Explanation:
(a) In order to calculate the the initial speed of the ball, you use the following formula:
[tex]y=y_o+v_ot-\frac{1}{2}gt^2[/tex] (1)
y: vertical position of the ball = 2.44m
yo: initial vertical position = 0m
vo: initial speed of the ball = ?
g: gravitational acceleration = 9.8m/s²
t: time on which the ball is at 2.44m above the ground = 5.00s
You solve the equation (1) for vo and replace the values of the other parameters:
[tex]v_o=\frac{y-y_o+1/2gt^2}{t}[/tex]
[tex]v_o=\frac{2.44m-0.00m+1/2(9.8m/s^2)(5.00s)^2}{5.00s}\\\\v_o=24.98\frac{m}{s}[/tex]
The initial speed of the ball is 24.98m/s
(b) To find the time the ball takes to arrive to the ground you use the equation (1) for y = 0m (ground) and solve for t:
[tex]0=24.98t-\frac{1}{2}(9.8)t^2\\\\t=5.09s[/tex]
The time that the ball takes to arrive to the ground is 5.09s
We have that for the Question, it can be said that the speed of ball and How much longer would it take the ball to reach the ground is
u=25.13m/sX=0.095sec
From the question we are told
A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come back down to the level of the crossbar. Assume the crossbar of a soccer goal is 2.44 m above the ground.
(a) How fast was the ball originally moving when it was kicked.
(b) How much longer would it take the ball to reach the ground?
a)
Generally the Newton equation for the Motion is mathematically given as
[tex]S=ut+1/2at^2\\\\Therefore\\\\2.44=ut+1/2(9.8)(5)^2\\\\u=25.13m/s\\\\[/tex]
b)
Generally the Newton equation for the Motion is mathematically given as
[tex]S=ut+1/2at^2\\\\Therefore\\\\t=\frac{-24}{a}\\\\t=\frac{-2*25.013}{9.81}\\\\t=5.095sec\\\\[/tex]
Therefore
[tex]X=5.095-5[/tex]
X=0.095sec
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A charging bull elephant with a mass of 5500 kg comes directly toward you with a speed of 4.70 m/s . You toss a 0.160-kg rubber ball at the elephant with a speed of 7.50 m/s(a) When the ball bounces back toward you, what is its speed? (b) How do you account for the fact that the ball's kinetic energy has increased?
Answer:
v2 = - 16.899 m/s
velocity of ball increases so that the kinetic energy of the ball increases.
Explanation:
given data
mass of elephant, m1 = 5500 kg
mass of ball, m2 = 0.160 kg
initial velocity of elephant, u1 = - 4.70 m/s
initial velocity of ball, u2 = 7.50 m/s
solution
we consider here final velocity of ball = v2
so collision formula is express as for v2
[tex]v_{2}=\left ( \frac{2m_{1}}{m_{1}+m_{2}} \right )u_{1}+\left ( \frac{m_{2}-m_{1}}{m_{1}+m_{2}} \right )u_{2}[/tex] .................1
put here value and we get
[tex]v_{2}=\left ( \frac{2\times 5500}{5500+0.160} \right )(-4.70)+\left ( \frac{0.16-5500}{5500+0.160} \right )(7.50)[/tex]
solve it we get
v2 = - 16.899 m/s
here negative sign shows that the ball bounces back towards you
and
here we know the velocity of ball increases so that the kinetic energy of the ball increases.
and due to this effect, it will gain in energy is due to the energy from the elephant mass
a wave with a high amplitude______?
. . . is carrying more energy than a wave in the same medium with a lower amplitude.
The first Leyden jar was probably discovered by a German clerk named E. Georg von Kleist. Because von Kleist was not a scientist and did not keep good records, the credit for the discovery of the Leyden jar usually goes to physicist Pieter Musschenbroek from Leyden, Holland. Musschenbroek accidentally discovered the Leyden jar when he tried to charge a jar of water and shocked himself by touching the wire on the inside of the jar while holding the jar on the outside. He said that the shock was no ordinary shock and his body shook violently as though he had been hit by lightning. The energy from the jar that passed through his body was probably around 1 J, and his jar probably had a capacitance of about 1 nF.A) Estimate the charge that passed through Musschenbroek's body.
B) What was the potential difference between the inside and outside of the Leyden jar before Musschenbroek discharged it?
Answer:
a) q = 4.47 10⁻⁵ C
b) ΔV = 4.47 10⁴ V
Explanation:
A Leyden bottle works as a condenser that accumulates electrical charge, so we can use the formula of the energy stored in a capacitor
U = Q² / 2C
Q = √ (2UC)
let's reduce the magnitudes to the SI system
c = 1 nF = 1 10⁻⁹ F
let's calculate
q = √ (2 1 10⁻⁹-9)
q = 0.447 10⁻⁴ C
q = 4.47 10⁻⁵ C
b) for the potential difference we use
C = Q / ΔV
ΔV = Q / C
ΔV = 4.47 10⁻⁵ / 1 10⁻⁹
ΔV = 4.47 10⁴ V
5. (10 points) Which of the following statements is(are) correct: A. Resistivity purely depends on internal properties of the conductor; B. Resistance purely depends on internal properties of the conductor; C. Resistivity depends on the size and shape of the conductor; D. Resistance depends on the size and shape of the conductor; E. A and D; F. B and C.
Answer:
B and D
Explanation:
Because
R= resistivity xlenght/ Area
Where R= resistance
A ranger needs to capture a monkey hanging on a tree branch. The ranger aims his dart gun directly at the monkey and fires the tranquilizer dart. However, the monkey lets go of the branch at exactly the same time as the ranger fires the dart. Will the monkey get hit or will it avoid the dart?
Answer:
Yes the monkey will get hit and it will not avoid the dart.
Explanation:
Yes, the monkey will be hit anyway because the dart will follow a hyperbolic path and and will thus fall below the branches, so if the monkey jumps it will be hit.
No, the monkey will not avoid the dart because dart velocity doesn't matter. The speed of the bullet doesn’t even matter in this case because a faster bullet will hit the monkey at a higher height and while a slower bullet will simply hit the monkey closer to the ground.
How much work will it take to lift a 2-kg pair of hiking boots 2 meters off the
ground and onto a shelf in your closet?
O A. 2.45 J
OB. 4J
C. 39.2 J
D. 20 J
Answer:
Option C - 39.2 J
Explanation:
We are given that;
Mass; m = 2 kg.
Distance moved off the floor;d = 10 m.
Acceleration due to gravity;g = 9.8 m/s².
We want to find the work done.
Now, the Formula for work done is given by;
Work = Force × displacement.
In this case, it's force of gravity to lift up the boots, thus;
Formula for this force is;
Force = mass x acceleration due to gravity
Force = 2 × 9.8 = 19.2 N
∴ Work done = 19.6 × 2
Work done = 39.2 J.
Hence, the Work done to life the boot of 2 kg to a height of 2 m is 39.2 J.
Answer:39.2J
Explanation: I just answered this question and this was the correct answer. 4J is the wrong answer.
What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150-kg deer initially running at 18.0 m/s in the same direction
What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150-kg deer initially running at 18.0 m/s in the same direction? Assume the deer remains on the car.
Answer:28.29m/s
Explanation:
In this situation, linear momentum is conserved. And since the deer remains on the car after collision, the linear momentum is given as;
([tex]m_{C}[/tex] x [tex]u_{C}[/tex]) + ([tex]m_{D}[/tex] x [tex]u_{D}[/tex]) = ([tex]m_{C}[/tex] + [tex]m_{D}[/tex]) v -----------------(i)
Where;
[tex]m_{C}[/tex] = mass of car
[tex]u_{C}[/tex] = initial velocity of car before collision
[tex]m_{D}[/tex] = mass of deer
[tex]u_{D}[/tex] = initial velocity of the deer before collision
v = common velocity with which the car and the deer move after collision
From the question;
[tex]m_{C}[/tex] = 900kg
[tex]u_{C}[/tex] = +30.0m/s (direction of the motion of the car taken positive)
[tex]m_{D}[/tex] = 150kg
[tex]u_{D}[/tex] = +18.0m/s (relative to the direction of the car, the velocity of the deer is also positive )
Substitute these values into equation (i) as follows;
(900 x 30.0) + (150 x 18.0) = (900 + 150)v
27000 + 2700 = 1050v
29700 = 1050v
v = [tex]\frac{29700}{1050}[/tex]
v = 28.29m/s
Therefore, the velocity of the car after hitting the deer is 28.29m/s. This is also the velocity of the deer after being hit by the car.