What is the acceleration of the object if a 300-N force acting on a 25 kg object
Given parameters:
Force on the object = 300N
Mass of object = 25kg
Unknown:
Acceleration = ?
Solution:
According to Newton's second law of motion, force is a product of mass and acceleration.
Force = mass x acceleration
Input the parameters and solve for the acceleration;
300 = 25 x acceleration
Acceleration = 12m/s²
The acceleration is 12m/s²
Pierce conducts an experiment in which waves collide in a way that the energy increases. What has occurred?
Answer:
The best and most correct answer among the choices provided by the question is the third choice. What he observed on the waves is the constructive interference. I hope my answer has come to your help. God bless and have a nice day ahead!
Explanation:
Answer:
Pierce conducts an experiment in which waves collide in a way that the energy increases. What has occurred?
refraction
reflection
constructive interference <<<---CORRECT
destructive interference
Explanation:
2021 EDGE
As part of a carnival game, a 5.00 kg target is freely hanging from a very long and very light wire. Contestants can use one of two 1.5 kg balls to try to hit the target and deflect it high enough to win a prize. Ball A will have an elastic collision and bounce back toward you while ball B will have a nearly perfectly inelastic collision, but rather than sticking to the target, the ball will just drop straight downward to the ground after the collision. You can throw each ball with a velocity of 12 m/s. You are the first to try the game and which ball should you throw? Calculate the expected height the target will reach after each is thrown.
Answer:
Height ball A will deflect the target is 2.65 m
Height ball A will deflectthe target is 2.2 m
Ball A, will deflect the target to greater height, thus i will throw ball A
Explanation:
Given;
mass of the target, m₁ = 5.00 kg
mass of the ball, m₂ = 1.5 kg
initial velocity of each throw, u = 12 m/s
Throwing ball A; apply the principle of conservation linear momentum for elastic collision;
m₁u₁ + m₂u₂ = v₁m₁ + v₂m₂
The initial velocity of the target, u₁ = 0
The ball bounced back at the same speed, v₂ = -12 m/s
the velocity of the target after collision, = v₁
0 + 1.5 x 12 = v₁(5) + (-12 x 1.5)
18 = 5v₁ - 18
18 + 18 = 5v₁
36 = 5v₁
v₁ = 36/5
v₁ = 7.2 m/s
The vertical height reached by the target is given by;
v₁² = u₁² + 2gh
v₁² = 0 + 2gh
v₁² = 2gh
h = v₁² / 2g
h = (7.2)² / (2 x 9.8)
h = 2.65 m
Throwing ball B; apply the principle of conservation energy for the inelastic collision;
the kinetic energy of the ball will be converted to the potential energy of the target.
¹/₂m₁u₂² = mgh
¹/₂(1.5)(12)² = (5 x 9.8)h
108 = 49h
h = 108 / 49
h = 2.2 m
Ball A will deflect the target to greater height, thus i will throw ball A.
During a baseball game, a hitter strikes the ball with a bat. When this happens, the ball
and the bat exert a force on each other. Why does the ball accelerate away from the bat
more than the bat accelerates away from the ball?
a. The ball has less mass, so it exerts less force on the bat.
b. The bat exerts more force than the ball because the batter is exerting a force
on the bat as it hits the ball.
c. The ball has less mass, so the equal force on the ball and the bat causes
greater acceleration of the ball.
d. The ball has a greater velocity before the collision, so the force affects the ball
more than it affects the bat.
Answer:
The answer is c
Explanation:
wavelength = 5.4 cm and frequency = 9.4 Hz
What is the speed
2) A boat travels 12.0 m while it reduces its velocity from 9.5 m/s to 5.5 m/s. What is the
boat's acceleration while it travels that distance?
Answer:
2.5m/s^2
Explanation:
Note that the boat is reducing its speed. It is having negative acceleration or deceleration.
V^2 = u^2 -2as ( minus sign is used because of speed is reduced)
Given that,
s = 12 m
v = 5.5 m/s
u = 9.5 m/s
a = (v^2 - u^2) ÷ (-2s)
a= ( 5.5^2 - 9.5^2) ÷ ( -2× 12)
a = 2.5 m/s^2
The acceleration of the boat is -2.5 m/s²
The given paramters;
distance traveled by the boat, d = 12 m
initial velocity of the boat, u = 9.5 m/s
final velocity of the boat, v = 5.5 m/s
The acceleration of the boat is calculated as;
[tex]v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(5.5)^2 -(9.5)^2 }{2(12)} \\\\a = -2.5 \ m/s^2[/tex]
Thus, the deceleration of the boat is 2.5 m/s²
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A cheetah can run at a maximum speed
91 km/h and a gazelle can run at a maximum speed of 72.7 km/h.
If both animals are running at full speed,
with the gazelle 87.5 m ahead, how long before
the cheetah catches its prey?
Answer in units of s.
Answer:
Approximately 17.21 seconds
Explanation:
With subtraction, we have the gazelle 18.3 km/h slower than the cheetah, which is about 5.08333 m/s. As the gazelle is 87.5 meters ahead of the cheetah, 87.5 divided by 5.083333333 is about 17.21 seconds.
is newton's first law true on earth?
Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
THIS LAW IS TRUE AS IT ALSO HAVE A REAL LIFE EXAMPLE.
Examples of Newton's 1st Law : If you slide a hockey puck on ice, eventually it will stop, because of friction on the ice. It will also stop if it hits something, like a player's stick or a goalpost.
A bus rolls to a stop along a horizontal road without the driver applying the brakes,
What answer choice BEST explains why this occurred?
A. The natural state of the bus is not to be in motion.
B. The bus ran out of momentum.
C. The force of gravity slowed the bus until it stopped.
D. The opposing force of friction stopped the train.
Answer:
should be d because friction allows things to go faster or slower
When a heavy football player and a light one run into each other, who exerts more force?
Answer:
When a heavy football player and a light one run into each other, does the lighter player really exert as much force on the heavy player s the heavy player exerts on the light one. Yes. The interaction between the two players, the force each exerts on the other have equal strength.
Explanation:
Which pairs of elements could not react to form an Ionic compound
Answer:
Sodium and calcium
Explanation:
Answer:
Ionic bond is the name given to one of the three ways in which atoms can interact with each other. The other forms of interaction between atoms are the covalent bond, which occurs between atoms of ametals, hydrogens, or ametal and hydrogen, and the metallic bond, which occurs only between atoms of the same metal.
Explanation
:The atoms of the chemical elements that participate in the ionic bond must present, necessarily, the nature of gaining or losing electrons, thus, the ionic bond can occur between:
a metal and an ametal;
a metal and hydrogen.
Sodium: metallic element, as it has the characteristic of losing electron; belonging to the IA family, atomic number 11, with an electron in the valence shell.
Chlorine: ametalic element, as it has the characteristic of gaining electrons; belonging to the VIIA family, with atomic number 17.
can you describe your own perspective whats Physical Science all about? PLS GUYS HELP
Answer:
Physical science is the study of the inorganic world. That is, it does not study living things. The four main branches of physical science are astronomy, physics, chemistry, and the Earth sciences, which include meteorology and geology.
PLEASE HELP
Discussion: If you put something like a piece of cardboard between a magnet and an iron nail, the magnet still holds the nail in place, even though the magnet is not touching the nail Explain how that happens. Use the words induce, magnetic field, permanent magnet and temporary magnet in your response.
PLZZZ HURRY!!! I will freind and make brainliest
What are the basic building blocks of matter?
electrons
atoms
molecules
compounds
Answer:
an atom
Explanation: atom then matter
Answer:
the basic building blocks that make up matter are called atoms.
Electricity & Magnetism
4
Electricity can be used to produce powerful forces.
What type of energy is electricity converted to in an electromagnet?
A. sound energy
B.
heat energy
C. light energy
D. magnetic energy
An archer shoots an arrow with vertical velocity of 10 m/s and horizontal velocity of
30m/s. What is the maximum height the arrow reaches?
Answer:
5.1 m
Explanation:
Given in the y direction:
v₀ = 10 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (10 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 5.1 m
I need help pls !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
figure d
Explanation:
A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a speed of vi = 2.60 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Figure e).
The right end of a horizontal spring labeled k is attached to a wall. Five images show five configurations as a block labeled m approaches, compresses, and then moves away from the spring.
In figure a, the block is to the left of the spring, and an arrow above the block points to the right.
In figure b, the block is just touching the uncompressed spring, and an arrow labeled vector vi above the block points to the right.
In figure c, the block has compressed the spring by a distance d, and a label indicates vector vf = 0.
In figure d, the block is just touching the uncompressed spring, and an arrow labeled vector v above the block points to the left.
In figure e, the block is a distance D away from the spring, and a label indicates vector v = 0.
(a)
Find the distance of compression d (in m).
m
(b)
Find the speed v (in m/s) at the unstretched position when the object is moving to the left (Figure d).
m/s
(c)
Find the distance D (in m) where the object comes to rest.
m
(d)
What If? If the object becomes attached securely to the end of the spring when it makes contact, what is the new value of the distance D (in m) at which the object will come to rest after moving to the left?
m
Answer:
(a) Approximately [tex]0.335\; \rm m[/tex].
(b) Approximately [tex]1.86\; \rm m\cdot s^{-1}[/tex].
(c) Approximately [tex]0.707\; \rm m[/tex].
(d) Approximately [tex]0.228\; \rm m[/tex].
Explanation:
[tex]v_i[/tex] denotes the velocity of the object in the first diagram right before it came into contact with the spring. Let [tex]m[/tex] denote the mass of the block. Let [tex]\mu[/tex] denote the constant of kinetic friction between the object and the surface. Let [tex]g[/tex] denote the constant of gravitational acceleration.Let [tex]k[/tex] denote the spring constant of this spring.(a)Consider the conversion of energy in this object-spring system.
First diagram: Right before the object came into contact with the spring, the object carries kinetic energy [tex]\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^2[/tex].
Second diagram: As the object moves towards the position in the third diagram, the spring gains elastic potential energy. At the same time, the object loses energy due to friction.
Third diagram: After the velocity of the object becomes zero, it has moved a distance of [tex]D[/tex] and compressed the spring by the same distance.
Energy lost to friction: [tex]\underbrace{(\mu \cdot m \cdot g)}_{\text{friction}} \cdot D[/tex]. Elastic potential energy that the spring has gained: [tex]\displaystyle \frac{1}{2}\,k\, D^2[/tex].The sum of these two energies should match the initial kinetic energy of the object (before it comes into contact with the spring.) That is:
[tex]\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^{2} = (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2[/tex].
Assume that [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex]. In the equation above, all symbols other than [tex]D[/tex] have known values:
[tex]m =1.10\; \rm kg[/tex].[tex]v_i = 2.60\; \rm m \cdot s^{-1}[/tex].[tex]\mu = 0.250[/tex].[tex]g = 9.81\; \rm m \cdot s^{-2}[/tex].[tex]k = 50.0\; \rm N \cdot m^{-1}[/tex].Substitute in the known values to obtain an equation for [tex]D[/tex] (where the unit of [tex]D\![/tex] is [tex]m[/tex].)
[tex]3.178 = 2.69775\, D + 25\, D^2[/tex].
[tex]2.69775\, D + 25\, D^2 + 3.178 = 0[/tex].
Simplify and solve for [tex]D[/tex]. Note that [tex]D > 0[/tex] because the energy lost to friction should be greater than zero.
[tex]D \approx 0.335\; \rm m[/tex].
(b)The energy of the object-spring system in the third diagram is the same as the elastic potential energy of the spring:
[tex]\displaystyle \frac{1}{2}\,k\, D^2 \approx 2.81\; \rm J[/tex].
As the object moves to the left, part of that energy will be lost to friction:
[tex](\mu \cdot m \cdot g) \, D \approx 0.905\; \rm J[/tex].
The rest will become the kinetic energy of that block by the time the block reaches the position in the fourth diagram:
[tex]2.81\; \rm J - 0.905\; \rm J \approx 1.91\; \rm J[/tex].
Calculate the velocity corresponding to that kinetic energy:
[tex]\displaystyle v =\sqrt{\frac{2\, (\text{Kinetic Energy})}{m}} \approx 1.86\; \rm m \cdot s^{-1}[/tex].
(c)As the object moves from the position in the fourth diagram to the position in the fifth, all its kinetic energy ([tex]1.91\; \rm J[/tex]) would be lost to friction.
How far would the object need to move on the surface to lose that much energy to friction? Again, the size of the friction force is [tex]\mu \cdot m \cdot g[/tex].
[tex]\displaystyle (\text{Distance Travelled}) = \frac{\text{(Work Done by friction)}}{\text{(Size of the Friction Force)}} \approx0.707\; \rm m[/tex].
(d)Similar to (a), solving (d) involves another quadratic equation about [tex]D[/tex].
Left-hand side of the equation: kinetic energy of the object (as in the fourth diagram,) [tex]1.91\; \rm J[/tex].
Right-hand side of the equation: energy lost to friction, plus the gain in the elastic potential energy of the spring.
[tex]\displaystyle {1.91\; \rm J} \approx (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2[/tex].
[tex]25\, D^2 + 2.69775\, D - 1.90811\approx 0[/tex].
Again, [tex]D > 0[/tex] because the energy lost to friction is greater than zero.
[tex]D \approx 0.228\; \rm m[/tex].
The energy transferred between the object and the spring as a closed system, therefore, conserved are;
(a) The distance of compression, d ≈ 0.3354 meters
(b) The speed in the un-stretched position wen the object is sliding to the left, v ≈ 1.8623 m/s
(c) The distance where the object comes to rest, D ≈ 0.7071 m
(d) The distance the object will come to rest attached to the spring, D ≈ 0.2278 m
The reason the above values are correct are as follows;
The known parameters are;
Mass of the object, m₁ = 1.10 kg
Coefficient of friction, μ = 0.250
The initial speed of the object, [tex]v_i[/tex] = 2.60 m/s
Force constant of the spring, K = 50.0 N/m
Distance the spring is compressed by the object = d
(a) Conservation of energy principle
[tex]Kinetic \ energy = \dfrac{1}{2} \cdot m\cdot v^2[/tex]
Work done = Force × Distance
Friction force, [tex]F_f[/tex] = W × μ
Weight, W = m·g
Weight = Mass × Acceleration
Energy transferred by object = Work done by spring + Work done by friction
[tex]Energy \ transferred \ by \ object = Kinetic \ energy = \dfrac{1}{2} \times 1.10\times 2.60^2 = 3.718[/tex]
Energy transferred by object = 3.718 J
[tex]Work \ done \ by \ spring = \dfrac{1}{2} \cdot k\cdot x^2[/tex]
[tex]Work \ by \ spring \ to \ bring \ object \ to \ rest, \ W_{spring} = \dfrac{1}{2} \times 50\times d^2[/tex]
[tex]W_{spring}[/tex] = 25·d²
Work done by friction, [tex]W_{friction}[/tex] = 1.10×9.81×0.250×d = 2.69775·d
Therefore;
3.718 = 25·d² + 2.69775·d
25·d² + 2.69775·d - 3.718 = 0
Solving gives
The distance of the compression d ≈ 0.3354 m
(b) The energy given by the spring = 25·d²
The work done by friction, [tex]W_{friction}[/tex] = 2.69775·d
Kinetic energy given to object = 0.55·v²
0.55·v² = 25·d² - 2.69775·d
0.55·v² = 25×0.3354² - 2.69775×0.3354
∴ v = √(3.4682) = 1.8623
The velocity of the object at the un stretched position, v ≈ 1.8623 m/s
(c) The kinetic energy, K.E. of the object on the way left is given as follows;
K.E. = 0.5 × 1.10 kg × 3.4682 m²/s² = 1.90751 J
The work done by friction before object comes to rest = 2.69775·D
[tex]D = \dfrac{1.90751 \, J}{2.69775 \, N} \approx 0.7071 \, m[/tex]
The distance where the object comes to rest, D ≈ 0.7071 m
(d) The work done on spring, [tex]W_{spring}[/tex] = 25·D'²
Work done on friction, [tex]W_{friction}[/tex] = 2.69775·D'
Kinetic energy of object, K.E. ≈ 1.90751 J
K.E. = [tex]W_{spring}[/tex] + [tex]W_{friction}[/tex]
1.90751 ≈ 25·D'² + 2.6775·D'
25·D'² + 2.6775·D' - 1.90751 = 0
Solving with a graphing calculator gives;
D' ≈ 0.2278 m
The new value of the distance D = 0.2278 m
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A block is attached to one end of a spring with the other end of the spring fixed to a wall. The block is vibrating horizontally on a frictionless surface. If the mass of the block is 4.0 kg, the spring constant is k
Complete Question
A block is attached to one end of a spring with the other end of the spring fixed to a wall. The block is vibrating horizontally on a frictionless surface. If the mass of the block is 4.0 kg, the spring constant is k = 100 N/m, and the maximum distance of the block from the equilibrium position is 20 cm, what is the speed of the block at an instant when it is a distance of 16 cm from the equilibrium position?
Answer:
The velocity is [tex]v = 0.6 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the block is m = 4.0 kg
The spring constant is k = 100 N/m
The maximum distance of the block from equilibrium position is d = 20 cm =0.20 m
The distance considered is [tex]d_k = 16 \ cm = 0.16 \ m[/tex]
Generally the maximum energy stored in the spring is mathematically represented as
[tex]E = \frac{1}{2} * k * d^2[/tex]
=> [tex]E = \frac{1}{2} *100 * 0.2^2[/tex]
=> [tex]E = 2.0 \ J[/tex]
Gnerally according to the law of energy conservation
The energy maximum energy of the spring = energy of the spring at [tex]d_k[/tex] + energy of the block at [tex]d_k[/tex]
Here energy of the block at [tex]d_k[/tex] is mathematically represented as
[tex]K_1 = \frac{1}{2} mv^2[/tex]
=> [tex]K_1 = \frac{1}{2} * 4* v^2[/tex]
=> [tex]K_1 = 2v^2[/tex]
Generally the energy of the spring at [tex]d_k[/tex] is mathematically represented as
[tex]E_2 =\frac{1}{2} * k * d_k^2[/tex]
=> [tex]E_2 =\frac{1}{2} * 100 * (0.16)^2[/tex]
=> [tex]E_2 =1.28 \ J[/tex]
So
[tex]2.0 = 1.28 + 2v^2[/tex]
=> [tex]v = 0.6 \ m/s[/tex]
The lungs are large organs which contain smaller, expandable sacs.
These sacs greatly increase the surface area of the lungs.
Why is a large surface area important to the function of the lungs? A. Large amounts of liquid wastes and fatty tissue must be stored in the lungs. B. Oxygen in air taken in by the lungs must move quickly through the lung tissue and into the blood. C. The lungs must be very sturdy and rigid so they cannot move. D. The lungs must be able to filter oxygen from water while the organism is swimming.
Answer:
B. It is B because without this much space for oxygen and blood to flow and go through, we would not survive.
Explanation:
A boy pushes a box with a force of 150 N at an angle of 40 with a flat floor. What component of his force is directed downward , or into the floor . PLEASE ANSWER!!!!!
Answer:
[tex]F_y=96.4N[/tex]
Explanation:
Hello.
In this case, considering the force diagram shown on the attached picture, we can see that the component of his force is directed downwards is:
[tex]F_y=F\times sin (\theta)[/tex]
Because the other component is the horizontal one:
[tex]F_x=F\times cos(\theta)[/tex]
In this case, the y-component force turns out:
[tex]F_y=150N\times sin (40\°)\\\\F_y=96.4N[/tex]
Moreover, the x-component force is also computed if required:
[tex]F_x=150N\times cos(40\°)\\\\F_x=114.9N[/tex]
Best regards.
An unbalanced force of 23 N is applied to a 13 kg mass. What is the acceleration of the
mass?
Answer:
1.77 m/s^2
Explanation:
since it is unbalanced there must be net force(resultant force)
[tex]net \: force \: = accleration \: \times mass[/tex]
net force = 23N
mass = 13kg
a = x
x = 23/13
= 1.769...
approximately = 1.77 m/s^2
A jet must reach a velocity of 75 m/s for takeoff. If the runway is
2100 meters long, what must the constant acceleration be?
Answer:
[tex]a=1.33\ m/s^2[/tex]
Explanation:
Given that,
Initialy, the jet is at rest, u = 0
Final velocity of the jet, v = 75 m/s
Distance, d = 2100 m
We need to find the acceleration of the jet. It is based on the concept of equation of kinematics. Using third equation of motion, w get :
[tex]v^2-u^2=2ad[/tex]
a is acceleration
[tex]75^2=2a\times 2100\\\\a=\dfrac{75^2}{2\times 2100}\\\\a=1.33\ m/s^2[/tex]
So, the acceleration of the jet is [tex]1.33\ m/s^2[/tex].
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Explain why your weight would change if you went to the moon, but your mass wouldn’t.
The moon's gravitation force is determined by the mass and the size of the moon. Since the moon has significantly less mass than the Earth, it will not pull objects toward itself at the strength that Earth will.
What are four metals other than iron that can be made to exhibit magnetic properties? Fill in the blank. Every magnet has _______ unlike poles. The _______ within a magnet lies between the north and south poles. If the north pole of a bar magnet is brought near the _______ pole of another magnet, the two magnets will repel one another. A material that’s attracted by a magnet but doesn’t necessarily become a magnet itself is called a/an _______ material.
Answer:
Four metals other than iron that can be made to exhibit magnetic properties are nickel, cobalt, manganese, and chromium.
two
neutral region
north
magnetic
Explanation:
PennFoster
The four metals other than iron that can be made to exhibit magnetic properties are:
NickelCobaltManganeseChromium.
Furthermore, the words which can be used to accurately complete the sentences below are:
Two Neutral region North MagneticThe magnet has two unlike poles and in the neutral region, there is the South and North poles.
The North and South poles repel each other and a magnet can attract any material which contains iron materials.
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The resultant force is equel to the.......of all the force
A) sum
B) product
C) subtraction
D) Division
Answer:
A) sum
Explanation:
That's the answer bro
Answer:
the answer is C subtraction
The nervous system has two distinct branches. They are the:
Answer:
central nervous system
peripheral nervous system
Explanation:
The nervous system has two main parts: The central nervous system is made up of the brain and spinal cord. The peripheral nervous system is made up of nerves that branch off from the spinal cord and extend to all parts of the body.Oct 1, 2018
Answer:
The central nervous system
The peripheral nervous system
Explanation:
The central nervous system (CNS) is the brain and spinal cord, and the peripheral nervous system (PNS) is everything else
I will give you branilest!
A race car exerts 18,344 N while the car travels at an acceleration of 92.54 m/s2. What is the mass of the car? (Use the formula m=F/a)
the answer is 198.23
Force(f) = 18344 N
Acceleration(a) = 92.54 m/s²
we know that m=f/a
m=18344/92.54</p><p>
ma 198.23 kg
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Engineers are using computer models to study train collisions to design safer
train cars. They start by modeling an elastic collision between two train cars
traveling toward each other. Car 1 is traveling east at 6 m/s and has a mass
of 3,154 kg, Car 2 is traveling west at 23 m/s and has a mass of 8,296 kg.
After the collision, car 1 has a final velocity of 7 m/s west. What is the final
velocity of car 2?
A. 18 m/s east
B. 23 m/s west
C. 23 m/s east
D. 18 m/s west
SUN
Answer:
18 m/s west
Explanation:
The correct option is C. 23 m/s east
What is law of conservation of momentum ?Conservation of momentum states that For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed.
using conservation of momentum
m1v1 + m2v2 = m1 v1' + m2 v2'
Given
m1 = 3154 kg
v1= 6m/s
m2 = 8296 kg
v2 = 23 m/s
v1' = 7m/s
v2' = ?
3154 * 6 + 8296 * 23 = 3154 * 7 + 8296 v2'
209732 - 22078 = 8296 v2'
187654 = 8296 v2'
v2' = 22.62 ≈ 23 m/s east
correct option is C. 23 m/s east
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Your car is initially at rest when your hit that gas and the car begins to accelerate at a rate of 2.857 m/s/s. The acceleration lasts for 15.5 s. What is the final speed of the car and how much ground does it cover during this acceleration?
Given parameters:
Initial velocity = 0m/s
Acceleration = 2.857m/s²
Time = 15.5s
Unknown:
Final speed of the car = ?
Solution:
We use one of the motion equations to solve this problem;
V = U + at
Where V is the final velocity
U is the initial velocity
a is the acceleration
t is the time taken
V = 0 + 2.857 x 15.5 = 44.28m/s