Answer:
This is the midline or the medium which is the exact middle of the graphs minimum and maximum points(which are the amplitude)
Car 1 of mass m1 is waiting at a traffic light.
Car 1 is struck from behind by Car 2 of mass m2.
The two cars stick together after the collision.
Car 2 was traveling at v2i = 30.0 m/s before the collision.
What is the kinetic, in [J], of the system after the collision if m1 = 2500 kg and m2 = 1000 kg?
Answer:
Explanation:
Conservation of momentum
2500(0) + 1000(30) = (2500 + 1000)v
v = 8.57 m/s
KE = ½(2500 + 1000)8.57² = 128,571.428... = 128 KJ
The kinetic energy of the system after the collision would be 128.5 KJ.
What is momentum?It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.
As given in the problem Car 1 of mass m1 is waiting at a traffic light.
Car 1 is struck from behind by Car 2 of mass m2. The two cars stick together after the collision. Car 2 was traveling at v2i = 30.0 m/s before the collision.
By using the conservation of the momentum,
2500(0) + 1000(30) = (2500 + 1000)v
v = 8.57 m/s
The final velocity of the system comes out to be 8.57 m/s.
The kinetic energy of the system after the collision,
KE =1/2×(2500 + 1000)×8.57² = 128,571.4
= 128.5 KJ
Thus, the kinetic energy of the system after the collision would be 128.5 KJ.
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#SPJ2
State the term used to describe the turning force exerted by the man
]A force called the effort force is applied at one point on the lever in order to move an object, known as the resistance force, located at some other point on the lever.
The way levers work is by multiplying the effort exerted by the user. Specifically, to lift and balance an object, the effort force the user applies multiplied by its distance to the fulcrum must equal the load force multiplied by its distance to the fulcrum. Consequently, the greater the distance between the effort force and the fulcrum, the heavier a load can be lifted with the same effort force.
Which of these is a push or a pull? Acceleration Force Mass Inertia
Answer:
the answer is force . force is applied as a push or pull
When a baseball curves to the right (a curveball) , air is flowing faster over the right side than over the left side. at the same speed all around the baseball, but the ball curves as a result of the way the wind is blowing on the field. faster over the left side than over the right side. faster over the top than underneath.
Answer:
faster over the left side than over the right side.
Explanation:
due to ball rotation, the right side is more closely matched to the speed of the air passing by as the ball progresses. This causes the air to stick more closely to the right side of the ball and that air stays with the ball surface as the spin moves it to the back of the ball and therefore leftward. As every action has an equal and opposite reaction the leftward force moving air causes the ball to experience an equal rightward force.
When the baseball curves to the right (a curveball), then the ball moves faster over the left side than over the right side.
What direction does a curveball move?The ball, which is thrown with a spin, is curve in the direction in which the front of the ball turns.
When a baseball curves to the right (a curveball),
For this condition, the pressure of air should be high on the left side than the pressure on right side.Molecules of the air on right side pushed backward by this spinning ball.The left side with high pressure push the ball towards right side where the pressure is low.Due to higher pressure, the ball move faster on the left side than the right side.Hence, when the baseball curves to the right (a curveball), then the ball moves faster over the left side than over the right side.
Learn more about the curveball movement here;
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The part of the circuit that converts electrical energy into other forms
Answer:
Load
Explanation:
The load in an electric circuit is any device that converts electrical energy into another form of energy.
Understanding what motivates anyone is not easy because each individual has different
Describe a vibration that is not periodic. NO LINKS PLEASE
Answer:
1)The position change of almost any manually operated room light switch.
2) Sunlight striking a point on the ground on a partly cloudy and windy day
Explanation:
An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (constant) acceleration of the vehicle during this time
Answer:
[tex]\huge\boxed{\sf a = 1200\ m/s\²}[/tex]
Explanation:
Given Data:
Initial Velocity = Vi = 40 m/s
Final Velocity = Vf = 80 m/s
Distance = S = 200 m
Required:
Acceleration = a = ?
Formula:
2aS = Vf² - Vi² (THIRD EQUATION OF MOTION)
Solution:
2a (200) = (80)² - (40)²
400a = 6400 - 1600
400a = 4800
Divide 400 to both sides
a = 4800 / 400
a = 1200 m/s²
[tex]\rule[225]{225}{2}[/tex]
Hope this helped!
~AH1807A racing car traveling with constant increases its speed from 10 m/s; 30 m/s over a distance of 60 mlong does this take? to
Answer:
Explanation:
constant acceleration???
assume it to be so
average speed is (10 + 30) / 2 = 20 m/s
t = d/v = 60/20 = 3 s
A 3.2 kg solid disk with a radius of 0.45 m has a tangential force of 420.4 N applied to it. What is the moment of inertia applied to the disk
Answer:
Explanation:
Your question makes no sense.
moment of inertia is a property of the disk and its geometry.
The moment of inertia of a uniform solid disk around an axis through its geometric center and perpendicular to its flat ends is
I = ½mR² = ½(3.2)0.45² = 0.324 kg•m²
the applied torque about the same axis would be
τ = FR = 420.4(0.45) = 189.18 N•m
and the angular acceleration about the same axis would be
α = τ/I = 189.18/0.324 = 583.9 rad/s²
if the momentum of a 1,400 kg car is the same as the truck in question 17, what is the velocity of the car?
Answer:
Explanation:
momentum is mass times velocity
p = mv
so take the momentum of the truck in question 17 and divide by the mass of this car
v = p/m = p / 1400
If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?
[tex]\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time, t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0 +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}[/tex]
25 gram saturated solution of potassium nitrate at 95 C is cooled down to 55 C then how much gram of crystals of potassium nitrate will be separated if the solubility of potassium nitrate at 95 c is 100 and 55 C is 25 correspondingly
The mass of potassium nitrate (KNO₃) crystals that will be separated is calculated as 6.25 g.
The given parameters:
Mass of KNO₃ = 25 gInitial temperature = 95 ⁰CFinal temperature = 55 ⁰CSolubility at 95 ⁰C = 100 MSolubility at 55 ⁰C = 25 MThe mass of KNO₃ at 95 ⁰C is calculated as follows;
[tex]m = \frac{25\ g \times 100\ g}{100\ g} \\\\m = 25 \ g[/tex]
mass of water = 100 g - 25 g = 75 g
The mass of KNO₃ at 55 ⁰C is calculated as follows;
[tex]m = \frac{75 \ g \times 25 \ g}{100 \ g} \\\\m = 18.75 \ g[/tex]
The mass of potassium nitrate (KNO₃) crystals that will be separated is calculated as;
[tex]m= 25\ g \ - \ 18.75 \ g\\\\m = 6.25 \ g[/tex]
Thus, the mass of potassium nitrate (KNO₃) crystals that will be separated is calculated as 6.25 g.
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A block of mass m = 3.0 kg is pushed a distance d = 2.0 m along a frictionless horizontal table by
a constant applied force of magnitude F= 20.0 N directed at an angle 0= 30.0° below the horizontal
as shown in Figure. Determine the work done by (a) the applied force, (b) the normal force exerted
by the table, and (d) the net force on the block.
Explanation:
We apply the definition of work by a constant force in the first three parts, but then in the fourth part we add up the answers. The total (net) work is the sum of the amounts of work done by the individual forces, and is the work done by the total (net) force. This identification is not represented by an equation in the chapter text, but is something you know by thinking about it, without relying on an equation in a list.
The definition of work by a constant force is W=FΔrcosθ.
(a) The applied force does work given by
W=FΔrcosθ=(16.0N)(2.20m)cos25.00=31.9J
(b), (c) The normal force and the weight are both at 900 to the displacement in any time interval. Both do 0 work.
(d) ∑W=31.9J+0+0=31.9J
Colloid - well ______ together but not ______________
Answer:Colloid - well compacted together but not one
What is the net force here?
11 N left
6 N right
1 N right
4 N right
answer = 6n to the right
Explanation:
2n plus 4n equals 6n
since 6n is more than 5n it goes 6n to the right
The mass of fifteen washers is _____ kg, which exerts a force of _____ N
Answer:
It could be related with the lesson from which this question belongs as far we did not read the lesson
Sorry
what is the pressure exerted by a force of 25 N on an area of 5m square
Answer:
pressure = force / area
then pressure = 25 / 5 = "5" N/m^2
In a Little League baseball game, the 145 g ball enters the strike zone with a speed of 17.0 m/s . The batter hits the ball, and it leaves his bat with a speed of 20.0 m/s in exactly the opposite direction. Part A What is the magnitude of the impulse delivered by the bat to the ball
Hi there!
Impulse = Change in momentum
I = Δp = mΔv = m(vf - vi)
Where:
m = mass of object (kg)
vf = final velocity (m/s)
vi = initial velocity (m/s)
Begin by converting grams to kilograms:
1 kg = 1000g ⇒ 145g = .145kg
Now, plug in the given values. Remember to assign directions since velocity is a vector. Let the initial direction be positive and the opposite be negative.
I = (.145)(-20 - 17) = -5.365 Ns
The magnitude is the absolute value, so:
|-5.365| = 5.365 Ns
Can someone help label these?
Name the energy possessed by hot air
Answer:
geothermal energy
Explanation:
the energy is obtained from the heat within the surface of earth
Answer:
heat energy
Explanation:
3. A 1500 kg car moving at 30 m/s strikes a 6000 kg van initially at rest. If the car
comes to a complete stop after the collision, what is the final velocity of the van?
Answer:
7.5m/s
Explanation:
Force= mass × velocity
Energy is conserved, the car and van should have the same overall force.
1500kg × 30m/s= 6000kg × final velocity
Final velocity = 7.5m/s
Can you solve this question?
Hi there!
In this instance, the object's centripetal force is provided by the horizontal component of the tension, so:
Tsinθ = mv²/r
**We use sine because in this situation, the angle is with the vertical**
We can plug in the known values for tension and theta:
60sin(60) = mv²/r
51.96 = mv²/r
The radius is equivalent to the sine of the string in respect to theta:
sin(60) = O/H = r/L
2sin(60) = 1.732 m
Now, solve for the velocity:
51.96 = mv²/r
51.96r / m = v²
51.96(1.732)/.400 = v²
v² = 225
v = 15 m/s
59. (II) The crate shown in Fig. 4-60 lies on a plane tilted at an angle A = 25.0° to the horizontal, with Mk 0.19. (a) Determine the acceleration of the crate as it slides down the plane. (b) If the crate starts from rest 8.15 m up along the plane from its base, what will be the crate's speed when it reaches the bottom of the incline?
Explanation:
a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:
[tex]x:\;\;\;\;\;mg\sin 25° - \mu_kN = ma\;\;\;\;\;\;(1)[/tex]
[tex]y:\;\;\;\;\;N - mg\cos 25° = 0\;\;\;\;\;\;\;\;\;(2)[/tex]
From Eqn(2), we see that
[tex]N = mg\cos 25°\;\;\;\;\;\;\;(3)[/tex]
so using Eqn(3) on Eqn(1), we get
[tex]mg\sin 25° - \mu_kmg\cos 25° = ma[/tex]
Solving for the acceleration, we see that
[tex]a = g(\sin 25° - \mu_k\cos 25°)[/tex]
[tex]\;\;\;\;= 2.45\:\text{m/s}^2[/tex]
b) Now that we have the acceleration, we can now solve for the velocity of the crate at the bottom of the plane. Using the equation
[tex]v^2 = v_0^2 + 2ax[/tex]
Since the crate started from rest, [tex]v_0 = 0.[/tex] Thus our equation reduces to
[tex]v^2 = 2ax \Rightarrow v = \sqrt{2ax}[/tex]
[tex]v = \sqrt{2(2.45\:\text{m/s}^2)(8.15\:\text{m})}[/tex]
[tex]\;\;\;\;= 6.32\:\text{m/s}[/tex]
A spring in a dart gun is compresscht a distance of 0.05 m. The spring has a spring constant
of 1,115 N/m. If the dart has a mass of 0.025 kg, determine the velocity of the dart as it
leaves the dart gun.
Answer:
Explanation:
ASSUMING that the dart is fired horizontally so that gravity potential energy considerations are not needed. Also ignoring friction work.
The spring potential will convert to kinetic.
KE = PS
½mv² = ½kx²
v = [tex]\sqrt{kx^2/m}[/tex]
v = [tex]\sqrt{1115(0.05^2)/0.025}[/tex]
v = 10.55935...
v = 11 m/s
what type of data do you need to collect in a ADI
define parking orbit?
Answer:
An orbit of a spacecraft from which the spacecraft or another vehicle may be launched on a new trajectory.
You are angry at Dr. Anderson for this exam, so you throw a 0.30-kg stone at his car with a speed of 44 m/s. How much kinetic energy does the stone have
Answer:
Explanation:
KE = ½mv²
KE = ½(0.30)44²
KE = 290 J rounded to 2 s.d.
A 5.0 m length of rope, with a mass of 0.52 kg, is pulled taut with a tension of 46 N. Find the speed of waves on the rope
Answer:
Speed of waves on the rope is 21 m/s
Explanation:
Length of the rope (l) = 5.0 m
Mass of the rope (m) = 0.52 kg
Tension in the rope (T) = 46 N
Formula of speed of waves on the rope:
[tex] \bold{v = \sqrt{\dfrac{T}{\mu}}} [/tex]
[tex] \mu [/tex] = Mass per unit length of the rope (m/l)
By substituting the values in the formula we get:
[tex] \implies \rm v = \sqrt{\dfrac{T}{ \dfrac{m}{l} }} \\ \\ \implies \rm v = \sqrt{\dfrac{Tl}{m}} \\ \\ \implies \rm v = \sqrt{ \dfrac{46 \times 5}{0.52} } \\ \\ \implies \rm v = \sqrt{ \dfrac{230}{0.52} } \\ \\ \implies \rm v = \sqrt{442.3} \\ \\ \implies \rm v = 21 \: m {s}^{ - 1} [/tex]
Speed of waves on the rope (v) = 21 m/s
A punter wants to kick a football so that the football has a total flight time of 4.70s and lands 56.0m away (measured along the ground). Neglect drag and the initial height of the football.
How long does the football need to rise?
What height will the football reach?
With what speed does the punter need to kick the football?
At what angle (θ), with the horizontal, does the punter need to kick the football?
Answer:
Explanation:
How long does the football need to rise?
4.70/3 = 2.35 s
What height will the football reach?
h = ½(9.81)2.35² = 27.1 m
With what speed does the punter need to kick the football?
vy = g•t = 9.81(2.35) = 23.1 m/s
vx = d/t = 56.0/4.70 = 11.9 m/s
v = √(vx²+vy²) = 26.0 m/s
At what angle (θ), with the horizontal, does the punter need to kick the football?
θ = arctan(vy/vx) = 62.7°