Answer:
Linus Pauling
Explanation:
Prize motivation: "for his fight against the nuclear arms race between East and West." Linus Pauling received his Nobel Prize one year later, in 1963.
BECAUSE OF THEIR A PAIR OF SUNGLEASSES ON THE DASHBOARD WILL CONTINUE MOVING FORWARD WHEN THE CAR TURNS SHARPLY
a. Acceleration
b. inertia
c. velocity
d. weight
Answer:
b. inertia
Explanation:
BECAUSE OF THEIR inertia A PAIR OF SUNGLEASSES ON THE DASHBOARD WILL CONTINUE MOVING FORWARD WHEN THE CAR TURNS SHARPLY.
Colloid - well ______ together but not ______________
Answer:Colloid - well compacted together but not one
A 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
Three particles are placed in the xy plane. A 30-g particle is located at (3, 4) m, and a 40-g particle is located at (-2, -2) m. Where must a 20-g particle be placed so that the center of mass of the three-particle system is at the origin?
Answer:
Explanation:
30(3) + 40(-2) + 20(x) = 0(20 + 30 + 40)
x = -0.5
30(4) + 40(-2) + 20(y) = 0(20 + 30 + 40)
y = -2
(-0.5, -2)
Physics!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
What is the formula for calculating distance?
QA: Speed x Time -- Speed/Time -- Time/Speed
Answer:
x=v.t
The answer: Distance= Speed x Time
And also
Time = Distance/Speed
Speed= Distance/Time
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 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°
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
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
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
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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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
write 2 situations in which the energy changes mentioned occur
Answer:
The types of energy is bond breaking and bond forming in chemical energy.
Explanation:
During Chemical reaction energy is required either for breaking up bonds in case of reactants and building bonds to form products.
The chemical reaction in which energy is released is called exothermic reactions, which is released due to making up the bonds.
The chemical reaction in which energy is absorbed is called endothermic reactions, in which energy is absorbed for breaking up the bonds.
Motion Velocity
Reference point Speed
1. An object is in __________ when its distance from a(n) ________ is changing.
2. Speed is given direction is called _______________
3. ____________ can be calculated if you know the distance that an object travels in one unit of time.
Answer:
1. An object is in motion when its distance from another object is changing.
2.Speed is given direction is called velocity.
Speed can be calculated.......
Mechanical energy conservation states that
The total amount of energy will eventually be destroyed.
Potential energy will be conserved, but kinetic energy will be destroyed.
The total amount of energy, kinetic plus potential, remains the same.
Kinetic energy will be conserved, but potential energy will be destroyed.
Pls hurry I’ll give 50 points
Answer:
The total amount of energy, kinetic plus potential, remains the same.
Explanation:
What does the horizontal line through the center of the wave on a graph represent?
Answer:
This is the midline or the medium which is the exact middle of the graphs minimum and maximum points(which are the amplitude)
3. A ball is dropped from the roof of a building 55 meters tall. What is the approximate time of fall?
(Neglect air resistance and round to 2 decimal places).
Answer:
3.35 seconds
Explanation:
Use one of the equations of accelerated motion:
Δd = v1Δt+1/2aΔt^2
and rearrange for Δt which is time
Δt = √(2Δd)/a
now we can substitute in the values
a= 9.8 (acceleration due to gravity) and Δd= 55 as that is the height of the building
Δt = √(2*55)/9.8
Δt = 3.3503s
Light and Reflection
Diagram Skills
E
STI
500
Mirrot
Flat Mirrors
1. The point of a 20.0 cm
D
pencil is placed 25.0 cm
from a flat mirror. Its
eraser is 15.0 cm from
the mirror. Three of the
light rays from the
pencil's point hit the
mirror with incident
angles of 0°, 20°, and
50° at points A, B, and C as shown.
a. Use a protractor to draw the reflected rays from points A, B, and C.
b. Where do reflected rays or their extensions intersect?
Mirror
B
c. What is the distance between the pencil's head and its image?
d. Would a person's eye located at point D perceive one of the reflected rays
drew? Will the person be able to see the image? Explain.
e. What if the eye is located at point E?
f. Draw incident rays from the eraser of the pencil to point A and to poin
The law of reflection allows to find the results for the questions about ray reflection in a plane mirror are:
a) Attachment we see a diagram of the incident and reflected rays, incident and reflected angles are equal.
b) The extension of the reflected rays is what forms the image.
c) The image's distance is 20 cm behind the flat mirror.
d) The point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C
e) the Rays at points A, B, C cannot perceive in the point E.
f) attachment we see the rays that come out from the pencil eraser.
g) The image is behind the mirror at 15 cm.
The geometric interaction describes the interaction of light rays with surfaces, looking for where the rays are directed, it is described by two phenomenological laws:
Refraction. Establishes a relationship between incident rays and those transmitted by material means. Reflection. It establishes that the angle of incidence and reflection of the rays is the same.[tex]\theta_i = \theta_{r}[/tex]
From these two general laws, geometric optics establishes a relationship for the formation of the image, called the constructor's equation.
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
Where f is the focal length, p and q are the distance to the object and the image, respectively.
In this exercise, the medium is a mirror, which is why it must comply with the law of reflection.
a) In the attachment we see a diagram of the incident and reflected rays for the three points.
According to the law of reflection, the incident and reflected angles are equal.
b) From the diagram we can see that the extension of the reflected rays is what forms the image, which is called virtual and is located behind the mirror.
c) In the diagram we see two rays to form the image, we see that the distance to the object is equal to the distance to the image.
From the constructor's equation a plane mirror has an infinite radius.
p = -q
Therefore the image's distance is 20 cm behind the flat mirror. Therefore the distance to the object and the image are the same, the negative sign indicates that the image is behind the mirror.
d) A person located at point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C since their angle of reflection is not equal to the incident angle.
To perceive a ray it must have an angle of incidence of 25º.
e) Point E is located very far from the pencil, so the incident angle increases as does the reflected angle.
the Rays at points A, B, C cannot perceive.
f) In the attachment we see the rays that come out from the pencil eraser, they indicate that the distance to the plane mirror is 15.0 cm,
g) The image is behind the mirror at 15 cm.
In conclusion using the law of reflection we can find the results for the questions are:
a) Attachment we see a diagram of the incident and reflected rays, incident and reflected angles are equal.
b) The extension of the reflected rays is what forms the image.
c) The image's distance is 20 cm behind the flat mirror.
d) The point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C
e) the Rays at points A, B, C cannot perceive in the point E.
f) attachment we see the rays that come out from the pencil eraser.
g) The image is behind the mirror at 15 cm.
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Understanding what motivates anyone is not easy because each individual has different
in a compoumd are atoms physically or chemically combined
Answer:
They are...if I'm correct Chemically combined, sorry if I'm wrong.
The 0.15kg baseball has a speed of v=30 m/s just before it is struck by the bat. It then travels along the trajectory shown before the outfielder catches it. Determine the magnitude of the average impulsive force imparted to the ball if it is in contact with the bat for 0.75 ms
The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N
The mass of the baseball, m = 0.15 kg
The speed at which it moves, v = 30 m/s
Time at which the baseball was in contact with the bat, t = 0.75 ms
t = 0.75/1000 s
t = 0.00075 s
The impulsive force is given by the formula:
[tex]F=\frac{mv}{t}[/tex]
Substitute m = 0.15 kg, v = 30, and t = 0.00075s into the formula above:
[tex]F=\frac{0.15 \times 30}{0.00075} \\\\F=6000N[/tex]
The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N
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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:
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
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.
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.
An object is released from height of 17m.
The object will hit the ground approximately in
[tex]\text{Given that,}\\\\\text{Height, h = 17 m}\\\\\\\text{We know that,}\\\\h = v_0t + \dfrac 12 gt^2\\\\\implies h = \dfrac 12 gt^2\\\\\implies t^2 = \dfrac{2h}g\\\\\implies t =\sqrt{\dfrac{2h}g} = \sqrt{\dfrac{2(17)}{9.81}} = 1.87 ~ \text{sec}[/tex]
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.
Can someone help label these?
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
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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