Answer:
Terminal velocity of object = 12.58 m/s
Explanation:
We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.
Gravitational force = mg = 80 * 9.8 = 784 N
Drag force = [tex]12.0v+4.00v^2[/tex]
Equating both, we have
[tex]784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0[/tex]
So v = 12.58 m/s or v = -15.58 m/s ( not possible)
So terminal velocity of object = 12.58 m/s
The terminal speed of the object is approximately [tex]\( 12.59 \, \text{m/s} \).[/tex]
To find the terminal speed of the object, we need to set the drag force equal to the gravitational force acting on the object. At terminal speed, the net force on the object is zero, meaning the drag force and the gravitational force are balanced.
The gravitational force [tex]\( F_g \)[/tex] acting on the object is given by the mass of the object m times the acceleration due to gravity g , which is approximately[tex]\( 9.81 \, \text{m/s}^2 \):[/tex]
[tex]\[ F_g = m \cdot g = 80.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \][/tex]
[tex]\[ F_g = 784.8 \, \text{N} \][/tex]
The drag force [tex]\( F_{\text{drag}} \)[/tex] is given by the equation:
[tex]\[ F_{\text{drag}} = (12.0 \, \text{N} \cdot \text{s/m})v + (4.00 \, \text{N} \cdot \text{s}^2/\text{m}^2)v^2 \][/tex]
At terminal speed,[tex]\( F_{\text{drag}} = F_g \)[/tex], so we have:
[tex]\[ (12.0 \, \text{N} \cdot \text{s/m})v_{\text{terminal}} + (4.00 \, \text{N} \cdot \text{s}^2/\text{m}^2)v_{\text{terminal}}^2 = 784.8 \, \text{N} \][/tex]
Now, we can use the quadratic formula to solve for [tex]\( v_{\text{terminal}} \):[/tex]
[tex]\[ v_{\text{terminal}} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 4.00 \, \text{N} \cdot \text{s}^2/\text{m}^2 \), \( b = 12.0 \, \text{N} \cdot \text{s/m} \)[/tex], and [tex]\( c = -784.8 \, \text{N} \).[/tex]
Plugging in the values:
[tex]\[ v_{\text{terminal}} = \frac{-12.0 \pm \sqrt{(12.0)^2 - 4 \cdot 4.00 \cdot (-784.8)}}{2 \cdot 4.00} \][/tex]
[tex]\[ v_{\text{terminal}} = \frac{-12.0 \pm \sqrt{144 + 12556.8}}{8} \][/tex]
[tex]\[ v_{\text{terminal}} = \frac{-12.0 \pm \sqrt{12700.8}}{8} \][/tex]
[tex]\[ v_{\text{terminal}} = \frac{-12.0 \pm 112.7}{8} \][/tex]
Since speed cannot be negative, we take the positive root:
[tex]\[ v_{\text{terminal}} = \frac{-12.0 + 112.7}{8} \][/tex]
[tex]\[ v_{\text{terminal}} = \frac{100.7}{8} \][/tex]
[tex]\[ v_{\text{terminal}} = 12.5875 \, \text{m/s} \][/tex]
As you climb a high mountain, the buoyant force exerted on you by the atmosphere _____.
Starting from rest, the Road Runner accelerates at 3 m/s for ten seconds. What is the final velocity of the Road Runner?
Answer
The answer to this question is [tex]30 ms^{-1}[/tex]
Explanation
As we know that accelartion is the rate of change of velocity. So, it can be write as
[tex]a = (V_f -V_i) /t[/tex]
where
[tex]V_f[/tex] is the final velocity
[tex]V_i[/tex] is the initial velocity
t is the time
a is the accelartion
as we konw
[tex]a = 3 ms^{-2}[/tex]
t = 10 s
From rest So,
[tex]V_i[/tex] = 0
[tex]V_f[/tex] = ?
Putting values
[tex]3 = (V_f - 0)/10[/tex]
[tex]3 * 10 = V_f[/tex]
[tex]30 = V_f[/tex]
[tex]V_f = 30 ms^{-1}[/tex]
So, the right answe is [tex]30 ms^{-1}[/tex]
(3 m/s²) x (10 s) = 30 m/s
His speed at the end of 10 sec is 30 m/s. We can't describe his velocity, because we have no information about the direction he's moving.
Eric decided to go for a walk from his home. He headed 6 meters east and x meters north. The angle formed by the displacement vector and the horizontal vector is 37°. What is the length of the displacement vector?
Answer:
Length of displacement vector = 7.51 m
Explanation:
Consider east as positive X axis and North as positive Y axis.
Eric headed 6 meters east, Displacement = 6 i
Then he headed X meters north, Displacement = X j
Total displacement = 6 i + X j
Angle it makes with horizontal axis, θ = tan⁻¹(X/6) = 37°
X/6 = tan 37
X = 4.52 m
Total displacement = 6 i + 4.52 j
Magnitude = [tex]\sqrt{6^2+4.52^2} =7.51m[/tex]
Length of displacement vector = 7.51 m
The answer is A. 7.5.
60 g of vinegar was used in a chemical reaction with baking soda. The total mass of the products is 140 g. What was the mass of the baking soda
The mass of the baking soda was 80 g
Answer: the mass is 80g
Explanation: In a solution of 140g, we knot that there are 60g of vinegar, the only other component in the solution is baking soda.
Now we want to find the mass of the baking soda in the solution.
Because there are only two components in the solution, if we remove the 60g of vinegar, the residue mass is the mass of the baking soda.
then we need to calculate:
Mass = 140g - 60g = 80g
So the mass of baking soda in the solution is 80g.
A hiker travels in a straight line for 40 minutes with an average velocity that has a magnitude of 1.2 m/s. How far is she from her starting point?
Given:
Time taken for the travel: 40 mins=2400 secs
Velocity: 1.2m/s
Speed and velocity have the same magnitude. Speed is a scalar where as velocity is a vector quantity.
Distance traveled= speed x time.
Distance traveled: 1.2 x 2400=2880m
Which of these statements about family relationships is true on Colonel Lloyd’s plantation?
Answer:
THE ANSWER IS D
Explanation:
I DID THE TEST
a race car accelerates uniformly from 18.5m/s to 46.1m/s in 2.4 seconds. determine the acceleration of the car and the distance traveled
The car's (average) acceleration would be
[tex]a=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.4\,\mathrm s}=11.5\,\dfrac{\mathrm m}{\mathrm s^2}[/tex]
The car's position over time would be given by
[tex]x=v_0t+\dfrac12at^2[/tex]
so that after 2.4 seconds, the car will have traveled a distance of
[tex]x=\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)(2.4\,\mathrm s)+\dfrac12\left(11.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm s)^2[/tex]
[tex]\implies x=77.5\,\mathrm m[/tex]
Hello!
A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.
* Determine the acceleration of the car....
We have the following data:
V (final velocity) = 46.1 m/s
Vo (initial velocity) = 18.5 m/s
ΔV (speed interval) = V - Vo → ΔV = 46.1 - 18.5 → ΔV = 27.6 m/s
ΔT (time interval) = 2.4 s
a (average acceleration) = ? (in m/s²)
Formula:
[tex]\boxed{a = \dfrac{\Delta{V}}{\Delta{T^}}}[/tex]
Solving:
[tex]a = \dfrac{\Delta{V}}{\Delta{T^}}[/tex]
[tex]a = \dfrac{27.6\:\dfrac{m}{s}}{2.4\:s}[/tex]
[tex]\boxed{\boxed{a \approx 11.5\:m/s^2}}\longleftarrow(acceleration)\:\:\:\:\:\:\bf\green{\checkmark}[/tex]
* The distance traveled ?
We have the following data:
Vi (initial velocity) = 18.5 m/s
t (time) = 2.47 s
a (average acceleration) = 11.5 m/s²
d (distance interval) = ? (in m)
By the formula of the space of the Uniformly Varied Movement, it is:
[tex]d = v_i * t + \dfrac{a*t^{2}}{2}[/tex]
[tex]d = 18.5 * 2.4 + \dfrac{11.5*(2.4)^{2}}{2}[/tex]
[tex]d = 44.4 + \dfrac{11.5*5.76}{2}[/tex]
[tex]d = 44.4 + \dfrac{66.24}{2}[/tex]
[tex]d = 44.4 + 33.12[/tex]
[tex]\boxed{\boxed{d = 77.52\:m}}\longleftarrow(distance)\:\:\:\:\:\:\bf\green{\checkmark}[/tex]
________________________________
[tex]\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}[/tex]
________ fuel is a renewable alternative to petroleum fuels that is made through either biologically or chemically driven production in a process called gasification.
Gasoline fuel is a renewable alternative to petroleum fuels that is made through either biologically or chemically driven production in a process called gasification.
Answer:
Hydrogen fuel is a renewable alternative to petroleum fuels that is made through either biologically or chemically driven production in a process called gasification.
How much work does an elephant do while moving a circus wagon 20 meters with a pulling force of 200n
Given:
s(distance)= 20 meters
F(force)=200N
Now we know that
work done= Force applied x distance
Substituting the given values in the above formula we get
Work done= 20 x 200= 4000J
The work done by an elephant while moving a circus wagon 20 meters with a pulling force of 200N is calculated using the formula: Work = Force x Distance. Plugging the values into the formula gives us 4000 Joules.
Explanation:The work done by the elephant pulling the circus wagon can be calculated using the formula for work, which is Work = Force x Distance. In this situation, the Force applied by the elephant is 200N and the Distance moved by the wagon is 20 meters.
So, the Work done would be: 200N x 20m = 4000 Joules.
This means the work done by the elephant is 4000 Joules.
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What are graphs used for? Select all that apply.
analyzing what data means
forming experiments
diagram showing the correlation between two quantities
predicting outcomes and patterns
Answer:
analyzing what data means
diagram showing the correlation between two quantities
Explanation:
Graph is the plot of two physical quantities and it describes the relation between both quantities. It gives a specific relation between the two physical quantities which is used to analyze the result.
Thus, the following options are correct,
analyzing what data means
diagram showing the correlation between two quantities
Answer: Analyzing what data means, diagram showing the correlation between two quantities, and predicting outcomes and patterns.
The first, third, and forth.
A 32.5-kg rock is traveling at a constant velocity through space. Then a force of 140 N is exerted on the rock. What acceleration does the force produce?
3.31 m/s^2
2.32m/s^2
4.31 m/s^2
45.5 m/s^2
10.7 m/s^2
As per Newton's II law we can say force applied on an object is product of mass and acceleration
[tex]F = ma[/tex]
given that
Force = 140 N
mass = 32.5 kg
now we can use above formula
[tex]140 = 32.5* a[/tex]
[tex]a = \frac{140}{32.5}[/tex]
[tex]a = 4.31 m/s^2[/tex]
so its acceleration must be 4.31 m/s^2
When a 60 g (=0.06 kg) tennis ball is served by a newly invented machine, it accelerates from zero to 50 m/s. The ball experiences a constant acceleration due to the impact with the racket over a distance of 0.5 m. What is the
here tennis ball is accelerated from initial speed zero to final speed 50 m/s
In this accelerated motion the distance moved by the ball is 0.5 m
So here we can use kinematics to find the acceleration of the ball during this distance
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
here given that
vf = 50 m/s
vi = 0
d = 0.5 m
[tex]50^2 - 0 = 2*a*0.5[/tex]
[tex]2500 m/s^2 = a[/tex]
now if we have acceleration during its motion so we can find the applied force by using Newton's II law
[tex]F = ma[/tex]
[tex]F = 0.06*2500[/tex]
[tex]F = 150 N[/tex]
now the applied force will be 150 N on the ball by racket
The colors of pigments is it a) are cyan,yellow,and magenta b) are the same as the secondary c) combine in equal amounts to provide black or d) all of the above
The answer would be A or D. The most appropriate answer would be D. Hope I helped :)
Consider a brick that is totally immersed in water, with the long edge of the brick vertical. The pressure on the brick is
A) greatest on the sides of the brick
B) greatest on the top of the brick
C) greatest on the bottom of the brick
D) greatest on the face with the largest area
E) the same on all surfaces of the brick
Explain your answer
The answer is C because only the bottom has the water pressure applied to it because it's the only thing emersed in the water.
The pressure on a brick that is totally immersed in water, with the long edge of the brick vertical is: C. greatest on the bottom of the brick.
Pressure can be defined as a measure of the external force acting upon a surface area. Mathematically, the pressure acting on an area is given by the formula:
[tex]Pressure = \frac{Force}{Area}[/tex]
In this scenario, the pressure on the brick with a long, vertical edge, that is totally immersed in water is greatest on its bottom because the long, vertical edge displaces water.
Read more: https://brainly.com/question/22480179
The force it would take to accelerate a 700-kg car at a rate of 5m/s2 is
mass of car =700kg
Required acceleration =5m/s^2
According to newton's second law
F=ma
F=(700)(5)
F=3500 Kgm/sec^2
Answer:3500
Explanation: mass of car =700 kg
Required acceleration =5 m/s^2
According to newton's second law
F=ma
F=(700)(5)
F=3500 K gm/sec^2
what does pascals principle state
Pascal's law (also Pascal's principle or the principle of transmission of fluid-pressure) is a principle in fluid mechanics that states that a pressure change occurring anywhere in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere.
Answer: The pressure at two pistons within an enclosed fluid system is always the same.
Explanation:
An object's true weight is 123 N. When it is completely submerged in water, its
apparent weight is 82.0 N. If the density of water is 1000 kg/m3
, then what is the density of the object?
Object true weight is given as
[tex]mg = 123 N[/tex]
now we know that g = 9.8 m/s^2
[tex]m* 9.8 = 123 [/tex]
[tex] m = \frac{123}{9.8} = 12.55 kg[/tex]
now when it is complete submerged in water its apparent weight is given as 82 N
apparent weight = weight - buoyancy force
apparent weight = 82 N
weight = 123 N
now we have
82 = 123 - buoyancy force
buoyancy force = 123 - 82 = 41 N
now we also know that buoyancy force is given as
[tex]F_b = p_{liq}Vg[/tex]
[tex]41 = 1000*V*9.8[/tex]
[tex]V = \frac{41}{1000*9.8}[/tex]
[tex]V = 4.18 * 10^{-3} m^3[/tex]
now as we know that mass of the object is 12.55 kg
its volume is 4.18 * 10^-3 m^3
now we know that density will be given as mass per unit volume
[tex]density = \frac{m}{V}[/tex]
[tex]density = \frac{12.55}{4.18*10^{-3}[/tex]
[tex]density = 3002.4 kg/m^3[/tex]
so here density of object is 3002.4 kg/m^3
Answer:
3000 kg/m^3
Explanation:
True weight = weight in air = 123 N
Apparent weight = weight in water = 82 N
Loss in weight of the object = true weight - apparent weight
Loss in weight = 123 - 82 = 41 N
According to the Archimedes principle, the loss in weight of the object is equal to the buoyant force acting on the object.
Let V be the volume of the object an d be the density of the object.
Buoyant force = volume of the object x density of water x gravity
41 = V x 1000 x g
[tex]V = \frac{41}{1000 g}[/tex] .... (1)
Now, true weight = Volume of the object x density of object x gravity
[tex]123 = \frac{41}{1000 g} \times d\times g[/tex] from (1)
d = 3000 kg/m^3
A certain metal has a coefficient of linear expansion of 2.00 × 10-5 K-1. It has been kept in a laboratory oven at
325°C for a long time. It is now removed from the oven and placed in a freezer at ‐145°C. After it has reached
freezer temperature, the percent change in its density during this process is closest to
A) +2.90%.
B) ‐2.90%.
C) +2.74%.
D) ‐2.74%.
E) It is not possible to tell without knowing the mass and original volume of the metal.
Hi, the answer is A) +2.90%.
A resource (like the Sun) that essentially never "runs out" is said to be _____.
A. reusable
B. recyclable
C. inexhaustible
D. nonrenewable
asap please
the answer youre looking for is (C
Just before snapping backup in the air, my speed is 50m/s. What is my kinetic energy just before snapping back up
Kinetic energy of anything is (1/2) (mass) (speed)² .
Your kinetic energy just before you snap back up (? ?) is
(1/2) (your mass in kilograms) (50 m/s)² =
(1/2) (your mass in kilograms) (2,500 m²/s²) =
(1,250 x your mass in kilograms) . The unit is Joules.
Match the following examples of energy with the primary form of energy exhibited.
friction:
nuclear power plant:
toaster element:
welding torch:
eraser sitting on a desk edge:
light bulb:
campfire:
moving car:
lump of coal in a storage bin:
stick of TNT:
Options are nuclear, heat, kinetic, potential, light, and solar energy
The primary energy exhibited is:
friction: Heat energy
Due to friction. there is heat loss in mostly all machines.
nuclear power plant: Nuclear energy
In nuclear power plant, fission or fusion of radio-active substances produces large amount of nuclear energy.
toaster element: Heat energy
Heat energy is produced in the toaster when electricity passes through the element. This is used to heat the toasts.
welding torch: Heat energy
The heat energy produced in the welding torch is used in welding.
eraser sitting on a desk edge: Potential energy
Potential energy is possessed by a body due to virtue of its height.
light bulb: Heat energy and light energy.
The element inside the bulb gets heated first and then light energy is produced.
campfire: Heat energy
Heat is produced and then light energy in case of campfire.
moving car: Kinetic energy
A moving body has kinetic energy.
lump of coal in a storage bin: Potential energy
A lump of coal in a storage bin has potential energy.
stick of TNT: potential energy
Stick of TNT is formed of explosive mixture. It contains potential energy.
Answer:
• Friction ( heat )
• Nuclear power plant ( nuclear )
• Toaster element ( heat )
• welding torch ( heat )
• eraser sitting on a desk edge (potential)
• light bulb ( light )
• campfire ( heat )
• moving car ( kinetic )
• lump of coal in a storage bin ( potential )
• stick of tent ( potential)
Explanation: hope this helps !
what gas is most abundant greenhouse gas ? a) ozone b) chlorofluorocarbon c) carbon dioxide d) methane e) water vapor
i believe the answer is water vapor
hope this helps
Answer:
WATER VAPOR
Explanation:
I got a 100% on my test
Calculate how much energy is needed to get a typical car with a mass of 1500 kg up to a speed of 40 mph ignoring friction or any other losses or inefficiencies. Show all work, and give your answer in kWh.
m = mass of the car = 1500 kg
we know that 1 mph = 0.45 m/s
hence
v = speed gained by the car = 40 mph = 40 x 0.45 m/s = 18 m/s
since the car is gaining speed here , hence the energy need is same as the kinetic energy gained by the car
kinetic energy is given as
KE = (0.5) m v² where m = mass , v = speed
inserting the above values in the formula
KE = (0.5) (1500) (18)²
KE = 2.43 x 10⁵ J
we know that , 1 J = 2.78 x 10⁻⁷ kwh
hence
KE = (2.43 x 10⁵) (2.78 x 10⁻⁷)
KE = 0.068 kwh
hence the energy needed is 0.068 kwh
A racing car whose mass is 1.2 X 10^3 kg is travelling at 8.9 m/s. It stops with a constant deceleration in a distance of 1.8X10^1 m. What force must the brakes apply to the car if the friction in the car’s engine is 1.8 X 10^3 N.
given that initial speed of the car is
[tex]v_i = 8.9 m/s[/tex]
now after travelling the distance d = 1.8 * 10^1 m the car will stop
so here we can use kinematics to find the acceleration of car
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 8.9^2 = 2 a d[/tex]
here we have
[tex]- 79.21 = 2*(18)*a[/tex]
[tex]a = -2.2 m/s^2[/tex]
net force applied due to brakes of car is given by Newton's II law
[tex]F = ma[/tex]
here we have
mass = 1.2 * 10^3 kg
[tex]F_{net} = 1.2 * 10^3 * 2.2[/tex]
[tex]F_{net} = 2.64 * 10^3 N[/tex]
now we can say
[tex]F_{net} = F_1 + F_2[/tex]
[tex]2.64 * 10^3 = 1.8 * 10^3 + F_2[/tex]
[tex]F_2 = 8.4 * 10^2 N[/tex]
So the force applied due to brakes is given as above
A 47-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37 degrees above the horizontal. If the tension in the rope is 142 N, how much work is done on the crate to move it 6.1m?
Tension in the rope due to applied force will be given as
[tex]F = 142 N[/tex]
angle of applied force with horizontal is 37 degree
displacement along the floor = 6.1 m
so here we can use the formula of work done
[tex]W = F d cos\theta[/tex]
now we can plug in all values above
[tex]W = 142 * 6.1 * cos37[/tex]
[tex]W = 691.8 J[/tex]
So here work done to pull is given by 691.8 J
Wire 1 has a resistance R1. Wire 2 is made of the same material, with length 1/2 as long and diameter 1/4. What is the resistance ratio R2/R1?
-- reduce the length of a wire to 1/2 . . . cut the resistance in half
-- reduce the diameter to 1/4 . . . reduce the cross-section area by (1/4²) . . . increase the resistance by 16x .
-- R2 = (R1) · (1/2) · (16) = 8 · R1
-- R2 / R1 = 8
Please Help Quick
Consider the average speed of a runner who jogs around a track four times. The distance (400m) remains constant for each lap. However, each lap is run 5 seconds slower than the first. The time for each lap increases. The average speed for each lap ______________. This is an example of a(n) _____________ relationship.
A) increases, direct
B) decreases, direct
C) increases, inverse
D) decreases, inverse
Note: If you have a question please do not ask it in the answer section of this question.
(D) decreases, inverse
Average speed of a body is defined as the ratio of the distance traveled to the time taken to travel the distance. The runner runs the same distance of 400 m during each lap but takes 5 seconds more for each lap.
Since [tex]v_a_v_g=\frac{d}{t}[/tex], as time increases, the average speed for each lap increases.
Thus, [tex]v\alpha \frac{1}{t}[/tex] when the distance traveled is the same. This is an example of an inverse proportionality relationship.
Cereal plants gain chemical potential energy as they grow. Which form of energy is converted to create the chemical potential energy? A. gravitational potential energy B. kinetic energy C. light energy D. thermal energy
Answer:
C. light energy
Explanation:
In all these plants the energy is converted from the photosynthesis process. As per this photosynthesis phenomenon plants get energy by the light received from the sun.
This will help to make the plants to grow and make fruits in the plant.
in this word photosynthesis we can see it is combination of photo + synthesis
which clearly means that this energy synthesis is with the help of photons received from the sun.
These photons are light photons
So here correct answer will be
C. light energy
Answer:
light energy
Explanation:
A father lifts a toddler 1.5m up in the air. The child gains 187.5J of gravitational potential energy as a result. What is the mass of the toddler? (Assume that the gravitational field strength is 10N/kg.)
Answer:
Mass of toddler = 12.5 kg
Explanation:
The potential energy of a body is given by the expression, PE = mgh, where m is the mass of the body, g is the acceleration due to gravity value and h is the height of the body.
Potential energy = 187.5 J
Height = 1.5 m
Substituting
187.5 = m * 10 * 1.5
m = 12.5 kg
So mass of toddler = 12.5 kg
What moon phase occurs 3-4 days after a waning gibbous?
The "waning gibbous" begins immediately after the Full Moon and lasts about a week.
At the end of the week, the moon is no longer gibbous, but it's still waning. At the instant it's exactly half-illuminated, it's called "Third Quarter", and then it continuous to wane for another week.
So 3 to 4 days after the END of the waning gibbous phase, it's a Waning Crescent. It still has another 3 to 4 days of waning to go, before it wanes away to nothing and we have the next New Moon.
The waning gibbous phase is the phase between the full and last quarter moon late at night or in the early morning . The waning crescent is the phase that occurs 3-4 days later a waning gibbous.
What is waning gibbous ?The waning gibbous phase is the phase between the full and last quarter moon late at night or in the early morning.
The waning gibbous phase moon appears more than half-lighted but less than the full moon. A full sunset is a time when the full moon rises. The rising time of the waning gibbous moon is late at night.
It looks red like a full moon when it’s near the horizon.
To learn more about the waning gibbous refer to the link ;
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