Answer:
The movement of energy
Explanation:
Answer:
the movement of energy
A boat travels at 15 m/s in a direction 45° east of north for an hour. The boat then turns and travels at 18 m/s in a direction 5° north of east for an hour.
Answer:
first one 31
second one 23
Explanation:
on edge ;))
Los muelles de un remolque están calibrados para soportar su peso, cuando se carga el remolque con 2100 kg, la fuerza ejercida es de 20580 N comprime los muelles de un remolque 5,5 cm a)Longitud que desciende el remolque si se carga con 28000 N b)Si ha descendido 4,2 cm la carga
Answer:
a) El remolque desciende 7.4 cm
b) La carga debe ser de 15715.6 N ó 1603.6 kg
Explanation:
Para los cálculos que involucren muelles, se aplica la Ley de Hooke, la cual relaciona el efecto de una Fuerza y el cambio de longitud que esta ejerce, en un resorte de elasticidad dada.
Escrito en fórmula:
[tex]F=-k \cdot \Delta L[/tex]
Donde:
F es la fuerza ejercida
k es la constante elástica del muelle
ΔL es la variación de longitud del muelle
El problema indica que al cargar 2100 kg se ejerce una fuerza de 20580 N
Esto se corrobora con la 2da ley de Newton y asumiendo una aceleración de gravedad de 9.8 [tex]\frac{m}{s^{2} }[/tex]
[tex]F_{1} =m \cdot a\\F_{1}=2100kg \cdot 9.8\frac{m}{s^2}\\F_{1}=20580N[/tex]
Esta fuerza comprime o reduce la longitud del muelle en 5.5 cm. Usando estos datos en la Ley de Hooke, podemos obtener la constante elástica k:
[tex]F=-k \cdot \Delta L\\20580N=-k \cdot (-0.055m)\\\\k=\frac{20580N}{0.055m}\\k= 374181\frac{N}{m}[/tex]
Ahora ya tenemos los datos para resolver las preguntas:
a) Longitud que desciende el remolque si se carga con 28000 NAplicando directamente la formula de la Ley de Hooke:
[tex]F=-k \cdot \Delta L\\\Delta L=\frac{F}{-k} \\\Delta L= \frac{28000N}{-374181\frac{N}{m}} \\\Delta L=-0.074 m = -7.4cm[/tex]
b) Si ha descendido 4,2 cm la cargaEn este caso debemos calcular la fuerza necesaria que haga descender el remolque 4.2cm. Nuevamente utilizando la Ley de Hooke con estos nuevos datos:
[tex]F=-k \cdot \Delta L\\F=-374181\frac{N}{m} \cdot (-0.042m)\\F=15715.6N[/tex]
Si queremos saber la carga en kilogramos:
[tex]F = m \cdot a\\m = \frac{F}{a} \\m = \frac{15715.6N}{9.8\frac{m}{s^2} }\\m= 1603.6 kg[/tex]
5. Describe the relationship between the buoyant force and the weight of an object if the object:
a) is floating
b) is sinking
c) is rising up through the water.
A) the state of deep mental and physical therapy or relaxation
B) falling or moving to a lower level
C) up to owns kness
[tex] \frac{1}{(1 - \sqrt{3) {}^{2} } } [/tex]
Answer:
\frac{2+\sqrt{3}}{2}
Explanation:
so we simplify tthe denominator first
[tex](1-\sqrt{3})^2=[/tex][tex]4-2\sqrt{3}[/tex]
so the new fraction is:[tex]\frac{1}{4-2\sqrt{3}}[/tex]
We'll ratinalize the fraction to get: [tex]\frac{2+\sqrt{3}}{2}[/tex]
What connects the colon to the anus?AColon B Anus C Rectum D Biceps
The answer is C. Rectum.
Answer:
You have a severe case.
Explanation:
My anus smells foul, like poop or diarrhea. Why? Like your anus might be tired of making the same old sensational brown. No fear needed! Anus Maker is here! With this app, you can post poop photos for help if you need! No more foul sensational browns! Start your free Anus Maker trial today!
which one?? please someone quick!
Answer:
i think its second law of motion.
Explanation:
Answer:
it’s the second law
Explanation:
What wavelength would a ripple in water have if the frequency is 1.8 Hz and a
wave speed of 825 m/s?
Explanation:
825m/s / 1.8Hz = 458.33m
λ=v/f
λ-wavelength
v-speed
f-frequency
λ=825/1.8=458.33m
A car of mass 1000 kg travelling at a velocity of 25 m/s collides with another car of mass 1500kg which is at rest. The two cars stick and move off together. What is the velocity of the two cars after the collision?
Answer:
The velocity of the two cars is 10 m/s after the collision.
Explanation:
Law Of Conservation Of Linear Momentum
The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is
P=m.v
If we have a system of bodies, then the total momentum is the sum of them all
[tex]P=m_1v_1+m_2v_2+...+m_nv_n[/tex]
If some collision occurs, the velocities change to v' and the final momentum is:
[tex]P'=m_1v'_1+m_2v'_2+...+m_nv'_n[/tex]
In a system of two masses, the law of conservation of linear momentum takes the form:
[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]
If both masses stick together after the collision at a common speed v', then:
[tex]m_1v_1+m_2v_2=(m_1+m_2)v'[/tex]
The car of mass m1=1000 Kg travels at v1=25 m/s and collides with another car of m2=1500 Kg which is at rest (v2=0).
Knowing both cars stick and move together after the collision, their velocity is found solving for v':
[tex]\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]
[tex]\displaystyle v'=\frac{1000*25+1500*0}{1000+1500}[/tex]
[tex]\displaystyle v'=\frac{25000}{2500}[/tex]
v' = 10 m/s
The velocity of the two cars is 10 m/s after the collision.
Please find attached photograph for your answer.
When electrons are moving freely between many positive ions, what type of bond is occuring?
a
metallic
b
ionic
c
covalent
d
crystalline
Answer:
a metallic
Explanation:
Metal atoms are joined together by metallic bonds. Metals can form cations (positive ions) with a sea of delocalized electrons.
when it comes to graphs I am no genius lol
Answer:
Option B
Explanation:
Displacement = Area under the velocity-time graph
So to find the displacement of a particle from 7 to 8 seconds we take the triangle and calculate its area.
[tex]Area\ of\ a\ right\ angled\ triangle = \frac{1}{2}*(base)*(height) \\\\Area\ of\ a\ right\ angled\ triangle = \frac{1}{2}*(1)*(3) \\\\Area\ of\ a\ right\ angled\ triangle = 1.5[/tex]
So the displacement is 1.5 meters
I have an uploaded image so you can understand it better hope it helps
HIII DROPPING COINS YALL
Answer:
tysm
Explanation:
Answer:
OMG THANK U SO MUCH I NEEDED THEMMMMM
A red 120 kg bumper car moving at 4 m/s collides with a green 100 kg bumper car moving at 3 m/s. The red bumper car bounces off at 2 m/s. What is the green car's final velocity?
Bob can run the 100 meter dash in 25 seconds. What is his speed?
Answer:
4 meters a second
Explanation:
100/25
plzz mark brainiest
Consider an elevator carrying Kermit the frog weighing 4000.0 N is held 5.00 m above a spring with a force constant of 8000.0 N/m. The elevator falls onto the spring while subject to a frictional force (brake) of 1000.0 N. Determine the maximum compression distance, x, of the spring.
Answer:
Maximum compression distance (x) = 2.236 m (Approx)
Explanation:
Given:
Weight of frog = 4,000 N
Height = 5 m
Constant force = 8,000 N/m
Frictional force = 1,000 N
Find:
Maximum compression distance (x)
Computation:
Using Law of conservation;
mgh = 1/2(k)(x)²
4,000(5) = 1/2(8,000)(x)²
Maximum compression distance (x) = 2.236 m (Approx)
HELP ASAP Which statement is true about magnetic field lines?
A. There is no consistent pattern in the lines, B. The lines form a loop from the north pole back to the north pole and from the south pole to the south pole. C. The lines point away from the south pole of a magnet and toward the north pole. D. The lines point away from the north pole of a magnet and toward the south pole.
Answer:
i think it's c
Explanation:
but I'm not sure
Answer:
The answer is D
Explanation:
the velocity of a body of mass 60kg reaches 15m/s from 0m/s in 12 second. calculate the kinetic energy and power of the body.
Answer:
KE=1/2m v^2
1/2*60*15*15
30*15*15
6750 joules
power=work/time
How should the amount of elastic potential energy when the spring is fully stretched compare to the amount of kinetic energy when the spring is relaxed?
If there are no dissipative forces acting on the string, than the principle of conservation of energy holds.
When the string is relaxed, it has zero elastic potential energy
If we were strech the she spring, we would increase it's potential energy. After releasing the spring, when it's relaxed again, all potential energy will have been converted into kinetic energy.
Therefore, the elastic potential energy of the stretched spring should be equal to the amount of kinetic energy when it's relaxed
SOMEONE PLEASE HELP WITH THIS QUESTION!!!!
Answer:
A 1
B 3
c 4
d 3
Explanation:
we are chaimpon boy i topped my school by cheating so donot study chill and watch movies like of sunny leon and Sabita bhabi
why is thouching the live wire of an appliane when it is connecfed to the mains dangerous
Answer: The live wire is the most dangerous one, since it is at 230 V. it should never touch the earth wire (unless the insulation is between them, of course!), because this would make a complete circuit from your mains supply to the ground (earth). A shock or fire would be highly likely.
Explanation:
What distance does a biker travel if he rides at a constant speed or 22 m/s for 45 seconds?
Answer:
it would be 990 m.
Explanation:
22 m/s x 45 seconds.
A substance will take on the shape of an open container if it is a
liquid.
plasma.
gas.
solid.
Answer:
liquid
Explanation:
Charles, the 75 kg trampoline artist, lands on a trampoline with a speed of 9.0 m/s.
If the trampoline behaves like a spring with a spring constant of 52,000 N/m, what maximum distance will Charles push down the trampoline before bouncing back up? (Hint: at maximum compression, Charles is not moving.)
Answer:
The maximum distance Charles will push down the trampoline ≈ 0.342 m
Explanation:
The given parameters are;
The mass of the trampoline artist, m = 75 kg
The speed with which the trampoline artist lands, v = 9.0 m/s
The value of the spring constant of the trampoline, k = 52,000 N/m
Let x represents the maximum distance Charles will push down the trampoline
Therefore, we have;
Kinetic energy = 1/2·m·v²
The kinetic energy with which the trampoline artist lands = 1/2 × 75 × 9.0² = 3037.5
The kinetic energy with which the trampoline artist lands = 3037.5 J
The potential energy stored in a spring = 1/2·k·x² = The kinetic energy with which the trampoline artist lands
∴ 1/2 × 52,000 × x² = 3037.5
∴ x = √(3037.5/(1/2 × 52,000)) ≈ 0.342
The maximum distance Charles will push down the trampoline = x ≈ 0.342 m
An object in motion will remain in motion and an object at rest will remain at rest until a greater force interrupts it. Explain this concept.
Answer:
A object, lets say a cup. This cup will never, ever move unless something or someone disturbs it. If something touches or hits this cup the cup will move. But, until the cup gets touched, nothing will EVER make it move.
Explanation:
I hope this helps!!
Integrated science please help ASAP!...
Answer:
SewageAgricultural pollutionOilRadioactive substanceRiver dumpingMarine dumpingLittering trash Industrial wasteMining activitiesChemical fertilizersExplanation:
I hope this helps
Where are alkaline earth metals found on the periodic table?
Group 1
Group 2
Groups 3–12
Group 17
Answer:
B
Explanation:
Answer:
B
Explanation:
I had a feeling
if the volume of a cube is 100cm3. what's the measurement of one of its length
Answer:
100 cm
Explanation:
Answer:
4.64
Explanation:
The cube root of 100 is 4.64.
A 1.0 kg ball has a potential energy of 10 J and falls to the ground. What is the velocity right before it hits the ground?
a)20 m/s
b)10 m/s
c)4.5 m/s
d)15 m/s
Answer:
10m/s
Explanation:
If a lever has a mechanical advantage of 5 and 50 N of force is used to lift a rock, what is the weight of the rock?
Circular motion requires the application of a constant force in which direction?
Outward from the circle
Toward the center of the circle
Toward the object in motion
Away from an object
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.