The answer is C. Rectum.
Answer:
You have a severe case.
Explanation:
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A 2450 kg stunt airplane accelerates from 120 m/s to 162 m/s in 2.10s. If the airplane is putting out an average force of 5.8810x10^4 N during this time, what is the average friction force exerted on the airplane by the air?
Given :
A 2450 kg stunt airplane accelerates from 120 m/s to 162 m/s in 2.10 s.
If the airplane is putting out an average force of [tex]5.8810\times 10^4 \ N[/tex].
To Find :
The average friction force exerted on the airplane by the air.
Solution :
Acceleration is given by :
[tex]a = \dfrac{162-120}{2.10}\ m/s^2\\\\a = 20 \ m/s^2[/tex]
Now, force equation is given by :
[tex]F - F_{friction} = ma\\\\F_{friction} = F-ma\\\\F_{friction} = 58810 - (2450\times 20 )\\\\F_{friction} = 9810\ N[/tex]
Therefore, frictional force exerted in the airplane by the air is 9810 N.
A book that weighs 5 N sits on a table. What force does the table apply to the book?
Answer:
E =F.d =[1/2]mv^2
mad = [1/2]mv^2
d= v^2/2a ,v=u+at , v^2 = [at]^2 since u=0
So d = at^2/2
F = ma= 20a= 50 , a=5/2 and t=2
so d = [5/2][2^2]/2=5
Explanation:
Every action has an equal and opposite reaction. It is an action-reaction principle. Therefore the table exerts a force of 5 N on the book in order to be in stable condition.
What is Newton's third law of motion?Newton's third law of motion state that every action has an equal and opposite reaction. It is an action-reaction principle. It stated that the force always exists in a pair.
Therefore the table exerts a force of 5 N on the book in order to be in stable condition.
The given data in the problem is ;
W is the weight of the book sits on table = 5N
N is the normal force on the book
From the equilibrium equation ;
Weight -Normal force on the book =0
Weight =Normal force on the book
The normal force on the book =5N
Hence the table exerts a force of 5 N on the book in order to be in stable condition.
To learn more about Newton's third law of motion refer to the link;
https://brainly.com/question/1077877
A spring has a spring constant of 120 N/m. How much energy is stored in the spring as it is stretched a distance of 0.20 meter?
Answer:
2.4J
Explanation:
Given parameters:
Spring constant = 120N/m
Extension = 0.2m
Unknown:
Amount of energy = ?
Solution:
The energy stored in this stretched spring is called the elastic potential energy.
It can be derived using the expression below:
Elastic Potential energy = [tex]\frac{1}{2}[/tex] ke²
k is the elastic constant
e is the extension
Insert the parameters;
Elastic potential energy = [tex]\frac{1}{2}[/tex] x 120 x 0.2² = 2.4J
Atoms of two different elements must have different
A. Electrical charges
B. Number of neutrons
C. Atomic numbers
D. Energy levels
Explanation:
C. Atomic numbers....
Answer:
atomic numbers
Explanation:
In order for two atoms to be different, they have to have a different number of protons. Protons are represented by the atomic number. Thus, atoms of two DIFFERENT elements must have different atomic numbers.
I took the test and got 100%
Hope this helps!
6th grade science I mark as brainliest.
Answer:
2m 13[tex]\frac{1}{3}[/tex]s
Explanation:
1.5m = 1s
200m = [tex]\frac{200}{1.5}[/tex] × 1s
= 133[tex]\frac{1}{3}[/tex]s
= 2m 13[tex]\frac{1}{3}[/tex]s
ANSWER QUICK!!
describe two uses for microwave radiation
heat food, warm water
A car accelerates from rest at a constant acceleration of 25.0 m/s^2. At some point, it then turns off its engine, letting the car decelerate slowly from the force of friction at a constant deceleration of 3 m/s^2 until it is at rest again. The total speed the car moves in this time is 200 meters. What is the minimum time needed for the car to move 200 meters given that it both starts and ends at rest?
Answer:
t = 9.14 s
Explanation:
We first analyze the accelerating motion by applying first equation of motion:
Vf₁ = Vi₁ + a₁t₁
where,
Vf₁ = Final Speed of Car before turning off engine
Vi₁ = Initial Speed of Car = 0 m/s
a₁ = acceleration of car = 25 m/s²
t₁ = time taken in accelerating motion
Therefore,
Vf₁ = 25t₁ ---------- equation (1)
Now, we apply second equation of motion:
s₁ = Vi₁ t₁ + (1/2)a₁t₁²
where,
s₁ = distance covered during accelerating motion
Therefore,
s₁ = (0)t₁ + (1/2)(25)t₁²
s₁ = 12.5 t₁² ----------- equation (2)
Now, we analyze the decelerating motion by applying first equation of motion:
Vf₂ = Vi₂ + a₂t₂
where,
Vf₂ = Final Speed of Car = 0 m/s
Vi₂ = Initial Speed of Car after turning off engine
a₂ = deceleration of car = - 3 m/s²
t₂ = time taken in decelerating motion
Therefore,
Vi₂ = 3t₂ ---------- equation (3)
Now, we apply second equation of motion:
s₂ = Vi₂ t₂ + (1/2)a₂t₂²
where,
s₂ = distance covered during decelerating motion
Therefore,
s₂ = (Vi₂)t₂ + (1/2)(-3)t₂²
s₂ = Vi₂ t₂ - 1.5 t₂²
using equation (3):
s₂ = 3 t₂² - 1.5 t₂²
s₂ = 1.5 t₂² ------------ equation (4)
Now, we know that the Final Velocity of accelerating motion (Vf₁) is equal to the initial velocity of decelerating motion (Vi₂):
Vf₁ = Vi₂
using equation (1) and equation (3):
25 t₁ = 3 t₂
t₁ = 0.12 t₂ ------------ equation (5)
Also, we know that sum of the distances is 200 m:
s₁ + s₂ = 200
using equation (2) and equation (4):
12.5 t₁² + 1.5 t₂² = 200
using equation (5):
12.5 (0.12 t₂²) + 1.5 t₂² = 200
3 t₂² = 200
t₂² = 200/3
t₂ = 8.16 s
substitute this in equation (5):
t₁ = 0.12(8.16 s)
t₁ = 0.97 s
Hence, the minimum time required for this motion is:
t = t₁ + t₂ = 0.97 s + 8.16 s
t = 9.14 s
A 2000 kg minivan runs into a 800 kg car that was at rest at a stop sign and they stick together. If the compact car does not stick to the van but, instead, causes the minivan to come to a complete stop. How fast will the car be going after the collision?
Answer:
v₂ = 2.5 v₁
car leaves at 2.5 times the speed of the minivan
Explanation:
This is an exercise of conservation of the momentum, to solve it we create a system formed by the minivan and the car, therefore during the crash the forces are internal and the momentum is conserved.
Initial instant. Before the crash
p₀ = M v₁+ 0
Final moment. After the crash
[tex]p_{f}[/tex] = M 0 + m v₂
how momentum is conserved
p₀ = p_{f}
M v₁ = m v₂
v₂ = [tex]\frac{M}{m}[/tex] v₁
let's calculate
v₂ = 2000/800 v₁
v₂ = 2.5 v₁
therefore the car leaves at 2.5 times the speed of the minivan
a block weighing (Fg) 50 N is resting on a steel table (us = 0.74). The minimum force to start this block moving is what N
Answer:37
Explanation:
How can you double the acceleration of an object if you cannot alter the object’s mass?
plz helpppp
Answer:
If you double the force, you double the acceleration, but if you double the mass, you cut the acceleration in half.
Explanation:
you can double the net force
According to the law of universal gravitation, gravity is the force keeping objects in the universe in their relative positions.
true
false
Answer true
Explanation
A car travels 3500 m in 200 seconds what is the car speed
Answer:
17.5 m/s
Explanation:
We can calculate the meters per second by dividing the distance by time. 3500 divided by 200 is 17.5, therefore the speed is 17.5 meters per second.
Integrated science please help ASAP!...
Answer:
SewageAgricultural pollutionOilRadioactive substanceRiver dumpingMarine dumpingLittering trash Industrial wasteMining activitiesChemical fertilizersExplanation:
I hope this helps
A fish is 4.7 cm from the front surface of a fish bowl of radius 21 cm. Where does the fish appear to be to someone in air viewing it from in front of the bowl? Do not forget the proper sign. (Give your answer in cm.)
______ cm
Where does the fish appear to be when it is 38.9 cm from the front surface of the bowl? (Give your answer in cm.)
______ cm
Answer:
Explanation:
From the information given:
We can properly determine the distance where the fish appear in the air viewing it from in front of the bowl by using the formula:
[tex]\dfrac{n_i}{d_o}+\dfrac{n_2}{d_1}= \dfrac{n_2-n_1}{r}[/tex]
where;
[tex]n_1[/tex] = refractive index in the air; = 1.33 &
[tex]n_2[/tex] = refractive index in water. = 1
[tex]\dfrac{n_2}{d_i}= \dfrac{n_2-n_1}{r}-\dfrac{n_1}{d_o}[/tex]
[tex]\dfrac{1}{d_i}= \dfrac{1-1.33}{-21 \ cm}-\dfrac{1.33}{4.7\ cm}[/tex]
[tex]\dfrac{1}{d_i}= - 0.26726 \ cm[/tex]
[tex]d_i =\dfrac{1}{ - 0.26726 \ cm}[/tex]
[tex]\mathbf{d_i }[/tex] = - 3.74 cm
2)
To determine where the fish appear to be when it is 38.9 cm from the front surface of the bowl by using the formula:
[tex]\dfrac{n_2}{d_i}= \dfrac{n_2-n_1}{r}-\dfrac{n_1}{d_o}[/tex]
[tex]\dfrac{1}{d_i}= \dfrac{1-1.33}{-21 \ cm}-\dfrac{1.33}{38.9\ cm}[/tex]
[tex]\dfrac{1}{d_i}=- 0.0184759 \ cm[/tex]
[tex]d_i = \dfrac{1}{- 0.0184759 \ cm}[/tex]
[tex]\mathbf{d_i = }[/tex] -54.12 cm
How much Tension force is required to pull a 1500 kg car (it is being towed) forward with an acceleration of 3 m/s^2 if the friction force on the towed car's tires is pulling backward with a force of 2500 N?
If the pull is done horizontally, then the net force on the car is
∑ F = T - f = (1500 kg) (3 m/s²)
where T is the magnitude of the tension in the towing cable, and f is the friction which points in the opposite direction. Then
T = f + (1500 kg) (3 m/s²)
T = 2500 N + 4500 N
T = 7000 N
plz help me with question ;-;
Answer:
C
Explanation:
Colette launches an air rocket in the upward, positive direction. It launches
with an initial velocity of 25.5 m/s. It accelerates in the downward, negative
direction at a rate of 9.81 m/s2. After 3.5 seconds, what is the magnitude of
the rocket's displacement?
A) 29 meters
B) 31 meters
C) –150 meters
D) 150 meters
Answer:
b
Explanation:
Answer:
The answer is A) 29 meters
Explanation:
I got this question right on the test! :)
A 8.0\,\text {kg}8.0kg8, point, 0, start text, k, g, end text box is released from rest at a height y_0 =0.25\,\text my 0 =0.25my, start subscript, 0, end subscript, equals, 0, point, 25, start text, m, end text on a frictionless ramp. The box slides from the ramp onto a rough horizontal surface. The box slides 2.0\,\text m2.0m2, point, 0, start text, m, end text horizontally until it stops.
Answer:
μ = 0.125
Explanation:
To solve this problem, which is generally asked for the coefficient of friction, we will use the conservation of energy.
Let's start working on the ramp
starting point. Highest point of the ramp
Em₀ = U = m h y
final point. Lower part of the ramp, before entering the rough surface
[tex]Em_{f}[/tex] = K = ½ m v²
as they indicate that there is no friction on the ramp
Em₀ = Em_{f}
m g y = ½ m v²
v = [tex]\sqrt{2gy}[/tex]
we calculate
v = √(2 9.8 0.25)
v = 2.21 m / s
in the rough part we use the relationship between work and kinetic energy
W = ΔK = K_{f} -K₀
as it stops the final kinetic energy is zero
W = -K₀
The work is done by the friction force, which opposes the movement
W = - fr x
friction force has the expression
fr = μ N
let's write Newton's second law for the vertical axis
N-W = 0
N = W = m g
we substitute
-μ m g x = - ½ m v²
μ = [tex]\frac{v^{2} }{2 g x}[/tex]
Let's calculate
μ = [tex]\frac{2.21^{2}}{2\ 9.8\ 2.0}[/tex]
μ = 0.125
A 8.0 kg box is released from rest at a height y0 = 0.25 m on a frictionless ramp. The box slides from the ramp onto a rough horizontal surface. The box slides 2.0 m horizontally until it stops.
What is the friction coefficient of the horizontal surface?
Answer: 0.125
A 0.050 kg ball starts from rest at some unknown height on a toy roller coaster.
At a later time, it travels through the top of a loop at 2 m/s and a height of 0.40 m.
Since this track is frictionless, what was the starting height of the ball?
Answer:
The starting height of the ball is approximately 0.604 m
Explanation:
The given parameters are;
The mass of the the ball = 0.050
The speed with which it travels through the top loop = 2 m/s
The given height at which the ball moves at 2 m/s = 0.40 m
Therefore, we have;
1/2·m·v² = m·g·h
1/2·v² = g·h
h = 1/2·v²/g = 1/2 × 2²/9.81 ≈ 0.204
The additional height = h = 0.204 m
Therefore;
The starting height of the ball ≈ The given height at which the ball moves at 2 m/s + h
The starting height of the ball ≈ 0.40 + 0.204 = 0.604 m
The starting height of the ball ≈ 0.604 m.
When any object in rest, then potential energy is present. But when object is in motion then object have kinetic energy.
Starting height of the ball is 0.604 m.
We know that, when any object is start from rest, then potential energy is converted into kinetic energy.
[tex]\frac{1}{2}mv^{2} =mgh[/tex]
Where m is mass of object, g is gravitational acceleration , h is height and v is velocity of object. (value of g = 9.81 m/ second square)
from above equation,
we get, extra height [tex]h=\frac{v^{2} }{2g} \\\\h=\frac{4}{2*9.81}\\\\h=0.204[/tex] meter
The starting height of the ball will be sum of the height at which ball moves 2 m/s and extra height.
Starting height = 0.40 + 0.204 = 0.604 meter.
Learn more:
https://brainly.com/question/18963960
What do scientists hope to learn from missions to visit asteroids?
Answer:
The mission will help scientists investigate how planets formed and how life began, as well as improve our understanding of asteroids that could impact Earth.
Explanation:
Hope this helps :)
Jack jumped off a diving board and hit the water at 15.68 m/s downward. How long was he falling for before he hit the water?
Answer:
1.6 s
Explanation:
From the question given above, the following data were:
Velocity (v) = 15.68 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
Thus, we can calculate the time taken for Jack to hit the water by using the following formula:
v = gt
15.68 = 9.8 × t
Divide both side by 9.8
t = 15.68 / 9.8
t = 1.6 s
Therefore, it took Jack 1.6 s to hit the water.
When a 0.622 kg basketball hits the floor, its velocity changes from 4.23 m/s down to 3.85 m/s up. If the average force was 72.9 N, how much time was it in contact with the floor?
(Unit = s)
Remember: up is +, down is -
Answer:
t = 0.0689 s
Explanation:
Given that,
Mass of a basketball, m = 0.622 kg
Initial velocity, u = 4.23 m/s (downward or negative)
Final velocity, v = 3.85 m/s (up of positive)
Average force, F = 72.9 N
We need to find the time it was in contact with the floor. The force is given by :
[tex]F=ma\\\\F=m\dfrac{v-u}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{0.622\times (3.85-(-4.23))}{72.9}\\\\t=0.0689\ s[/tex]
So, the time of contact is 0.0689 s.
Why was basketball invented?
bold completed over 100m in 9.58
seconds what is his average speed
Answer:
At the Berlin 2009 World Championships, Bolt set a world record time of 9.58 seconds for the 100m race, notching a top speed of 27.8 miles per hour (44.72 kilometers per hour) between meters 60 and 80, with an average speed of 23.5 mph.
Explanation:
Jason rides his bicycle for 3 hours at a speed to 40 km/h. What distance does he
travel? -
Answer:
120km
Explanation:
The density of an object is dependent upon the object’s mass and ---
Answer:Volume
Explanation:
Density = mass/ Volume
Answer:
Volume
Explanation:
Which of the following charts correctly compares plant and animal cells?
Answer:
Wheres the charts??
Explanation:
A parallel-plate capacitor is constructed using adielectric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00X108V/m. The desired capacitance is 0.250 μF, and the capacitor must withstand a maximum potential difference of 4.00 kV. Find the minimum area of the capacitor plates.
Answer:
A = 0.188 m²
Explanation:
First we find the distance between the plates by using the formula of electric field intensity:
E = ΔV/d
d = ΔV/E
where,
d = distance between plates = ?
ΔV = Potential Difference = 4 KV = 4000 V
E = Electric Field = 2 x 10⁸ V/m
Therefore,
d = 4000 V/(2 x 10⁸ V/m)
d = 2 x 10⁻⁵ m
Now, we find the Area of Plates by using formula of capacitance:
C = A∈₀∈r/d
where,
C = Capacitance = 0.25 μF = 0.25 x 10⁻⁶ F
A = Area of Plates = ?
∈₀ = Permittivity of free space = 8.85 x 10⁻¹² C/N.m²
∈r = Dielectric Constant = 3
Therefore,
0.25 x 10⁻⁶ F = A(8.85 x 10⁻¹² C/N.m²)(3)/(2 x 10⁻⁵ m)
A = (0.25 x 10⁻⁶ F)(2 x 10⁻⁵ m)/(3)(8.85 x 10⁻¹² C/N.m²)
A = 0.188 m²
Which of the following equations accurately defines acceleration?
Plz help ASAP
Answer:
Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.
Explanation:
Answer: i think its b
sorry if im wrong
Explanation:
Students are asked to design an experiment about Newton’s 2nd Law. One student decides to roll a marble down a ramp into a pile of sand to measure the force impact.
Which variable should she manipulate to best exemplify the relationship explained by this law?
A.She should use a heavier marble, because the marbles will roll at the same rate of acceleration but more mass will produce a larger impact force.
B.She should increase the slope of the ramp by propping it up to higher height, because a steeper ramp will cause a greater rate of acceleration and a larger impact force.
C.She should use a heavier marble, because a bigger marble will accelerate more quickly down the ramp and cause a greater impact force.
D.She should decrease the slope of the ramp, because a ramp with a smaller slope will allow the ball more time to build up speed and cause a greater impact force.
Answer:
A
Explanation:
Trust me I just took it !
