Answer:
the acceleration of the airplane is 5.06 x 10⁻³ m/s²
Explanation:
Given;
initial velocity of the airplane. u = 34.5 m/s
distance traveled by the airplane, s = 46,100 m
final velocity of the airplane, v = 40.7 m/s
The acceleration of the airplane is calculated from the following kinematic equation;
v² = u² + 2as
[tex]2as= v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(40.7)^2 -(34.5)^2}{2 \times 46,100} \\\\a = 5.06 \ \times \ 10^{-3} \ m/s^2[/tex]
Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²
Find the Horizontal (x) vector for these forces.
Answer:
See the explanation below.
Explanation:
The force is a vector therefore we can decompose the force into components x & y. as we need the horizontal component of the force, we must use the cosine function of the angle.
[tex]F_{1x}=30.8*cos(20)\\F_{1x}=28.94[N]\\F_{2x}=34.3*cos(20)\\\\F_{2x}= 32.23[N][/tex]
Two surfaces in contact are moving slowly past each other. As the relative speed between the two surfaces in contact increases, what happens to the magnitude of their coefficient of kinetic friction
Answer:
It begins to increase
Explanation:
As the surfaces move slowly and then increase their relative speeds, the coefficient of kinetic friction begins to increase until it reaches it terminal value in which, it can no longer increase.
You're driving your new sports car at 80 mph over the top of a hill that has a radius of curvature of 540 m. What fraction of your normal weight is your apparent weight as you crest the hill?
Answer:
75.84%
Explanation:
We were given Speed of the sports car, v as 80 mph , we can convert to m/s for unit consistency.
v=80mph= 35.76 m/s
The radius of curvature is given as , r = 540 m
✓ the normal weight can be denoted as Wn
✓ the apparent weight of the person can be denoted as Wa
Wn= normal weight= mg
Wa=apparent weight = (mg - mv^2/r)
g= acceleration due to gravity= 9.8m/s^2
The apparent weightand normal weight has a ratio of
Mn/Ma= [mg - mv^2/r]/mg ........eqn(1)
If we simplify eqn(1) we have
Mn/Ma=[g - vr^2/g].............eqn(2)
Then substitute the given values
Mn/Ma=9.8 - [(35.76^2)/540]/ 9.8
=0.758×100%
Mn/Ma=75.84%
Hence, the required fraction is 75.84%
please genuinely help. 15 pts
:(
rank in terms of momentum
A. A 10,000 kg train car at rest
B. A 100 kg person running at 5 m/s
C. A 1200 kg car going 15 m/s
D. A 15 kg meteor going at a speed of 1000 m/s
Answer:
See the explanation below,
Explanation:
Momentum is defined as the product of mass by Velocity. In this way, we will replace the following equation in each of the cases.
[tex]P=m*v[/tex]
where:
P = momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
A)
[tex]P = 10000*0\\P=0[/tex]
B)
[tex]P=100*5\\P=500[kg*m/s][/tex]
C)
[tex]P=1200*15\\P=18000[kg*m/s][/tex]
D)
[tex]P=15*1000\\P=15000[kg*m/s][/tex]
From higher to lower momentum
C,D,B,A
What happens to the amplitude of the resultant wave when two sound waves with equal amplitude constructively interfere?
Answer:
For two waves of equal amplitude interfering constructively, the resulting amplitude is twice as large as the amplitude of an individual wave. For 100 waves of the same amplitude interfering constructively, the resulting amplitude is 100 times larger than the amplitude of an individual wave.
6.A nurse pushes a woman in a wheelchair at a constant acceleration up a ramp for a period of 17.5 s. At the base of the ramp, the wheel chair is at rest, and at the top of the ramp, the
wheelchair has a velocity of 0.50 m/s. The woman and her wheelchair have a combined
mass of 65.0 kg. What is the magnitude of the net force on the woman and her wheelchair
as they acceferate up the ramp?
(A) 0.27 N
(B) 1.9 N
(C) 2.3 N
Answer:
The answer is :(B) 1.9 N
please make my answer brainlist
What does a negative velocity indicate about an object's motion? Plz help
Answer:
The sign of the velocity depends on the coordinate system chosen to define the position. A positive velocity simply means that the object is moving in the positive direction, as defined by the coordinate system, while a negative velocity means the object is traveling in the other direction.
- Hope this helps!
2. A rock is shot straight up into the air with a slingshot that had been stretched 0.30 m. Assume
no air resistance.
a. Qualitatively complete the energy flow diagram and the energy bar graphs.
m= 500 g
A
• Y=0
Position A
Position
System/Flow
O
k=100N
- 100
a 옝
&x=0.30m
Om
KE PE, PE
KE PE, PEe Ethermal
Initial
Final
b. Quantitative Energy Conservation Equation
C
Determine the greatest height the rock could reach.
Answer:
e
Explanation:
What needs to happen to make electrical current flow in a generator?
a projectile is fired so that when it reaches the maximum height h it has speed vtop when the projectile is at h/2 the speed is 2vtop at what angle is the projectile fired
Answer:
Explanation:
Let the velocity of firing be u at angle θ
At maximum height velocity will be equal to horizontal component of initial velocity or vcosθ
So , vtop = v cosθ
At height h/2
vertical component of velocity v₂
v₂² = (usinθ)² - 2 g . h/2
v₂² = u²sin²θ - gh
horizontal component of velocity at height h/2 = u cosθ
velocity at height h / 2
= √ ( u²sin²θ - gh + u² cos²θ)
Given
√ ( u²sin²θ - gh + u² cos²θ) = 2 vtop
u²sin²θ - gh + u² cos²θ = 4 v²top = 4 u² cos²θ
u²sin²θ - gh = 3 u² cos²θ
At height h , vertical component of velocity is zero
0 = u²sin²θ - 2gh
gh = u²sin²θ / 2
u²sin²θ - u²sin²θ / 2 = 3 u² cos²θ
u²sin²θ / 2 = 3 u² cos²θ
Tan²θ = 6
Tanθ = 2.45
θ = 68⁰ .
Why don't we use x-ray to communicate information?
Answer: The difficulty comes in the modulation, and in the fact that X-rays can't be transmitted via wires or fiber.
Explanation: Mark me as brainlisti
PLEASE HELP (100)
A block M, with mass of 4 kg, is released from rest and slides down a ramp that is angled at 28° above the horizontal (from the bottom of the ramp). The length of the ramp, d, is 3 meters. How much work is done by the gravitational force as the block slides down the ramp?
Answer:
The work done will be [tex]W=55.27\: J[/tex]
Explanation:
The work equation is given by:
[tex]W=F\cdot x[/tex]
Where:
F is the force due to gravity (weight = mg)
x is the length of the ramp (3 m)
Now, the force acting here is the component of weight in the ramp direction, so it will be:
[tex]F_{x-direction}=mgsin(28)[/tex]
Therefore, the work done will be:
[tex]W=mgsin(28)*3[/tex]
[tex]W=4*9.81*sin(28)*3[/tex]
[tex]W=55.27\: J[/tex]
I hope it helps you!
When cold milk
is added to hot
coffee, the milk
warms up and the
coffee cools down.
What can be said
about the energy
of the milk and
the energy of
the coffee when
this happens?
Answer:
See explanation below.
Explanation:
The internal energy (kinetic energy) of the milk particles increases thus warming the milk up, while the internal energy of the coffee particles reduce their initial average speed as they are imparting energy to the slow milk particles.
Fill in the chart with the correct values of F for each change in the system described in questions
Answer:
We know that the gravitational force F between two masses P and Q, that are separated by a distance R is:
[tex]F = G*\frac{P*Q}{R^2}[/tex]
Where G is the gravitational constant.
a) Mass P is doubled, then we have 2*P instead of P, the new force is:
[tex]F' = G*\frac{(2*P)*Q}{R^2} = 2*(G*\frac{P*Q}{R^2} ) = 2*F[/tex]
b) Now R is doubled, then instead of R, we have 2*R:
[tex]F' = G*\frac{P*Q}{(2*R)^2} = G*\frac{P*Q}{4*R^2} = G*\frac{P*Q}{R^2}*(1/4) = F/4[/tex]
c) Now we replace P by 2*P, and Q by 3*Q
[tex]F' = G*\frac{(2*P)*(3*Q)}{R^2} = 2*3*(G*\frac{P*Q}{R^2} ) = 6*F[/tex]
d) The entire mass of the system is increased by a factor of 4, then both of the individual masses are increased by a factor of 4.
Then we need to replace P by 4*P, and Q by 4*Q.
[tex]F' = G*\frac{(4*P)*(4*Q)}{R^2} = 4*4(G*\frac{P*Q}{R^2} ) = 16*F[/tex]
e) Now we replace R by R/2.
[tex]F' = G*\frac{P*Q}{(R/2)^2} = G*\frac{P*Q}{R^2/4}= 4*G\frac{P*Q}{R^2} = 4*F[/tex]
when someone pushes a large crate across floor. what forces act upon the crate
friction, applied, gravity or air resistance?
Answer:
Friction, Applied Force and Gravity
At t = 0, a car registers at 30 miles/hr. Forty seconds later, the car’s velocity is now at 50 miles/hr. Assuming constant acceleration, what is the distance covered after 40s?
Select one:
a. 0.442 mi
b. .33 mi
c. 0.11 mi
d. 10.33
e. None of the choices given.
Answer: Choice A) 0.44 miles
=================================================
Work Shown:
1 hour = 60 minutes
1 minute = 60 seconds
1 hour = 60*60 = 3600 seconds
40 seconds = 40*(1/3600) = 40/3600 = 1/90 hours
The car travels 1/90 of an hour. Let t = 1/90.
The initial velocity is vi = 30 mph.
The final velocity is vf = 50 mph.
Apply one of the kinematics equations as given below.
x = distance traveled
x = 0.5*(vi+vf)*t
x = 0.5*(30+50)*(1/90)
x = 0.44444444444444 which is approximate
x = 0.44
The car traveled roughly 0.44 miles. This matches with choice A.
I'm assuming choice A is supposed to say 0.44 instead of 0.442; otherwise, I think your teacher made a typo by putting that 2 in there. Each of the other answer choices are accurate to 2 decimal places, so it would make sense that choice A is also accurate to 2 decimal places as well.
If 0.442 was intended by your teacher, and it's not a typo, then the answer would be E) None of the choices given.
How does water move from the surfaces of lakes, rivers, and the ocean directly back into the atmosphere?
A.precipitation
B.evaporation
C.condensation
Answer:
B. Evaporation
Explanation:
Because I am absolutely positive
Please give Brainlyest
A 110 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.220 m/s. How much work must be done on the hoop to stop it
Answer:
the work that must be done to stop the hoop is 2.662 J
Explanation:
Given;
mass of the hoop, m = 110 kg
speed of the center mass, v = 0.22 m/s
The work that must be done to stop the hoop is equal to the change in the kinetic energy of the hoop;
W = ΔK.E
W = ¹/₂mv²
W = ¹/₂ x 110 x 0.22²
W = 2.662 J
Therefore, the work that must be done to stop the hoop is 2.662 J
The amount of work that must be done to stop a hoop of mass 110 kg moving with s speed of 0.220 m/s is 2.662 J
Work can be defined as the ability or the capacity to perform work. Work is also the measure of energy of a body. The S.I unit of work is Joules (J)
From the question, The kinetic energy of the hoop = The energy required to stop the hoop.Formular for kinetic energy of the hoop is
EK = mv²/2.................... Equation 1
Where EK = Kinetic energy of the hoop, m = mass of the hoop, v = velocity of the hoop.
Given: m = 110kg, v = 0.220 m/s.
Substitute these values into equation 1
EK = 110(0.22)²/2
Ek = 2.662 J.
Therefore, The amount of work that must be done on the hoop to stop it is
Learn more about Work here: https://brainly.com/question/12106685
Help me and get points..
When a basketball player jumps to make a shot, once his or her feet are off the ground, the jumper's acceleration
A) Depends on the time in the air.
B) Depends on launch speed.
C) Is usually greater for taller players (but not always).
D) All the above are true
E) Is 10 m/s/s
Answer:
B) Depends on launch speed.
Explanation:
This is true when considering the basketball player in a given basketball game. The speed at which the player jumps up is a strong factor which determines the acceleration of his acceleration. The direct co-relation show that, speed and acceleration of the basketball player are interrelated.
A stretched string has a mass per unit length of 5.12 g/cm and a tension of 19.3 N. A sinusoidal wave on this string has an amplitude of 0.143 mm and a frequency of 76.9 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, and (c) ω, and (d) the correct choice of sign in front of ω?
Answer:
a. 0.143 mm b. 77.6 rad/m c. 483.18 rad/s d. +1
Explanation:
a. ym
Since the amplitude is 0.143 mm, ym = amplitude = 0.143 mm
b. k
We know k = wave number = 2π/λ where λ = wavelength.
Also, λ = v/f where v = speed of wave in string = √(T/μ) where T = tension in string = 19.3 N and μ = mass per unit length = 5.12 g/cm = 5.12 ÷ 1000 kg/(1 ÷ 100 m) = 0.512 kg/m and f = frequency = 76.9 Hz.
So, λ = v/f = √(T/μ)/f
substituting the values of the variables into the equation, we have
λ = √(T/μ)/f
= √(19.6 N/0.512 kg/m)/76.9 Hz
= √(38.28 Nkg/m)/76.9 Hz
= 6.187 m/s ÷ 76.9 Hz
= 0.081 m
= 81 mm
So, k = 2π/λ
= 2π/0.081 m
= 77.6 rad/m
c. ω
ω = angular frequency = 2πf where f = frequency of wave = 76.9 Hz
So, ω = 2πf
= 2π × 76.9 Hz
= 483.18 rad/s
d. The correct choice of sign in front of ω?
Since the wave is travelling in the negative x - direction, the sign in front of ω is positive. That is +1.
What is it known as when a force is applied to an object for an amount of time?
momentum
impulse
inertia
acceleration
Answer:
Impulse. Impulse is when a force is applied to an object for an amount of time.
Explanation:
It can't be momentum because momentum is the motion of a moving body, that is measured as a product of its mass and velocity.
It' cant be inertia because inertia is when an object doesn't move/ it does nothing as well as an object remaining unchanged.
And it can't be acceleration because acceleration is when how fast an object can go/ there velocity.
And that leaves us with impulse, so that's the correct one.
Hope this helped!
Answer:
impulse
Explanation:
took the test
What forces are related to a person pushing a large crate accross the floor?
Answer:
98
Explanation:
Answer:
friction, air resistance
Explanation:
Which common characteristic of planets do Saturn and Earth share?
They have rings.
They have moons.
They are made of rock.
They have thick atmospheres.
The common characteristics of planets do Saturn and Earth share is they both have moon. Saturn have 13 moons and earth have only one moon.
What is moons of planets?Moons of planets are spatial objects which revolves around a planet and experiences gravitational pull from the planet. Both saturn and earth are planets of solar system.
The earth that is the only planet planet with living things have only one moon which is called moon itself. Earth is entirely different from the composition and nature of all other planets.
Saturn have 13 moons namely, enceladus, janus, mimas etc. Saturn have some rings made of dust and gases which are not found for any other planet. The composition and gaseous matter in saturn is different from that of earth.
To find more on saturn, refer here:
https://brainly.com/question/12181523
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Suppose you have a string of length LaTeX: LL clamped down on both sides. Is it possible for the wavelength of a standing wave on this string to equal LaTeX: L/3L / 3
Answer:
Explanation:
For a string of length L clamped down on both ends , for fundamental mode of vibration
2 . λ / 4 = L
= λ / 2 = L
λ = 2 L
For other modes , the formula is
n . λ/2 = L where n is an integer .
λ = 2L / n
if n = 6
λ = 2L / 6
= L/3
So wavelength = L/3 is possible .
find the length of the smallest standing wave that can be formed with light having frequency 600 THz
Answer:
[tex]L=2.5*10^-^7m[/tex]
Explanation:
From the question we are told that
Frequency of light [tex]f= 600 THz[/tex]
Generally the equation for wave is mathematically given as
[tex]\lambda=\frac{c}{\gamma}[/tex]
Where
[tex]1THz=10^1^2[/tex]
[tex]\lambda=\frac{3*10^8}{600*10^1^2}[/tex]
[tex]\lambda=0.5*10^-^6m[/tex]
Generally the equation for a fundamental Frequency is mathematically given as
[tex]L=\frac{\lambda}{2}[/tex]
[tex]L=\frac{0.5*10^-^6}{2}[/tex]
Therefore the length of the smallest standing wave
[tex]L=2.5*10^-^7m[/tex]
glucose+oxygen--->carbon dioxide+water+energy= What reaction??
Fill in the answers to describe this chemical equation.
Name the reactants in this reaction.
Name the products in this reaction.
How many oxygen atoms are involved in this reaction?
Is this equation balanced, and how can you tell?
Is this reaction endothermic or exothermic, and how can you tell?
I REALLY NEED HELP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
combustion reaction
glucose,oxygen
carbon dioxide,water,energy
8
no. required balanced equation is:
C6H12O6+6O2--->6CO2+6H2O+energy
endothermic as it needs heat in the form of sunlight.
A football is thrown from the edge of a cliff from a height of 22 m at a velocity of 18 m/s [39degrees above the horizontal]. A player at the bottom of the cliff is 12 m away from the base of the cliff and runs at a maximum speed of 6.0 m/s to catch the ball.
Is it possible for the player to catch the ball? Support your answer with calculations.
Please and thankyou!
Answer:
The ball will get to the bottom of the cliff in 1.26 s, while the player will reach the bottom of the cliff 2 s later. Thus, it is not possible for the player to catch the ball.
Explanation:
Given;
vertical height of the cliff, h = 22 m
velocity of the ball, u = 18 m/s at an angle 39⁰
vertical component of the velocity, [tex]u_y = u \ sin \theta[/tex]
The time for the ball to get to the bottom of the cliff is calculated as;
h = ut + ¹/₂gt²
h = (u sinθ)t + ¹/₂ x 9.8 x t²
22 = (18 sin 39)t + 4.9t²
22 = 11.328t + 4.9t²
4.9t² + 11.328t - 22 = 0
Solve the above equation with formula method;
a = 4.9, b = 11.328, c = -22
[tex]t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-11.328\ \ +/- \ \ \sqrt{(-11.328)^2 - 4(4.9\times -22)} }{2(4.9)}\\\\t = 1.26 \ s[/tex]
The time for the player to get to the bottom of the cliff;
Given maximum speed, Vx = 6.0 m/s and horizontal distance, X = 12 m;
[tex]t = \frac{X}{V_x} \\\\t = \frac{12}{6} \\\\t = 2 \ s[/tex]
The ball will get to the bottom of the cliff in 1.26 s, while the player will reach the bottom of the cliff 2 s later. Thus, it is not possible for the player to catch the ball.
A 0.5 m3 container is filled with a fluid whose specific volume is 0.001 m3/kg. At standard gravitational acceleration, the contents of this container weigh:
Answer: the contents of this container weighs 4905 kg.m/s²
Explanation:
Given that;
volume of a container V = 0.5 m³
we know that standard gravitational acceleration g = 9.81 m/s²
specific volume of liquid filled in the container v = 0.001 m³/kg
now we express the equation for weight of the container.
W = mg
W = (pV)g
W = Vg / ν
so we substitute
W = (0.5 m³)(9.81 m/s ) / 0.001 m³/kg
W = 4.905 / 0.001
W = 4905 kg.m/s²
Therefore, the contents of this container weighs 4905 kg.m/s²
A sculpture is suspended in equilibrium by two cables, one from a wall and the other
from the ceiling of a museum gallery. Cable 1 applies a horizontal force to the right of
the sculpture and has a tension, Fn. Catble 2 applies a force upward and to the left at an
angle of 37.0° to the negative x-axis and has a tension, Fn. The gravitational force on
the sculpture is 5.00 '10`N. What is Fn?
Answer:
[tex]T_1=6655.295917 \approx 6655.3N[/tex]
Explanation:
From the question we are told that
Angle of cable 2 [tex]\theta=37.0\textdegree[/tex]
Weight of sculpture [tex]W=5000 N[/tex]
Generally the Tension from cable 2 [tex]T_2[/tex] is mathematically given by
[tex]T_2sin37\textdegree=5000N[/tex]
[tex]T_2=5000N/sin37\textdegree[/tex]
[tex]T_2=8308.2N[/tex]
Generally the Tension from Cable 1 [tex]T_1[/tex] is mathematically given by
[tex]T_1=T_2 cos37\textdegree[/tex]
[tex]T_1=8308.2* cos 37\textdegree[/tex]
[tex]T_1=6655.295917 \approx 6655.3N[/tex]
