Answer: 55n
Explanation:
Es muy común que cuando se viaja hacia un río o lago se juegue "ranita", el cual consiste en lanzar una piedra horizontalmente hacia adelante para que cuando ésta toque la superficie del agua haga varios "saltos" sobre el agua. Durante un juego de estos, un desocupado nota que una de las piedras que arroja se demora 0,4 s en tocar la superficie del agua y la toca a 2,5 m de la orilla del lago, desde donde fue lanzada. Encuentre: a) La altura de la que fue lanzada la piedra. b) La velocidad con la que fue lanzada.
Answer:
a) La piedra es lanzada desde una altura de 0,785 metros.
b) La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.
Explanation:
a) Dado que la piedra es lanzada horizontalmente, tenemos que la piedra experimenta un movimiento horizontal a velocidad constante y uno vertical uniformemente acelerado debido a la gravedad. La altura de la que fue lanzada la piedra se puede determinar mediante la siguiente ecuación cinemática:
[tex]y = y_{o}+v_{o,y}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}[/tex] (1)
Donde:
[tex]y[/tex] - Altura final, medida en metros.
[tex]y_{o}[/tex] - Altura inicial, medida en metros.
[tex]v_{o,y}[/tex] - Componente vertical de la velocidad inicial, medida en metros por segundo.
[tex]t[/tex] - Tiempo, medido en segundos.
[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo cuadrado.
Si sabemos que [tex]y = 0\,m[/tex], [tex]v_{o,y} = 0\,\frac{m}{s}[/tex], [tex]t = 0,4\,s[/tex] y [tex]g = -9,807\,\frac{m}{s^{2}}[/tex], entonces la altura inicial de la piedra es:
[tex]y_{o} = y-v_{o,y}\cdot t -\frac{1}{2}\cdot g\cdot t^{2}[/tex]
[tex]y_{o} = 0\,m-\left(0\,\frac{m}{s} \right)\cdot (0,4\,s)-\frac{1}{2}\cdot \left(-9,807\,\frac{m}{s^{2}} \right) \cdot (0,4\,s)^{2}[/tex]
[tex]y_{o} = 0,785\,m[/tex]
La piedra es lanzada desde una altura de 0,785 metros.
b) Ahora, obtenemos el componente horizontal de la velocidad inicial a partir de la siguiente ecuación cinemática:
[tex]v_{o,x} = \frac{x-x_{o}}{t}[/tex] (2)
Donde:
[tex]x_{o}[/tex], [tex]x[/tex] - Posiciones horizontales iniciales y finales, medidas en metros.
[tex]t[/tex] - Tiempo, medido en segundos.
Si tenemos que [tex]x_{o} = 0\,m[/tex], [tex]x = 2,5\,m[/tex] y [tex]t = 0,4\,s[/tex], entonces el componente horizontal de la velocidad inicial es:
[tex]v_{o,x} = \frac{2,5\,m-0\,m}{0,4\,s}[/tex]
[tex]v_{o,x} = 6,25\,\frac{m}{s}[/tex]
La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.
What travels by vibrating particles? Mechincal Waves or ElecrtoMagnetic Waves.
Answer:mechanical waves.
Explanation:
Mechanical waves require the particles of the medium to vibrate in order for energy to be transferred. For example, water waves, earthquake/seismic waves, sound waves, and the waves that travel down a rope or spring are also mechanical waves.
A friend is coming to Tim’s house to study after school. What directions would Tim give for reaching his house from the entrance of the school?
(I WILL GIVE BRAINLIEST)
Answer:
go up the street as you exit the house and make a right and keep going up for 3 blocks and you should see the school
A man speeding at 40m/s decides to outrun the cops and starts to
accelerate at a rate of 2.5m/s2 for 12 seconds. What is the criminal's new
speed?
Answer:
70 m/s.
Explanation:
Given that,
Initial speed, u = 40 m/s
Acceleration = 2.5 m/s²
Time, t = 12 s
We need to find criminal's new speed. Let it is v. Using equation of motion to find it as follows :
v = u +at
Substitute all the values
v = 40 + 2.5(12)
v = 70 m/s
So, the new speed is 70 m/s.
A 250-kg moose stands in the middle of the railroad tracks in Sweden, frozen by the lights of an oncoming 10,000kg train traveling at 20m/s. Even though the engineer attempted in vain to slow the train down in time to avoid hitting the moose, the moose rides down the remaining track sitting on the train’s cowcatcher. What is the final velocity of the train and moose after the collision?
(Momentum & Impulse)
Answer:
The final velocity of the train and the moose after collision is approximately 19.51 m/s
Explanation:
The given mass of the moose, m₁ = 250 kg
The velocity of the moose, v₁ = 0
The mass of the oncoming train, m₂ = 10,000 kg
The velocity of the train, v₂ = 20 m/s
The velocity of the moose and the train after collision = v₃
By the principle of conservation of linear momentum, the total initial momentum before the collision = The total final momentum after collision
m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v₃
Therefore, by substitution, we have;
250×0 + 10,000× 20 = (10,000 + 250) × v₃
200,000 = 10,250 × v₃
v₃ = 200,000/10,250 ≈ 19.51 m/s
The final velocity of the train and the moose after collision = v₃ ≈ 19.51 m/s
A hose on the ground projects a water current upwards at an angle 40 to the horizontal at velocity 20 m/s find height at which water hits a wall at 8 m away from the hose (consider that acceleration due to gravity =9.8 m/s2)
Answer:
The water hits the wall at a height of 5.38 m
Explanation:
Projectile Motion
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
The object describes a parabolic path given by the equation:
[tex]{\displaystyle y=\tan(\theta )\cdot x-{\frac {g}{2v_{0}^{2}\cos ^{2}\theta }}\cdot x^{2}}[/tex]
Where:
y = vertical displacement
x = horizontal displacement
θ = Elevation angle
vo = Initial speed
The hose projects a water current upwards at an angle of θ=40° at a speed vo=20 m/s.
The height at which the water hits a wall located at x=8 m from the hose is:
[tex]{\displaystyle y=\tan40^\circ\cdot 8-{\frac {9.8}{2*20^{2}\cos ^{2}40^\circ }}\cdot 8^{2}}[/tex]
Calculating:
y = 5.38 m
The water hits the wall at a height of 5.38 m
what are the effects of moon rotation and revolution
Answer:
The effects of the Moon's rotation includes;
1) The Moon rotation and revolution gives the appearance of a perfectly still Moon to observers of the Moon from the Earth
2) The Moon has two sides, the near side that continuously faces the Earth and the "back" or far side, which is also known as the dark side of the Moon
The effect of the Moon's revolution
1) The tides in the ocean and water bodies, due to the gravitational forces from the Moon
2) The changes in the observed shape of the Moon due to the amount of Sunlight that is able to be reflected from the Moon as a result of the relative position of the Moon, the Earth and the Sun, at a given point in time
3) Lunar and Solar eclipse, when the Earth and the Moon blocks the light coming from the Sun respectively, due to their combined revolution
Explanation:
The duration of the Moon's orbit round the Earth = 27.322 days
The time it takes the Moon to rotate round its axis = 27 days
The Moon is the closest cosmic body to the Earth.
show your work. john uses a 25N force to push a boulder off a cliff that is 312m tall. What is the work done on the boulder?
Answer:7800
work=force x distance
Force in Newtons
Distance in Meters
Work in Joules
a squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy
What is the squirrels mass
Answer:
yeet yeet yeet yeet
Explanation:
Kinetic energy (K.E):-
So, the Mass of the Squirrel is 0.51 Kg (or) 510 grams.
A squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy.
What is the squirrel’s mass?
Answer: 0.51 kg
A golf ball is sitting on a tee. The ball is struck with a golf club and flies
through the air. How does the force on the club compare with the force on the
ball when momentum is transferred between the club and ball?
Answer:
c i kn now it is
Explanation:
Can u anser 5,6 on the picture
Answer: Number 6 is Periods
Explanation:
A car traveling with 500,000 J of kinetic energy is brought to a kinetic energy of
100,000 J in 12 seconds. What is the force acting on the car to accomplish this?
A:41,666.67
B:3x10^-5N
C:33,333.33N
D:8,333.33N
Answer:
33,333.33 N
Explanation:
Given that :
Initial kinetic energy = 500,000 J
Final kinetic energy = 100,000 J
Using the relation :
Force * time = change in momentum (Newton's law)
Force (F) * 0.12 = (500,000 - 100,000)
0.12F = 400,000 J
Force = (400,000 J) / 0.12s
Force = 33333.333
Force = 33,333.33 N
Describe what happens to the moving boat when the oars are out of the water and the forward thrust is zero
Answer:
The boat won't be able to move if the oars were out and there was no thruster. If there was a flow of the water then yes there would be a moving boat.
please answer asignment due today
Answer:
Give me some time okkkk
In raising an object vertically at a constant speed of 2 m / s 10 watts of power is developed the weight of the object is
Answer:
5NExplanation:
Power = Workdone/Time
Since workdone = Force * distance = mgh
Power = mgh/t
Given
Power = 10watts
speed = 2m/s
Power = Weight * velocity
Substitute
10 = Weight * 2
Weight = 10/2
Weight = 5N
Hence the weight of the object is 5N
Under what circumstances will the projectile have the greatest velocity when it hits the ground? Explain please
(Blank) is caused by plate motion.
It’s please help.
Answer:
heat
Explanation:
Which electron dot diagram shows the bonding between 2 chlorine atoms?2 dots then C l with 2 dots above and 1 dot below then 2 dots then 2 dots then C l with 2 dots above and 1 dot below then 2 dots.2 dots then C l with 2 dots above and 2 dots below then 2 dots then C l with 2 dots above and 2 dot below then 2 dots.2 dots then C l with 2 dots above and 2 dots below then 1 dot then C l with 2 dots above and 2 dots below then 2 dots.2 dots then C l with 2 dots above and 1 dot below then 3 dots then 3 dots then C l with 2 dots above and 1 dot below then 2 dots.
Answer:
It is B
Explanation:
Answer: 2nd answer
Explanation: took exam
NEED AWNSER NOW! WILL MARK BRAINLY! Which term is defined as the ratio of the speed of light in a vacuum to the speed of light in the material it is passing through?
index of reflection
index of refraction
angle of reflection
angle of incidence
Answer:
Index of refraction.
Answer:
index of refraction
Explanation:
I just took the k12 quiz.
Jerry is pushing a 50-kg box across a moth floor with an acceleration of 0.6 m/s2. What force is he applying to the box? *
83.3 N
0.012 N
0
30 N
Answer:
30 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 50 × 0.6
We have the final answer as
30 NHope this helps you
The magnitude of vector vector A is 84.9 m and it points in the +y axis direction. The magnitude of vector vector B is 195.0 m and it points at an angle of 41.0° counterclockwise from +x axis. The magnitude of vector vector C is 126.2 m and it points in the +x axis direction.
Solution:
The magnitude of A vector is 84.9 m in the positive y-axis direction.
So the X component of A =0
the Y component of A = 84.9 m
Now the magnitude of B vector is 195 m and it makes an angle of 41° in the direction from the positive x-axis direction.
So the X component of B = B cos 41°
= 195 x cos 41°
= 195 x 0.75 = 146.25 m
the Y component of B = B sin 41°
= 195 x sin 41°
= 195 x 0.65 = 126.75 m
Now it is given that vector C has a magnitude of 126.2 m and it makes a direction towards the positive x-axis.
So the X component of C =126.2 m
the Y component of C = 0
Comparing all these, we get
1. B vector has the largest X component
2. B vector has the largest Y component
When two ocean plates come together, one ocean plate __________________
under the other, causing a chain of ________________ __________________
to form.
Answer:
A subduction zone is also generated when two oceanic plates collide — the older plate is forced under the younger one — and it leads to the formation of chains of volcanic islands known as island arcs.
Explanation:
draw position time graph when speed is increasing
Explanation:
We need to draw position-time graph when the speed is increasing.
The slope of position-time graph gives the speed of an object.
Position means distance covered.
When the speed of an object is increasing with time. It means it is moving with increasing speed.
The attached figure shows the position -time graph when speed is increasing.
A force of 30 N stretches a very light ideal spring 0.73 m from equilibrium. What is the force constant (spring constant) of the spring
The forces constant (spring constant) of the spring will be 41.09 N/m.
What is spring force?The force required to extend or compress a spring by some distance scales linearly concerning that distance is known as the spring force. Its formula is;
F = kx
The given data in the problem is;
F is the spring force = 30 N
K is the spring constant= ?
x is the displacement of spring = 0.73 m
The spring constant is;
K =F/x
K=30/0.73
K=41.09 N/m
Hence the force constant (spring constant) of the spring will be 41.09 N/m.
To learn more about the spring force refer to the link;
https://brainly.com/question/4291098
#SPJ1
an object has an mass of 15 kg and is falling at a rate of 2.0 m/s what is the momentum?
Answer:
30 kg.m/sExplanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 15 × 2
We have the final answer as
30 kg.m/sHope this helps you
a metallic cube whose each side is 10 cm is subjected to a shearing force of 100 kg. The top force is displaced through 0.25 cm with respect to the bottom. calculate the shearing stress strain and modulus
Answer:
9.8×104Nm−2,0.025,3.92×106Nm−2
Solution :
Here, L=10cm=10×10−2m
F=100kgf=100×9.8N
ΔL=0.25cm=0.25×10−2m,
Shearing stress =FL2=100×9.8(10×10−2) Sheraing strain =ΔLL=0.25×10−210×10−2 = 0.025 Shear Modulus of elasticity, G=Shearing stressShearing strain=9.8×1040.025
=3.92×106Nm−2
Explanation:
It takes a crane 59s to lift a flagstone using 342 W of power. How much work is done on the flagstone?
Answer: The work done on the flagstone is 20178 J
Explanation:
Power is the rate at which work is done . It is equal to the amount of work done divided by the time it takes to do the work.
[tex]Work=Power\times time[/tex]
Given : work = ?
Power = 342 W = 342J/s
Time = 59 s
[tex]Work=342J/s\times 59s=20178J[/tex]
Thus the work done on the flagstone is 20178 J
The circumference of the Earth is
approximately 40,075 kilometers.
How many significant figures are
there in the numerical value?
Answer:
5 sig figures
Explanation:
A(n) ____________ stretch is one done where antagonist muscles are used to stretch the muscles. But the _____________ stretch is one that is done that the muscle needs help from something or someone to stretch the muscle.
Answer: Dynamic - Static Flexibility
Explanation:
A wagon having a mass of 42 kg is accelerated across a level road at 1.50 m/s2. What net force acts on the wagon horizontally?
Answer:
The net force acting on the wagon is 63 N
Explanation:
Net Force
According to the second Newton's law, the net force exerted by an external agent on an object of mass (m) is:
F = m.a
Where (a) is the acceleration of the object.
The wagon has a mass of m=42 Kg is accelerated at [tex]a=1.5\ m/s^2[/tex]. The net force is:
F = 42*1.5
F = 63 N
The net force acting on the wagon is 63 N