Answer:
4 kg → +4 m/s
5 kg → -5 m/s
Explanation:
The law of conservation of momentum states that:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'left side → velocities before collisionright side → velocities after collisionYou'll notice that we have two missing variables: v₁' & v₂'. Assuming this is a perfectly elastic collision, we can use the conservation of kinetic energy to set the initial and final velocities of the individual bodies equal to each other.
v₁ + v₁' = v₂ + v₂'Let's substitute all known variables into the first equation.
(4)(-6) + (5)(3) = (4)v₁' + (5)v₂' -24 + 15 = 4v₁' + 5v₂' -9 = 4v₁' + 5v₂'Let's substitute the known variables into the second equation.
(-6) + v₁' = (3) + v₂' -9 = -v₁' + v₂' 9 = v₁' - v₂'Now we have a system of equations where we can solve for v₁ and v₂.
-9 = 4v₁' + 5v₂' 9 = v₁' - v₂'Use the elimination method and multiply the bottom equation by -4.
-9 = 4v₁' + 5v₂' -36 = -4v₁' + 4v₂'Add the equations together.
-45 = 9v₂'-5 = v₂'The final velocity of the second body (5 kg) is -5 m/s. Substitute this value into one of the equations in the system to find v₁.
9 = v₁' - v₂'9 = v₁' - (-5) 9 = v₁' + 5 4 = v₁'The final velocity of the first body (4 kg) is 4 m/s.
We can verify our answer by making sure that the law of conservation of momentum is followed.
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂' (4)(-6) + (5)(3) = (4)(4) + (5)(-5)-24 + 15 = 16 - 25 -9 = -9The combined momentum of the bodies before the collision is equal to the combined momentum of the bodies after the collision. [✓]
(only answer if you know the answer or I'll report) Solve it w the steps tysm
Answer:
v_f = 20 m / s
Explanation:
For this exercise let's use the relationship between momentum and moment
I = Δp
F t = m v_f - m v₀
as the body starts from rest v₀ = 0
F t = m v_f
v_f = [tex]\frac{F t}{m}[/tex]
let's calculate
v_f = 4 2 / 0.4
v_f = 20 m / s
ayuda porfa es de actividades fisicas
Translate to English please
Who's Buzz Aldrin?
______________
Answer:
Buzz Aldrin was American astronaut, engineer and fighter pilot. he did three spacewalks as pilot of the 1966 Gemini 12 mission, and as the lunar modulbe pilot on the 1969 Apollo 11 mission, he and mission commander Neil Armstrong were the first two people to land on the Moon.
I hope it helps
have have a great day
?????? ?????help please
As the wavelength of light decreases,
What happens
Answer:
Waves. Refraction is an effect that occurs when a light wave, incident at an angle away from the normal, passes a boundary from one medium into another in which there is a change in velocity of the light. ... The wavelength decreases as the light enters the medium and the light wave changes direction.
Explanation:
As a wavelength increases in size, its frequency and energy (E) decrease. From these equations you may realize that as the frequency increases, the wavelength gets shorter. ... Mechanical and electromagnetic waves with long wavelengths contain less energy than waves with short wavelengths.
a car moving with a speed of 10 metre per second is accelerator at the rate of 2 metre per second square find its velocity after 6 seconds
a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )
2 = ( v(2) - 10 ) ÷ ( 6 - 0 )
2 × 6 = v(2) - 10
v(2) = 12 + 10
v(2) = 22 m/s
why does a child look different to both parents?
Answer:
Each parent has a set of genes, and a combination of the genes from the mom and the dad could come together to make the child look totally different.
A child looks different to both parents because children usually look like a "combination" of both their parents. Each parent gives their child their genes. So a child wouldn't look exactly like one parent.
Please Help!!!!
A scientist defects an earthquake wave traveling at a speed of 6.5 km/s in rock with a density of 2.8 g/cubic cm. Based on the data, a scientist can BEST conclude that the wave is traveling through which type of rock?
Answers choices:
A. limestone
B. shale
C. anhydrite
D. dolomite
[tex]what \: is \: refraction \: of \: light \: {?} [/tex]
When light passes through one transparent medium to another transperent medium it bends and that beding of light is know as refraction of light !
what is force? and give its formula
Force is a push or pull that changes or attempts to change the shape or size of an object's position.
Formula Of ForceF = m × a
Here
F= Force
m= mass of object
a= Acceleration
Hope This Helps You ❤️A bird in flight pushes itself upward with
a 7.28 N force. If the bird is climbing at a
constant rate of 1 m/s (no acceleration),
what is the weight of the bird?
[?] N
Answer:
The weight of the bird is equal to 7.28 N.
Explanation:
The upward force acting on the bird = 7.28 N
The bird is climbing at a constant rate of 1 m/s.
We need to find the weight of the bird.
We know that the weight of an object is the force of gravity acting on it. It can be calculated as follows :
W = mg
In this case, 7.28 N of force is acting on the object. Hence, the weight of the bird is equal to 7.28 N.
11. In a new science fiction movie an enemy spaceship explodes. Will an observer on a nearby
moon hear the explosion, see the explosion, or both? Explain
Answer:
Sound is transmitted by longitudinal waves thru air - the observer cannot hear the explosion because no air is present between the moon and the ship
Light is transmitted by electromagnetic waves so the light from the explosion can travel between the moon and the observer.
What is the pressure on your feet if you weigh 800 N and the area covered by your feet is 0.67 m^2?
Answer:
Pressure = 1194.03 N/m
Explanation:
Given the following data;
Force = 800 N
Area = 0.67 m²
To find the pressure on your feet;
Mathematically, pressure is given by the formula;
Pressure = force/area
Pressure = 800/0.67
Pressure = 1194.03 N/m
An object is pushed from rest across a sheet of ice, accelerating at 8.0 m/s^2 [E] over a displacement of 1.05 m [E]. The object then slides at a constant velocity for 6.0 s until it reaches a rough section that causes the object to stop in 2.5 s.
Answer:
[tex]D_T=18.567m[/tex]
Explanation:
From the question we are told that:
Acceleration [tex]a=8.0 m/s^2[/tex]
Displacement [tex]d=1.05 m[/tex]
Initial time [tex]t_1=6.0s[/tex]
Final Time [tex]t_2=2.5s[/tex]
Generally the equation for Velocity of 1.05 travel is mathematically given by
Using Newton's Law of Motion
[tex]V^2=2as[/tex]
[tex]V=\sqrt{2*6*1.05}[/tex]
[tex]V=4.1m/s[/tex]
Generally the equation for Distance traveled before stop is mathematically given by
[tex]d_2=v*t_1[/tex]
[tex]d_2=3.098*4[/tex]
[tex]d_2=12.392[/tex]
Generally the equation for Distance to stop is mathematically given by
Since For this Final section
Final velocity [tex]v_3=0 m/s[/tex]
Initial velocity [tex]u_3=4.1 m/s[/tex]
Therefore
Using Newton's Law of Motion
[tex]-a_3=(4.1)/(2.5)[/tex]
[tex]-a_3=1.64m/s^2[/tex]
Giving
[tex]v_3^2=u^2-2ad_3[/tex]
Therefore
[tex]d_3=\frac{u_3^2}{2ad_3}[/tex]
[tex]d_3=\frac{4.1^2}{2*1.64}[/tex]
[tex]d_3=5.125m[/tex]
Generally the Total Distance Traveled is mathematically given by
[tex]D_T=d_1+d_2+d_3[/tex]
[tex]D_T=5.125m+12.392+1.05 m[/tex]
[tex]D_T=18.567m[/tex]
If a 70 kg man is standing on the surface of the earth, then what is his weight on the moon
Answer:
P = 113.4 N
Explanation:
We know that if it is on the Moon it has another gravity. Therefore, it is known through the International System that the gravity of the Moon is 1.62 m/s², so..
Data:
m = 70 kgg = 1.62 m/s²P = ?Use formula:
[tex]\boxed{\bold{P=m*g}}[/tex]Replace and solve:
[tex]\boxed{\bold{P=70\ kg*1.62\frac{m}{s^{2}}}}[/tex][tex]\boxed{\boxed{\bold{P=113.4\ N}}}[/tex]The man's weight is 113.4 Newtons.
Greetings.
A ball of 10kg falls from rest from a height of 150m, Neglating air resistance, calculate its kinetic energy after falling a distance of 40m. Take g=10m/s
Answer: [tex]3920\ J[/tex]
Explanation:
Given
mass of ball m=10 kg
It is placed at a height of 150 m
It is dropped from the height and allowed to free fall for 40 m
Velocity acquired by the ball during this fall is given by [tex]v^2-u^2=2as[/tex]
Insert u=0, a=g
[tex]\Rightarrow v^2-0=2\times 9.8\times 40\\\Rightarrow v=\sqrt{784}\\\Rightarrow v=28\ m/s[/tex]
Kinetic energy at this instant
[tex]K.E.=\dfrac{1}{2}\times 10\times 28^2\\\\\Rightarrow K.E.=3920\ J[/tex]
What is the net force acting on the airplane?
740 N right -->
700 N right -->
100 N left <--
760 N right -->
[tex]\huge{ \mathrm{ \underline{ Answer }\: \: ✓ }}[/tex]
Total force acting on right side = 800 N
Total force acting on left side :
60 N + 40 N100 NNow, equivalent force acting on the plane is :
greater force - minor force 800 N - 100 N 700 NewtonsAnd the direction of equivalent force will be the direction of greater force, that is right direction.
Hence, Correct option is :
700 N right -->_____________________________
[tex]\mathrm{ \:TeeNForeveR\:}[/tex]
77. The first law of motion applies to
a. Only objects that are moving
b. Only objects that are not moving
C. All objects whether moving or not
No obiect, whether moving or not
A ball of 5.2 kg is swings around on a rope with radius 0.86
meters. If the ball is traveling at 1.8 m/s, what is its angular
momentum? (Round to the nearest hundredth).
Answer:
Explanation:
Angular momentum is L = mvr:
L = (5.2)(1.8)(.86) so
L = 8.05 kg*m/s
Give your answer to 2 dp
When taking off a plane accelerates at 2.7m/s2 down the runway. It accelerates from a stationary position for 25 seconds before leaving the ground. What
is the planes speed when it leaves the ground?
Answer:
67.5
Explanation:
The plane accelerates at 2.7m/s,^2
Time is 25 seconds
The velocity can be calculated as follows
= 25×2.7
= 67.5
Hence the speed f the plane is 67.5
What are the four components to a workout program?
Answer:
Aerobic Fitness. Aerobic fitness improves overall health and well-being. ...
Muscular Fitness. Strength training improves your muscle and bone health and helps with weight loss. ...
Flexibility. Flexibility allows you to move your body freely. ...
Stability and Balance
Explanation:
UP
PLEASE HELP WILL MARK BRAINLIEST *
Damian is texting while driving and ends up slamming into the back end of the car in front of him, which then strikes the car in front of it. By texting while driving, Damian has created a
The answer for this question is negative externality
If 150 Joules of work is needed to move a box 1000 cm, calculate the force that was used?
Answer:
[tex] \large{ \tt{☄ EXPLANATION}} : [/tex]
We're provided : Work ( W ) = 150 J & Displacement ( D ) = 1000 cm & We're asked to find out the force that was used. Firstly, Notice that we're provided the unit of Displacement as centimetre so we have to convert 1000 cm into m. Displacement ( D ) = [tex] \frac{1000}{ 100} = 10[/tex] m[tex] \large{ \tt{♡ \: LET'S \: GET \: STARTED}} : [/tex]
Work is defined as the product of force and Displacement. By definition ,[tex] \large{ \boxed{ \tt{❃ \: WORK ( \: W \: ) = FORCE\: ( \: F) \times DISPLACEMENT \: ( \: D \: )}}}[/tex]
- Plug the values and then simplify !
[tex] \large{ \bf{↦ \: 150 = F \times 10}}[/tex]
[tex] \large{ \bf{↦F\times 10 = 150}}[/tex]
[tex] \large{ \bf{↦ \: F = \frac{150}{10}}} [/tex]
[tex] \large {\bf{↦F = \boxed{ \bf{15} \: N}}}[/tex]
[tex] \boxed{ \boxed{ \large{ \tt{✤ \: OUR \: FINAL \: ANSWER : \boxed{ \bf{15 \: n}}}}}}[/tex]
KEEP READING , KEEP STUDYING , KEEP WORKING , KEEP PUSHING , KEEP TAKING CARE OF YOURSELF. YOUR HARD WORK WILL PAY OFF ! ♪ツ Hope I helped ! ۵
☪ Have a wonderful day / evening ! ☼
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A skipper on a boat notices wave crests passing the anchor chain every 6.0 seconds. The skipper estimates the distance between crests at 30.0
m. What is the speed of the water waves?
Explanation:
given Time Period = 5 Sec
Amplitude = 1m, Wave length = 15 m
using time period = 1/ frequency , Frequency = 1/5 cycles per sec = 0.2 sec = 12 per min
wave speed= Frequency * Wave length
speed = 0.2*15 = 3m/s
13. When you say that something is a factor, you mean that it is an) (factor)
a. problem.
b. obstacle.
c. ingredient.
d. fact
Answer:
C.
A factor of something is an ingredient, or it makes up that thing. For example, in math, we always try to find the factors of a number. We are basically trying to find the "ingredients" of a number.
A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time did the skateboarder spend in the air and how far will be his final position from the bridge?
Answer:
t = 0.714 s and x = 5.0 m
Explanation:
This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed
vₓ = 7.0 m / s
Let's find the time it takes to get to the river
y = y₀ + v_{oy} t - ½ g t²
the initial vertical speed is zero and when it reaches the river its height is zero
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{\frac{2y_o}{g} }[/tex]
t = ra 2 2.5 / 9.8
t = 0.714 s
the distance traveled is
x = vₓ t
x = 7.0 0.714
x = 5.0 m
After being assaulted by flying cannonballs, the knights on the castle walls (12.0 m above the ground) respond by propelling flaming pitch balls at their assailants. One ball lands on the ground at a distance of 81.1 m from the castle walls. If it was launched at an angle of 53.0° above the horizontal, what was its initial speed?
Answer:
28.6 m/s
Explanation:
Using the equation for the range of a projectile,
R = U²sin2θ/g where U = initial speed of flaming pitch balls, θ = launch angle = 53° and g = acceleration due to gravity = 9.8 m/s²
Making U subject of the formula, we have
U = √(gR/sin2θ)
substituting the values of the variables into the equation given that R = 81.1 m, we have
U = √(9.8 m/s² × 81.1 m/sin2(53°))
U = √(794.78 m²/s²/sin106°)
U = √(794.78 m²/s²/0.9613)
U = √(826.78 m²/s²)
U = 28.75 m/s
U ≅ 28.6 m/s
A forensics investigator discharged an assault rifle-replica such that the bullet fired at an angle of 30 (degrees) off the horizontal with an initial velocity
of 28
m/s northwest. What is the maximum height the bullet will reach?
O 14 m/s
10 m
O 30 km
O 0.4351 seconds
Answer:
Initial y-component of speed
Vy = 28 * sin 30 = 14 m/sec vertically
1/2 m Vy^2 = 2 m g h conservation of energy of y-component
h = Vy^2 / (2 * g) = 14^2 / (2 * 9.8) = 10 m
Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by dE dt = q2a2 6pP0c3 where c is the speed of light. (a) Verify that this equation is dimensionally correct. (b) If a proton with a kinetic energy of 6.0 MeV is traveling in a particle accelerator in a circular orbit of radius 0.750 m, what fraction of its energy does it radiate per second? (c) Consider an electron orbiting with the same speed and radius. What fraction of its energy does it radiate per second?
Answer:
a) [tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]
b) [tex]x=\frac{6.0}{4.3*10^{-32}}[/tex]
c) [tex]x'=\frac{6.0}{2.22*10^{-3}}[/tex]
Explanation:
From the question we are told that:
Kinetic energy of Proton[tex]K.E_p= 6.0 MeV[/tex]
Radius [tex]r=0.750[/tex]
Energy [tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]
a)
[tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]
b)
Generally the equation for acceleration of proton is mathematically given by
[tex]a_p=\frac{v^2}{r}[/tex]
Where
Speed of Proton particle is
[tex]V_p=2.12 *10^3 m/s[/tex]
[tex]a_p=\frac{(2.12 *10^3)^2}{0.750}\\a_p=1.27*10^{10}m/s^2[/tex]
Therefore
[tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]
[tex]\frac{dE}{dt}=\frac{9*10^9*(1.6*10^{-19}^2*(1.27*10^{10})^2)}{(3*10^8)^3}[/tex]
[tex]\frac{dE}{dt}=6.881810^{-51}J/s[/tex]
Energy radiated per sec
[tex]mev=\frac{6.881810^{-51}}{1.6*10^{-19}}\\mev=4.3*10^{-32}ev[/tex]
Therefore the Fraction of its energy it radiates per second is given as
[tex]x=\frac{ K.E_p}{mev}\\[/tex]
[tex]x=\frac{6.0}{4.3*10^{-32}}[/tex]
c)
Generally the equation for acceleration of proton is mathematically given by
[tex]a_p=\frac{v^2}{r}[/tex]
Where
Speed of Proton particle is
[tex]V_p=2.5 *10^7 m/s[/tex]
[tex]a_p=\frac{(2.5 *10^7)^2}{0.750}\\a_p=0.33*10^{14}m/s^2[/tex]
Therefore
[tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]
[tex]\frac{dE}{dt}=\frac{9*10^9*(1.6*10^{-19}^2*(0.33*10^{14})^2)}{(3*10^8)^3}[/tex]
[tex]\frac{dE}{dt}=59.26^{-25}J/s[/tex]
Energy radiated per sec
[tex]mev=\frac{59.26^{-25}}{1.6*10^{-19}}\\mev=2.22*10^{-3}ev[/tex]
Therefore the Fraction of its energy it radiates per second is given as
[tex]x'=\frac{ K.E_p}{mev}\\[/tex]
[tex]x'=\frac{6.0}{2.22*10^{-3}}[/tex]