Answer:
Actually the answer is C, To align the coordinate axes with the acceleration of the object on the plane.
Describe Kinetic Energy.
A rally car accelerates from 10 m s−1 to 58 m s−1 in 8 seconds as it moves along a straight road. Given that the acceleration is constant, what is the acceleration of the car?
Answer:
a = 6 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics
[tex]v_{f} =v_{o} +a*t[/tex]
where:
Vf = final velocity = 58 [m/s]
Vo = initial velocity = 10 [m/s]
a = acceleration [m/s²]
t = time = 8 [s]
Now replacing:
58 = 10 + (a*8)
58 - 10 = 8*a
48 = 8*a
a = 6 [m/s²]
Compare the momentum of a 7160 kg truck moving at 5.00 m/s to the
momentum of a 2240 kg car moving at 15.0 m/s.
Answer:
2,200
Explanation:
truck=35800,car=33600
A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s the instant she leaves the cliff, determine the following.
Her gravitational potential energy relative to the water surface when she leaves the cliff
Her kinetic energy when she leaves the cliff
Her total mechanical energy relative to the water surface when she leaves the cliff
Her total mechanical energy relative to the water surface just before she enters the water.
The speed at which she enters the water.
Answer:
1) 2197.44 J
2) 0 J
3) 2197.44 J = Constant
4) 2197.44 J
5) Approximately 8.86 m/s
Explanation:
The given parameters are;
The mass of the diver, m = 56 kg
The height of the cliff, h = 4.0 m
The speed with which the diver is moving, vₓ = 8.0 m/s
The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g
1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J
2) The kinetic energy = 1/2·m·u²
Where;
u = Her initial velocity = 0 when she just leaves the cliff
Therefore;
Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J
3) The total mechanical energy = Kinetic energy + Potential energy
The total mechanical energy is constant
Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant
4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J
5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h
Where;
u = The initial velocity at the top of the cliff before she jumps= 0 m/s
∴ v² = 0² + 2 × 9.81 × 4 = 78.48
v = √78.48 ≈ 8.86 m/s
The speed with which she enters the water, v ≈ 8.86 m/s
1) Her gravitational potential energy relative to the water surface when she leaves the cliff is; GPE(leaves cliff) = 2195.2 J
2) Her kinetic energy when she leaves the cliff is; KE = 0J
3) Her total mechanical energy relative to the water surface when she leaves the cliff is; ME_total = 2195.2 J
4) Her total mechanical energy relative to the water surface just before she enters the water is; ME_total = 2195.2 J
5) The speed at which she enters the water is; v = 8.85 m/s
We are given;
Mass of the diver; m = 56 kg
Height of the cliff; h = 4 m
Speed at which she is moving; vₓ = 8 m/s
1) Formula for gravitational potential energy is;
GPE = mgh
where;
m is mass
g is acceleration due to gravity
h is height
Thus;
GPE = 56 × 4 × 9.8
GPE(leaves cliff) = 2195.2 J
2) The formula for kinetic energy when she leaves the cliff is;
KE = ¹/₂mu²
Where;
m is mass
u = initial velocity = 0 m/s
Thus;
KE = ¹/₂ × 56 × 0²
KE(leaves cliff) = 0 J
3) The total mechanical energy relative to the water surface when she leaves the cliffis;
ME_total = GPE(leaves cliff) + KE(leaves cliff)
Thus;
ME_total = 2195.2 + 0
ME_total = 2195.2 J
4) Her total mechanical energy relative to the water surface just before she enters the water is same as that when she leaves the cliff = 2195.2 J
5) The speed with which she enters the water, v, is gotten from newtons third equation of motion;
v² = u² + 2gh
Thus;
v² = 0² + (2 × 9.8 × 4)
v² = 78.4
v = √78.4
v = 8.85 m/s
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Which terms describe the motion of most objects in our solar system?
This is a Earth Science question
Answer:Geography, Earth Science, Astronomy
Rotation describes the circular motion of an object around its center.
Explanation:
Answer:
it is Cyclic and predictable
What type of substances is an alloy?
A) Moleclue?
B) Mixture?
C) Compound?
D) Element?
In a tug of war game, team A is very strong and is pulling team B with a speed of 1 foot per second. In this situation… *
A. Only team A is doing work
B.Only team B is doing work
C. Team A and team B are doing work
D. Neither team A nor team B is doing work
Which is one limitation to technological design?
being innovative
designing a solution
having product appeal
identifying a problem
Answer:
c. Having product appeal
Explanation:
Answer:
c
Explanation:
Which does not contain a lens?
a mirror
binoculars
an eye
a camera
Help Me get A good grade PLZ
a supersonic aircraft travels faster than the speed of sound. What might be the top speed of such an aircraft? 200 Kilometers per hour 500 Kilometers per hour O 1000 Kilometers per hour O 1500 Kilometers per hour
1500 kpm
When blocking in football, why does a defending lineman often attempt to get his body under that of his opponent and push forward?
Answer:
because he is trying to break past you and sack your quarterback
Explanation:
Which statement best describes the difference between acceleration and velocity?
A. acceleration only shows a change in direction, but velocity shows a change in speed.
B .acceleration depends on direction, but velocity does not.
C. acceleration is a change in velocity, but velocity is speed in a given direction.
D. acceleration only shows change in speed, but velocity only shows a change in direction.
Two positively charged particles are 0.03 m apart. The first
particle has a charge of 8.1 C, and the second a charge of 2.6
C. Which change would lead to the smallest increase in the
- electric force between the two particles?
A. Reduce the charge on the first particle by one half.
O B. Double the distance between the particles.
O C. Double the charge on the second particle.
OD. Reduce the distance between the particles by one
half
The correct option is option C.
Doubling the charge on the second particle produces the smallest increase in force.
Electrostatic force:Let the given charges be Q and q such that:
Q = 8.1C and q = 2.6C
separated by a distance of R = 0.03m
Now, we have to calculate the minimum increase in electric force.
The electrostatic force is given by:
F = kQq/R²
Option A is wrong since reducing the charge will reduce the force.
Option B is wrong since the force is inversely proportional to the distance, so increasing the distance will decrease the force.
Option C ⇒ Double the charge on the second particle:
F = 2kQq/R²
Option D ⇒ Reduce the distance between the particles by one half:
So, the new distance between the charges is R/2, then
F = kQq/(R/2)²
F = 4kQq/R²
Clearly, the operation in option C gives the smallest increase in the force.
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What is the Range of the projectile motion?
Answer:
Range of projectile, R
For projection above ground surface, the range of the angle of projection with respect to horizontal direction, θ, is 0° ≤ θ ≤ 90° and the corresponding range of 2θ is 0° ≤ 2θ ≤ 180°.S
hope you're looking for this.
this is also for a Digital Electronics class
Answer:
Rt = 908.25 [ohm]
Explanation:
In order to solve this problem, we must remember that the resistors connected in series are added up arithmetically.
In this case, R2 and R3 are in series therefore.
R₂₃ = 200 + 470
R₂₃ = 670 [ohm]
Now this new resistor (R₂₃) is connected in parallel with the resistor R4. therefore we must use the following arithmetic expression, to add resistances in parallel.
[tex]\frac{1}{R_{4-23} }= \frac{1}{R_{4}}+\frac{1}{R_{23} } \\\frac{1}{R_{4-23} }=\frac{1}{1800}+\frac{1}{670} \\R_{4-23}=488.25[ohm][/tex]
In this way R₁, R₅ and R₄₋₂₃ are connected in series.
Rt = R₁ + R₅ + R₄₋₂₃
Rt = 150 + 270 + 488.25
Rt = 908.25 [ohm]
Whenever an action and reaction occur, momentum is what?
Answer:
I'm not sure what your asking for
reaction and law
Explanation:
the newton third law of motion
PLS HELP WILL GIVE 50 POINTS
How is energy sent from Earth to space?
conduction and convection
radiation and conduction
reflection and conduction
radiation and reflection
Answer:
I'm pretty sure its radiation and reflection. Some of the Sun's energy gets reflected off into space. (I could be wrong though)
Answer:
radiation and reflection
hope helps you
have a nice day
Explanation:
hope this will surely be helpful to you
A boxer hits punching bag and gives it a change in momentum of 12 kg multiplied by m divided by s over 7.0ms what is the magnitude of the net force on the punching bag
Answer: 1700
Explanation:
A tough kid on a tricycle has combined mass 32 kg. She starts from rest and travels 3.5 m in 2.5 s. What is the net force used to accelerate. (Hint: d=(1/2)at2)
Answer:
F = 35.84NExplanation:
Force = mass * acceleration (Newton's second law of motion)
Given
Mass = 32kg
Get the acceleration using the expression d = 1/2at^2
3.5 = 1/2 (a)2.5^2
3.5 = 3.125a
a = 3.5/3.125
a = 1.12m/s^2
Get the net force;
F = 3.2 * 1.12
F = 35.84N
Hence the net force used to accelerate is 35.84N
explain the difference in how particles are arranged in a solid,liquids and gases
Answer:
hope this helps
Explanation:
solid particles are arranged according to the size of the object
while liquid particles are arranged how they move freely together in the object they are in.
while gas particles are arranged scatterly like they spread.
A race is held between a sports car and a motorcycle. The sports car can accelerate at 5.0 m/s^2 and the motorcycle can accelerate at 8.0 m/s^2. The two vehicles start the race at the same time and accelerate from rest. After 5.0 s, how fast is the sports car going? After 6.0 s, how distance will the motorcycle have gone? To make the race fair, the sports car starts 50.0 m ahead of the motorcycle. If the course is 200.0 m long, which vehicle wins the race?
Answer:
Vf₁ = 30 m/s
s₂ = 90 m
car wins the race.
Explanation:
To find the speed of car after 5 s, we use 1st equation of motion:
Vf₁ = Vi₁ + a₁t₁
where,
Vf₁ = Final Speed of Car = ?
Vi₁ = Initial Speed of Car = 0 m/s
a₁ = acceleration of car = 5 m/s²
t₁ = time = 5 s
Therefore,
Vf₁ = 0 m/s + (5 m/s²)(5 s)
Vf₁ = 25 m/s
To find the distance of motorcycle after 6 s, we use 2nd equation of motion:
s₂ = Vi₂ t₂ + (1/2)a₂t₂²
where,
s₂ = distance covered by the motorcycle = ?
Vi₁ = Initial Speed of motorcycle = 0 m/s
a₂ = acceleration of motorcycle = 8 m/s²
t₂ = time = 6 s
Therefore,
s₂ = (0 m/s)(6 s) + (1/2)(5 m/s²)(6 s)²
s₂ = 90 m
We can use 2nd equation of motion to find time taken by each car and motorcycle to reach the finish point:
For Car:
s₁ = Vi₁ t₁ + (1/2)a₁t₁²
where,
s₁ = 200 m - 50 m = 150 m (since, car starts 50 m ahead)
Therefore.
150 m = (0 m/s)(t₁) + (1/2)(5 m/s²)t₁²
t₁ = √60 s²
t₁ = 7.74 s
For Motorcycle:
s₂ = Vi₂ t₂ + (1/2)a₂t₂²
where,
s₁ = 200 m
Therefore.
200 m = (0 m/s)(t₂) + (1/2)(5 m/s²)t₂²
t₂ = √80 s²
t₂ = 8.94 s
Since, the car takes less time to reach finish line.
Therefore, car wins the race.
Just helping as usual what about you
Answer:
same i just love helping people out
god bless you have a blessed day
Explanation:
Answer:
I do too
Explanation:
I like to help others especially if it has been filled in by struggling in a certain concept of a subject.
Do u agree?
A 12.0 kg box is being pulled along level ground at constant velocity by a horizontal force of 38.0 N. What is the coefficient of kinetic friction between the box and the floor?
The coefficient of kinetic friction between the box and the floor is 0.323.
By Newton's Law of Motion we understand that an object is at equilibrium if and only if it is at rest or it is moving at constant velocity. If a horizontal force is applied on the box and it is moving, then a force with equal magnitude and opposed to that force must exists, which corresponds to the kinetic friction. Let suppose that the box is moving on a horizontal ground.
The equation of equilibrium for the box is described below:
[tex]\Sigma F = P - \mu_{k}\cdot m\cdot g = 0[/tex] (1)
Where:
[tex]P[/tex] - External force, in newtons.[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.[tex]m[/tex] - Mass, in kilograms.[tex]g[/tex] - Gravitational constant, in meters per square second.An expression for the kinetic coefficient of friction is derived by clearing the variable in (1):
[tex]\mu_{k} = \frac{P}{m\cdot g}[/tex] (2)
If we know that [tex]P = 38\,N[/tex], [tex]m = 12\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the kinetic coefficient of friction is:
[tex]\mu_{k} = \frac{38\,N}{(12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{k} = 0.323[/tex]
The coefficient of kinetic friction between the box and the floor is 0.323.
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A ferry boat is 4.0 m wide and 6.0 m long. When a truck pulls onto it,
the boat sinks 4.00 cm in the water, What is the weight of the boat?
what is the slope of the line
Answer:
[tex]m=\frac{1}{3}[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDASAlgebra I
Slope Formula: [tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]Explanation:
Step 1: Define
Point (0, -1)
Point (3, 0)
Step 2: Find slope m
Substitute: [tex]m=\frac{0+1}{3-0}[/tex]Add/Subtract: [tex]m=\frac{1}{3}[/tex]linear expansivity?
Linear expansivity, area expansivity and volume or cubic expansivity are
The target heart rate zone is:
a
the level your recovery heart rate should reach.
b
where you want your exercise heart rate.
c
where you want your resting heart rate.
d
your maximum heart rate.
The target heart rate zone is where you want your exercise heart rate to be. Option B is correct.
The target heart rate zone is a specific range of heart rates that you should aim to achieve during exercise to maximize the benefits of your workout. This range is typically calculated based on a percentage of your maximum heart rate. For example, if your maximum heart rate is 180 beats per minute, your target heart rate zone might be between 120 and 150 beats per minute.
The objective pulse is 50 to 85 percent of your greatest pulse. It is the level at which your heart is pulsating with moderate to extreme focus. To decide your greatest pulse, take 220 and deduct your age. Supporting an exercise at this speed works on cardiorespiratory perseverance.
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a runner makes one lap around a 200m track in 25s, what is the runners (a) average speed and (b) average velocity
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what does this quote mean?
Answer:
Basically, it means that you never know until you try.
Explanation:
A bus initially moving at 20 m/s with an acceleration of -4m/s² for 5
seconds. What is the displacement AND final velocity?
Answer:
50m; 0m/s.
Explanation:
Given the following data;
Initial velocity = 20m/s
Acceleration, a = - 4m/s²
Time, t = 5secs
To find the displacement, we would use the second equation of motion;
[tex] S = ut + \frac {1}{2}at^{2}[/tex]
Substituting into the equation, we have;
[tex] S =20*5 + \frac{1}{2}*(-4)*5^{2}[/tex]
[tex] S =100 + (-2)*25[/tex]
[tex] S =100 - 50[/tex]
S = 50m
Next, to find the final velocity, we would use the third equation of motion;
[tex] V^{2} = U^{2} + 2aS [/tex]
Where;
V represents the final velocity measured in meter per seconds. U represents the initial velocity measured in meter per seconds. a represents acceleration measured in meters per seconds square.Substituting into the equation, we have;
[tex] V^{2} = 20^{2} + 2(-4)*50 [/tex]
[tex] V^{2} = 400 - 400[/tex]
[tex] V^{2} = 0[/tex]
V = 0m/s
Therefore, the displacement of the bus is 50m and its final velocity is 0m/s.
