Answer:
a. Capacitance
b. Charge on the plates
e. Energy stored in the capacitor
Explanation:
Let A be the area of the capacitor plate
The capacitance of a capacitor is given as;
[tex]C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\[/tex]
where;
V is the potential difference between the plates
The charge on the plates is given as;
[tex]Q = \frac{V\epsilon _0 A}{d}[/tex]
The energy stored in the capacitor is given as;
[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2[/tex]
Thus, the physical variables listed that will change include;
a. Capacitance
b. Charge on the plates
e. Energy stored in the capacitor
As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock. As the waves recede, they carry the sediment away. In this scenario, which process represents weathering, and which process represents erosion?
Answer:
WEATHERING is represented by the scenario (As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock).
Erosion is represented by the scenario (As the waves recede, they carry the sediment away).
Explanation:
A wave is a disturbance which travels through a medium and transfers energy from one point to another. When wind blows over a water body like the ocean, ocean waves are formed. As the generated energy from the wind is transported through the water by the waves, the can hit against rocks on the shores leading to its break down with time. WEATHERING occurs when tiny bit of sediments from rocks are loosened due to the impact of ocean waves.
Erosion can be described as the wearing away of the earth's surface due to the impact of wind, rainfall ( water) or waves. There are different types of erosion which is classified according it's cause of formation.
Wave erosion occurs when sediments such as sand, shell and other materials are carried to the shoreline by ocean waves. This erodes the shore over time as the sediments act like sandpapers.
In addition to the sources of error mentioned above, since the caliper jaws squeezed the flexible, rubber surface of the ball slightly, the measured diameters were slightly __________________ compared to what a non-contact method of measuring would provide. This represents an ________________ ________________ error in ________________. This additional source of error ________________ ________________ significant. When the caliper jaws closed, the zero mark on the sliding Vernier scale, ________________ line up with the zero mark on the measuring scale. This means the caliper ________________ calibrated correctly.
Answer:
MINORS, SYSTEMATIC, STATISTICAL, BOTOM LINE, ZERO MATCHES
Explanation:
In general the sources of error or uncertainty can be classified
* Statistics. Which are those that describe the statistical formulas, for example: average, standard deviation, absolute error, etc.
* Systematic. That they occur due to an inappropriate measurement or to an interaction between the system and the instrument that cannot be quantified, in general this error shifts the measurements towards an explicit side
* Random, so errors that sometimes occur in the measurement and sometimes not, for example temperature changes during the medical process
In this case, you are asked to complete the sentences with the appropriate word
the measured diameters were slightly ___ MINORS________ compared to what a non-contact method of measuring would provide
. This represents an ____SYSTEMATIC_______ error in ________________.
This additional source of error ________STATISTICAL________ significant.
When the caliper jaws closed, the zero mark on the sliding Vernier scale, BOTOM LINE AND THE __________ line up with the zero mark on the measuring scale.
This means the caliper ___ZERO MATCHES_____________ calibrated correctly.
which experimental result led to a revision of Thomas's plum pudding model of the atom?
A. electrons were found to have higher energy the farther they are from the nucleus
B. the beam in a cathode ray tube was moved by an electric force
C. A few alpha particles bounced off a thin sheet of gold foil
D. most alpha particles passed straight through a thin sheet of gold foil
Answer: C. A few alpha particles bounced off a thin sheet of gold foil.
what changes when unbalanced forces are applied to a moving bike along the direction its moving?
A. inertia
B.friction
C. mass
D.velocity
Answer:
D.velocity
Explanation:
When unbalanced forces are applied to a moving bike along the direction its moving, the velocity of the bike changes.
Unbalanced forces causes a body to accelerate. Acceleration is the rate of change of velocity with time.
When unbalanced forces acts on a body, the velocity changes.
An unbalanced force is a force whose resultant is not zero. Such a force causes a body to change motion and then acceleration. For a body to accelerate, the velocity must change per unit of time.An RC car is carrying a tiny slingshot with a spring constant of 85 N/m at 0.2 m off the ground at 5.6 m/s. The sling shot is pulled back 3.5 cm from a relaxed state and shoots a 25 g steel pellet in the same direction the car is moving. What is the velocity of the steel pellet relative to the ground as it leaves the sling shot
Answer:
The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.
Explanation:
Let suppose that RC car-slingshot-steelpellet is a conservative system, that is, that non-conservative forces (i.e. friction, air viscosity) can be neglected. The velocity of the steel pellet can be found by means of the Principle of Energy Conservation and under the consideration that change in gravitational potential energy is negligible and that the RC car travels at constant velocity:
[tex]\frac{1}{2}\cdot (m_{C}+m_{P})\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{C}\cdot v_{o}^{2} + \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]
[tex]\frac{1}{2}\cdot m_{P}\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]
[tex]m_{P}\cdot v_{o}^{2} + k\cdot x^{2} = m_{P}\cdot v^{2}[/tex]
[tex]v^{2} = v_{o}^{2} + \frac{k}{m_{P}}\cdot x^{2}[/tex]
[tex]v = \sqrt{v_{o}^{2}+\frac{k}{m_{P}}\cdot x^{2} }[/tex] (1)
Where:
[tex]v_{o}[/tex] - Initial velocity of the steel pellet, measured in meters per second.
[tex]v[/tex] - Final velocity of the steel pellet, measured in meters per second.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]m_{P}[/tex] - Mass of the steel pellet, measured in kilograms.
[tex]m_{C}[/tex] - Mass of the RC car, measured in kilograms.
[tex]x[/tex] - Initial deformation of the spring, measured in meters.
If we know that [tex]v_{o} = 5.6\,\frac{m}{s}[/tex], [tex]k = 85\,\frac{N}{m}[/tex], [tex]m_{P} = 0.025\,kg[/tex] and [tex]x = 0.035\,m[/tex], then the velocity of the steel pellet relative to the ground when it leaves the sling shot is:
[tex]v = \sqrt{\left(5.6\,\frac{m}{s} \right)^{2}+\frac{\left(85\,\frac{N}{m} \right)\cdot (0.035\,m)^{2}}{0.025\,kg} }[/tex]
[tex]v \approx 5.960\,\frac{m}{s}[/tex]
The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.
What is the displacement of an object during a specific unit of time.
Answer:
velocity
Explanation:
the displacement of an object during a specific unit of time.
In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 2820 J of work is done on the gas.
Required:
a. How much heat flows into or out of the gas?
b. What is the change in internal energy of the gas?
c. Does its temperature rise or fall?
Answer:
[tex]Q=0[/tex][tex]U=2820[/tex]Energy increasesExplanation:
From the question we are told that
Work done [tex]W=2820[/tex]
a)Generally the heat flow for an adiabatic process is 0 (zero)
[tex]Q= U + W =>0[/tex]
[tex]Q=0[/tex]
b)Generally Change in internal energy of gas is mathematically given by
Since [tex]W=-2820J[/tex]
Therefore
[tex]U=2820[/tex]
Giving
[tex]Q= 2820 -2820[/tex]
[tex]Q=O[/tex]
c)With increases in internal energy brings increase in temperature
Therefore
Energy increases
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.30 s for the boat to travel from its highest point to its lowest, a total distance of 0.660 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart. How fast are the waves traveling
Answer:
v = 1.2 m/s
Explanation:
The wavelength of the waves is given as the horizontal distance between the crests:
λ = wavelength = 5.5 m
Now, the time period is given as the time taken by boat to move from the highest point again to the highest point. So it will be equal to twice the time taken by the boat to travel from highest to the lowest point:
T = Time Period = 2(2.3 s) = 4.6 s
Now, the speed of the wave is given as:
[tex]v = f\lambda[/tex]
where,
v= speed of wave = ?
f = frequency of wave = [tex]\frac{1}{T} = \frac{1}{4.6\ s} = 0.217\ Hz[/tex]
Therefore,
[tex]v = (0.217\ Hz)(5.5\ m)\\[/tex]
v = 1.2 m/s
a car accelerates from 2 m/s to 28m/s at a constant rate of 3 m/s^2. How far does it travel while accelerating?
Answer:
Distance, S = 130m
Explanation:
Given the following data;
Initial velocity = 2m/s
Final velocity = 28m/s
Acceleration = 3m/s²
To find the distance, we would use the third equation of motion.
V² = U² + 2aS
Substituting into the equation, we have;
28² = 2² + 2*3*S
784 = 4 + 6S
6S = 784 - 4
6S = 780
S = 780/6
Distance, S = 130m
A 5.7 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4.5 + 13.7 x − 1.5 x 2 , where Fx is in Newtons and x is in meters. Find the work done by this force on the particle as the particle moves from x = 0 m to x = 1.9 m. Answer in units of J.
Answer:
The work done by the force on the particle is 29.85 J.
Explanation:
The work is given by:
[tex] W = ^{x_{2}}_{x_{1}}\int F_{x} dx [/tex]
Where:
x₁: is the lower limit = 0 m
x₂: is the upper limit = 1.9 m
Fₓ: is the force in the horizontal direction = (4.5 + 13.7x - 1.5x²)N
[tex]W = ^{1.9}_{0}\int (4.5 + 13.7x - 1.5x^{2}) dx[/tex]
[tex] W = 4.5x|^{1.9}_{0} + \frac{13.7}{2}x^{2}|^{1.9}_{0} - \frac{1.5}{3}x^{3}|^{1.9}_{0} [/tex]
[tex] W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3} [/tex]
[tex] W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3} [/tex]
[tex]W = 29.85 J[/tex]
Therefore, the work done by the force on the particle is 29.85 J.
I hope it helps you!
"45 meters north" is an example of
Answer:
Displacement
Explanation:
The quantity 45m north is a typical example of displacement.
Displacement is the distance traveled by a body in a specific direction. Displacement is a vector quantity with both magnitude and direction.
When we are specifying the displacement of a body, the direction must be indicated accurately. Therefore, the quantity given is displacementTwo objects travel the same distance. The one that is moving faster will:
Take more time to go the distance
Take less time to go the same distance
Take the same time as the slower object
None of the above
Answer: take less time to go the same distance
Explanation:
Because if it is going faster let’s say mph 60 mph is 60 miles per hour if you are going 40 miles per hour it will take you longer to get to your destination.
A constant force FA is applied to an object of mass M, initially at rest. The object moves in the horizontal x-direction, and the force is applied in the same direction. After the force has been applied, the object has a speed of vf. Which mathematical routines can be used to determine the time in which the force is applied to the object of mas
Answer:
t = [tex]\frac{ v \ F}{ m}[/tex]
Explanation:
The question is a bit strange, for this exercise we must use the mathematical relationship of Newton's second law to find the acceleration of the body
F = m a
a = F / m (1)
with this acceleration the mathematical relations of kinematics of accelerated motion must be used
v = v₀ + a t
with the body starting from rest its initial velocity is zero
v = a t
t = v / a (2)
if we substitute the equation 1 in 2
t = [tex]\frac{ v \ F}{ m}[/tex]
this is the final mathematical expression that allows to find the time based on the data of the problem
a block of mas \( m \) = 4.8 kg slides head on into a spring of spring constant \( k \) = 430 N/m. When the block stops, it has compressed the spring by 5.8 cm. The coefficient of kinetic friction between block and floor is 0.28. \( (g =9.8m/s^2) \)
Answer:
See explanation below
Explanation:
The question is incomplete. The missing part of this question is the following:
"While the block is in contact with the spring and being brought to rest, what are (a)the work done by the spring force and (b) the increase in thermal energy of the blockfloor system? (c) What is the blocks speed just as it reaches the spring?"
According to this we need to calculate three values: Work, Thermal Energy and Speed of the block when it reaches the spring.
Let's do this by parts.
a) Work done by the spring:
In this case, we need to apply the following expression:
W = -1/2 kx² (1)
We know that k = 430 N/m, and x is the distance of compressed spring which is 5.8 cm (or 0.058 m). Replacing that into the expression:
W = -1/2 * 430 * (0.058)²
W = -0.7233 Jb) Increase in thermal energy
In this case we need to use the following expression:
ΔEt = Fk * x (2)
And Fk is the force of the kinetic energy which is:
Fk = μk * N (3)
Where μk is the coeffient of kinetic friction
N is the normal force which is the same as the weight, so:
N = mg (4)
Let's calculate first the Normal force (4), then Fk (3) and finally the chance in the thermal energy (2):
N = 4.8 * 9.8 = 47.04 N
Fk = 0.28 * 47.04 = 13.1712 N
Finally the Thermal energy:
ΔEt = 13.1712 * 0.058
ΔEt = 0.7639 Jc) Block's speed reaching the spring
As the block is just reaching the speed, the initial Work is 0. And the following expression will help us to get the speed:
V = √2Ki/m (5)
And Ki, which is the initial kinetic energy can be calculated with:
Ki = ΔU + ΔEt (6)
And ΔU is the same value of work calculated in part (a) but instead of being negative, it will be positive here. So replacing the data first in (6) and then in (5), we can calculate the speed:
Ki = 0.7233 + 0.7639 = 1.4872 J
Finally the speed:
V = √(2 * 1.4872) / 4.8
V = 0.7872 m/sHope this helps
car is moving at 40 m/s. At 10 meters the driver spots a deer on the road and instantly steps on the brakes. If the car is 400 kg how much force must the breaks exert to stop the car in time?
Answer:
32000 N
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Mass (m) of car = 400 Kg
Force (F) =?
Next, we shall determine the acceleration of the the car. This can be obtained as follow:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Acceleration (a) =?
v² = u² + 2as
0² = 40² + (2 × a × 10)
0 = 1600 + 20a
Collect like terms
0 – 1600 = 20a
–1600 = 20a
Divide both side by –1600
a = –1600 / 20
a = –80 m/s²
The negative sign indicate that the car is decelerating i.e coming to rest.
Finally, we shall determine the force needed to stop the car. This can be obtained as follow:
Mass (m) of car = 400 Kg
Acceleration (a) = –80 m/s²
Force (F) =?
F = ma
F = 400 × –80
F = – 32000 N
NOTE: The negative sign indicate that the force is in opposite direction to the motion of the car.
What is the difference between a wave and energy?
Answer:
The higher the amplitude, the higher the energy. To summarise, waves carry energy. The amount of energy they carry is related to their frequency and their amplitude. The higher the frequency, the more energy, and the higher the amplitude, the more energy
Explanation:
The main difference between a wave and energy is: wave is oscillation of energy whereas energy is ability of doing work.
What is wave?A wave is an energetic disturbance in a medium that doesn't include any net particle motion. Elastic deformation, a change in pressure, an electric or magnetic intensity, an electric potential, or a change in temperature are a few examples.
What is energy?The capacity to do work is energy. Energy can only be changed from one form to another; it cannot be created or destroyed. Energy is measured in Joules, the same unit used to measure work. There are several sorts of energy since it is present in many different things.
There are two types of energy: kinetic and potential. Kinetic energy is the energy that is in motion, whereas potential energy is the energy that is stored in an object and is determined by the amount of work that is required.
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A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 8.11 m below its highest point?
Answer:
a) the kinetic energy of the ball at its highest point is 69.58 J
b) its speed when it is 8.11 m below its highest point is 55.97 m/s
Explanation:
Given that;
mass of golf ball m = 46.8 g = 0.0468 kg
initial speed of the ball v₁ = 58.8 m/s
height h = 24.7 m
acceleration due to gravity = 9.8 m/s²
the kinetic energy of the ball at its highest point = ?
from the conservation of energy;
Kinetic energy at the highest point will be;
K.Ei + P.Ei = KEf + PEf
now the Initial potential energy of the ball P.Ei = 0 J
so
1/2mv² + 0 J = KEf + mgh
K.Ef = 1/2mv² - mgh
we substitute
K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]
K.Ef = 80.904 - 11.3284
K.Ef = 69.58 J
Therefore, the kinetic energy of the ball at its highest point is 69.58 J
b) when the ball is 8.11 m below the highest point, speed = ?
so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m
so our velocity will be v₂
also using the principle of energy conservation;
K.Ei + P.Ei = KEh + PEh
1/2mv² + 0 J = 1/2mv₂² + mgh'
1/2mv₂² = 1/2mv² - mgh'
multiply through by 2/m
v₂² = v² - 2gh'
v₂ = √( v² - 2gh' )
we substitute
v₂ = √( (58.8)² - 2×9.8×16.59 )
v₂ = √( 3457.44 - 325.164 )
v₂ = √( 3132.276 )
v₂ = 55.97 m/s
Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s
CAN YOU GUYS PLEASE ANSWER THIS QUICKLY THIS IS DUE IN AN HOUR AND IM GETTING WORRIED
I'm pretty sure I the third option C.
Explanation:
sorry if I'm wrong
A chocolate chip cookie is an example of a (2 points) a homogeneous mixture b heterogeneous mixture c suspension d colloid
Answer:
I think it is heterogeneous mixture. have a good day
Answer:
heterogeneous mixture
Explanation:
i took the test
Does changing the height of point C affect the speed of the coaster car at point D?
Without friction, NO.
The speed at D depends only on the difference in height between A and D. Whatever happens between them doesn't matter.
The speed of the coaster car at point D will be affected if the height of point C is changed.
Potencial Energy:
It is the enrgy in a body due to the position of differnt part of the object or system.
As we increase the the hight of the car the potetial enrgy increase, the gravitational acceleration on car will be more due to the high of the point C.
Therefore, the speed of the coaster car at point D will be affected if the height of point C is changed.
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What is displacement?
a. The distance an object travels.
b. The distance between the starting point and the ending point of an object's
journey.
C. The amount of time it takes an object to travel to a destination.
d. The path in which an object travels.
Answer:
displacement is the distance between the starting point and the ending point of an object's journey
Write the properties of Non Metals and the families containig non Metals.
Non-Malleable and Ductile: Non-metals are very brittle, and cannot be rolled into wires or pounded into sheets. Conduction: They are poor conductors of heat and electricity. Luster: These have no metallic luster and do not reflect light.
Group 15, the nitrogen family, contains two nonmetals: nitrogen and phosphorus. These non-metals usually gain or share three electrons when reacting with atoms of other elements. Group 16, the oxygen family, contains three nonmetals: oxygen, sulfur, and selenium.
Elements: Nitrogen; Oxygen; Phosphorus; Selenium...
A stretched string has a mass per unit length of 5.00 g/cm and a tension of 10.0 N. A sinusoidal wave on this string has an amplitude of 0.12 mm and a frequency of 100 Hz and is traveling in the negative direction of an x axis. If the wave equation is of the form y(x, t) = ym sin(kx ± ωt), what are (a) ym, (b) k, (c) ω, and (d) the correct choice of sign in front of ω?
Answer:
0.12 mm ; 140.50 rad/m ; 628.32 rad/sec ; +
Explanation:
Given the wave equation of the form :
y(x, t) = ym sin(kx ± ωt)
Mas per unit length (u) = 5 g/cm = (5÷1000)kg / 0.01m) = 0.005kg/0.01m = 0.5kg/m
Tension, T = 10 N
Amplitude, A = 0.12 mm
Frequency, F = 100 Hz
Comparing with the general wave equation :
y = Asin(kx ± ωt)
A = amplitude = ym = 0.12 mm
2.) k = 2π / λ
Recall :
v = fλ
v = sqrt(T/u) = sqrt(10/0.5) = sqrt(20) = 4.472
λ = v/ f = 4.472 / 100 = 0.04472
Hence,
k = (2 * π) / 0.04472
k = 140.50 rad/m
3.) Angular frequency, ω
ω = 2πf = 2 * 3.14 * 100 = 628.32 rad/sec
4.) sign is +ve
Direction of wave propagation as given is in the negative x axis
A ratio is another name for a decimal true or false
A 0.5kg football thrown by Tony Romo with a velocity of 15 m/s is caught by a stationary receiver and brought to rest in 0.02 seconds. a) what impulse is delivered to the ball? b) how much force must be exerted in order to stop the ball?
The answers are -7.5kg m/s and F = -375N, but I don't know how they got them. Steps please! Thank you!!
Answer:
a.-7.5 kg m/s
b.-375 N
Explanation:
We are given that
Mass of football, m=0.5 kg
Initial velocity ,u=15m/s
Final speed ,v=0
Time, t=0.02 s
a. We have to find the impulse delivered to the ball.
We know that
Impulse=Change in momentum
I=m(v-u)
Using the formula
[tex]I=0.5(0-15)=-7.5 kg m/s[/tex]
Hence, the impulse delivered to the ball=-7.5 kg m/s
(b)
We know that
Force,[tex]F=\frac{|mpulse}{time}[/tex]
Using the formula
[tex]F=\frac{-7.5}{0.02}=-375 N[/tex]
effects of heat on matter
Answer:
it can melt orcan put them past their boiling point
Explanation:
a 14n force is applied for 0.33 seconds, calculate the impulse
Answer:
4.62 N-s
Explanation:
recall that the formula for impulse is given by
Impulse = Force x change in time
in our case, we are given
Force = 14 N
change in time = 0.33s
Simply substituting the above into the equation for impulse, we get
Impulse = Force x change in time
Impulse = 14 x 0.33
= 4.62 N-s
[tex]\\ \sf\longmapsto Impulse=Force(Time)[/tex]
[tex]\\ \sf\longmapsto Impulse=14(0.33)[/tex]
[tex]\\ \sf\longmapsto Impulse=4.62Ns[/tex]
The _______ changes light energy into nerve signals using receptors called rods and cones. A. retina B. lens C. iris D. pupil
Answer:
A. Retina
Explanation:
A ball of mass m makes a head-on elastic collision with a second ball (at rest) and rebounds in the opposite direction with a speed equal to one-fourth its original speed. what is the mass of the second ball?
When a ball of mass m makes a head-on elastic collision with a second ball (at rest) and rebounds in the opposite direction with a speed equal to one-fourth its original speed, then mass of the second ball having v/3 is velocity after collision is 9m/4.
What is momentum ?Momentum is defined as mass times velocity of body. it is denoted by p and its SI unit is Kg.m/s. It has both magnitude and direction. it is a vector quantity. it tells about the moment of the body. it is denoted by p and expressed in kg.m/s. mathematically it is written as p = mv. A body having zero velocity or zero mass has zero momentum. its dimensions is [M¹ L¹ T⁻¹]. Momentum is conserved throughout the motion.
initial momentum = final momentum
Given,
mass of first body m₁ = m
initial velocity of first body = v₁' = v
final velocity of first body = v₁'' =v/4
mass of second body m₂ = ?
initial velocity of second body = v₂' = 0
final velocity of second body = v₂'' = v/3
According to conservation of momentum,
initial momentum = final momentum
m₁v₁' + m₂v₂' = m₁v₁'' + m₂v₂''
putting al above values
m₁v + 0 = m₁v/4 + m₂v/3
m₁v - m₁v/4 = m₂v/3
m (1 - 1/4)v = m₂v/3
3m/4 = m₂/3
m₂ = 9m/4
Hence mass of the second body is 9m/4.
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Which landform is produced at location E where the Mississippi River enters the Gulf of
Mexico?
a delta a drumlin an out wash an escarpment
Answer:
a delta
Explanation:
The landform produced at the location E where the Mississippi River enters the Gulf of Mexico is a delta.
A delta is a depositional landform where a smaller body of water enters into a larger one.
The Gulf of Mexico contains a larger body of water and as the Mississippi river enters into it, it splits up into many distributaries.
So, this feature is a delta.