What is the average SPEED/VELOCITY of a car that traveled 1 complete lap around an oval track that’s 5000m long in 1000s
Answer:
5 m/s
Explanation:
5000/1000=5 m/s
:))
Can someone PLEASE help me??
A car is driving 12m/sec, has to stop suddenly because a pedestrian dashes out in front of the car. If the coefficient of kinetic friction between the tires and parking lot is ∪=60
what is the time, after the breaks are applied, before the car comes to a stop? Sketch the velocity time graph for the car's motion from the instant the breaks are applied until the car comes to a stop.
Answer:
Approximately [tex]2\; \rm s[/tex], assuming that the floor of this parking lot is level, [tex]\mu_{\rm k} = 0.60[/tex], and [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].
Explanation:
Let [tex]m[/tex] denote the mass of this vehicle. Weight of this vehicle: [tex]m\, g[/tex].
If the floor of this parking lot is level, the normal force on this vehicle would be equal to its weight: [tex]N = m \, g[/tex].
Given that [tex]\mu_{\rm k}[/tex], the kinetic friction between this vehicle and the ground would be consistently [tex]\mu_{\rm k} \, N = \mu_{\rm k} \, m \, g[/tex] until the vehicle comes to a stop.
Assuming that all forces on this vehicle other than friction are balanced. The net force of this vehicle during braking would be [tex](-\mu_{\rm k} \, m \, g)[/tex] (negative because this force is opposite to the direction of the motion.)
By Newton's second law of motion, the acceleration of this vehicle would be:
[tex]\begin{aligned}a &= \frac{F_\text{net}}{m} \\ &= \frac{-\mu_{\rm k} \, m \, g}{m} \\ &= -\mu_{\rm k}\, g \\ &= -0.60 \times 9.81\; \rm m\cdot s^{-2} \\ &= -5.886\; \rm m\cdot s^{-2}\end{aligned}[/tex].
In other words, braking would reduce the velocity of this vehicle by a constant [tex]5.886\; \rm m\cdot s^{-1}[/tex] every second until the vehicle comes to a stop. Calculate the time it would take to reduce the velocity of this vehicle from [tex]v_{0} = 12\; \rm m\cdot s^{-1}[/tex] to [tex]v_{1} = 0\; \rm m\cdot s^{-1}[/tex]:
[tex]\begin{aligned}t &= \frac{v_{1} - v_{0}}{a} \\ &= \frac{0\; \rm m\cdot s^{-1} - 12\; \rm m\cdot s^{-1}}{-5.886\; \rm m \cdot s^{-2}} \\ &\approx 2.0\; \rm s \end{aligned}[/tex].
Acceleration is the slope of the velocity-time graph. Since the acceleration here is constant, the velocity-time graph of this vehicle would be a line with a negative slope.
PLEASE HELP ON THIS QUESTION
[tex]r = 1.29×10^8\:\text{m}[/tex]
Explanation:
According to Newton's law of universal gravitation, the gravitational force between Uranus and Miranda is
[tex]F_G = G\dfrac{M_UM_M}{r^2}[/tex]
where [tex]M_U[/tex] is the mass of planet Uranus, [tex]M_M[/tex] is the mass of its satellite Miranda, r is the distance between their centers and G is the universal gravitational constant. Moving the variable r to the left side, we get
[tex]r^2 = G\dfrac{M_UM_M}{F_G}[/tex]
Taking the square root of the equation above, we get
[tex]r = \sqrt{G\dfrac{M_UM_M}{F_G}}[/tex]
Plugging in the values, we get
[tex]r = \sqrt{(6.67×10^{-11}\:\text{N-m}^2{\text{/kg}}^2)\dfrac{(8.68×10^{25}\:\text{kg})(6.59×10^{19}\:\text{kg})}{2.28×10^{19}\:\text{N}}}[/tex]
[tex]\:\:\:\:\:=1.29×10^8\:\text{m}[/tex]
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1- a car speeds up to get onto the freeway. it goes from 21 m/s to 39 m/s in 4.1 seconds. How far did it travel??
2- a boulder fell off a cliff and fell for 4.1 seconds. How tall was the cliff?
Answer:
Explanation:
1) average velocity is
v = (21 + 39)/2 = 30m m/s
d = vt 30(4.1) = 123 = 120 m
2) d = ½gt²
d = ½(9.8)(4.1²)
d = 82.369 = 82 m
when rounding to the two significant digits of the question numerals.
calculate the mass of a block of ice having volume 5m³. (density of ice≈920 kg/m³)
Answer:
4600kg
Explanation:
Density=mass÷volume
920=m/5
m=920×5=4600kg
Disk A, with a mass of 2.0 kg and a radius of 40 cm , rotates clockwise about a frictionless vertical axle at 50 rev/s . Disk B, also 2.0 kg but with a radius of 20 cm , rotates counterclockwise about that same axle, but at a greater height than disk A, at 50 rev/s . Disk B slides down the axle until it lands on top of disk A, after which they rotate together.
After the collision, what is magnitude of their common angular velocity (in rev/s)?
Hi there!
For this problem, we must use the conservation of angular momentum. This is an example of an inelastic "collision", so:
I₁w₁ + I₂w₂ = (I₁ + I₂)wf
We know that the moment of inertia of a disk is 1/2mR², so we can calculate the moments of inertia for both disks:
Disk 1: 1/2(2)(0.40²) = .16 kgm²/s
Disk 2: 1/2(2)(0.20²) = .04 kgm²/s
Plug in the values. Let counterclockwise be positive.
.16(-50) + .04(50) = (.16 + .04)wf
Solve:
wf = -30 rev/s
An object is travels 50 m in 4 s. It had no initial velocity and experiences constant acceleration. What is the magnitude of the acceleration?
Free-fall Acceleration is -10 m/s^2
I also need the Formula
Answer:
Explanation:
s = s₀ + v₀t + ½at²
50 = 0 + 0(4) + ½a(4²)
50 = 8a
a = 50/8 = 6.25 m/s²
Some students conduct an experiment to prove conservation of momentum. They use two objects that collide Measurements
are taken before and after the collision.
Which two quantities will the students multiply together before and after the collision?
A. mass and velocity
B. distance and time
C. mass and acceleration
D. velocity and time
This question involves the concepts of the law of conservation of momentum, velocity, and mass.
The two quantities, the students should multiply before and after the collision are "A. mass and velocity".
According to the law of conservation of momentum, In an isolated system, the total momentum of the system before the collision is always equal to the total momentum of the system after the collision.
To prove the law of conservation of momentum, consider two balls of masses ‘m₁’ and ‘m₂’, moving with velocities ‘u₁’ and ‘u₂’, respectively, such that u₁ is greater than u₂. After some time, these balls collide with each other and their velocities become ‘v₁’ and ‘v₂’, respectively.
This situation is illustrated in the attached picture.
So, according to the law of conservation of momentum:
Total Momentum Before Collision = Total Momentum After Collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
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Sorry this is a year late, but here it is for those of you who are stuck on the same thing.
======================================Proving Conservation of Momentum Quick Check - 5/5NOTE: Please Check and Confirm That You Are On The Same Assignment with The Same Questions and Number of Questions. Thank You and Good Luck!
=======================================1. Mass & velocity
2. The total momentum after the collision is the same as the total momentum before the collision.
3. 0.54 kg⋅m/s
4. The system has external forces, such as friction and air resistance, acting on it.
5. 3.0 m/s
A thin piece of semiconducting silicon will be used to fabricate an electrical device. This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. Electrical contacts are placed at opposite ends of its length. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.
Required:
a. If the application of 1.0 volt to the contacts results in a current of 0.019 amps, what is the resistivity in (ohm-cm) of the material?
b. If the material's conductivity is due to doping with aluminum to a level of [Al]= 1x10^17 atoms/cm^3, what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material (assuming that the aluminum is fully ionized - i.e. all Al atoms donated electrons).
We have that for the Question "a)what is the resistivity in (ohm-cm) of the material? b) what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material"
Answer:
Resistivity = [tex]1.754 ohm-cm[/tex]Conductivity = [tex]6.25*10^{25} cm^3/V-s[/tex]
From the question we are told
This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.
A) Resistivity is given as,
[tex]p = \frac{RA}{l}[/tex]
where,
[tex]R = \frac{V}{I}[/tex]
Therefore,
[tex]p = \frac{VA}{Il}\\\\p = \frac{1*(0.1*0.5)}{0.019*1.5}\\\\p = 1.754 ohm-cm[/tex]
B) Conductivity is given as,
[tex]U = \frac{\rho}{pe}\\\\U = \frac{10^{17}}{10^{10}*1.6*10^{-19}}\\\\U = 6.25*10^{25} cm^3/V-s[/tex]
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What is the angle of incidence when incident ray is reflected backwards along the same path
Answer:
The angle of incidence = The angle of reflection
Explanation:
For instance, the formula is <i = <r which means if the surface is smooth then the reflacted anglw will be equal according to the normal (dotted 90°) line.
Upon being reflected backwards the angle of incidence is simply the same, except from the other side
If the penny is thrown horizontally at 25 m/s from the 170 meter building, how long will it take for the penny to hit the ground?
9514 1404 393
Answer:
about 5.89 seconds
Explanation:
The penny will hit the ground at the same time it would if it were simply dropped. The equation for the vertical motion is ...
h(t) = -4.9t^2 +170 . . . . . where 170 is the initial height in meters
h(t) = 0 when ...
4.9t^2 = 170
t = √(170/4.9) ≈ 5.89
The penny will hit the ground in about 5.89 seconds.
The elevation at the base of a ski hill is 350 m above sea level. A ski lift raises a skier (total mass=72 kg, including equipment) to the top of the hill. If the skier's gravitational potential energy relative to the base of the hill is now 9.2 x 105 J, what is the elevation at the top of the hill?
The elevation at the top of the hill is 1,653.85 m.
The given parameters;
initial height of the skier, h₁ = 350 mlet the final height of the skier at the hill top, = h₂total mass, m = 72 kggravitational potential energy of the skier, P.E = 9.2 x 10⁵ JThe elevation at the top of the hill is calculated as follows;
[tex]P.E = mg\Delta h\\\\P.E = mg(h_2 -h_1)\\\\h_2 -h_1 = \frac{P.E}{mg} \\\\h_2 = \frac{P.E}{mg} + h_1\\\\h_2 = \frac{9.2 \times 10^5 }{72 \times 9.8} \ + \ 350 \ m\\\\h_2 = 1,653.85 \ m[/tex]
Thus, the elevation at the top of the hill is 1,653.85 m.
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How do I resolve moments about the point P?
Answer:
By applying the definition of torques ( [tex]\vec \tau = \vec r \times \vec F[/tex] ) and them remembering a few tricks.
Namely: if you wrap your RIGHT hand fingers around something and stick your thumb out, the direction your finger wraps gives you the verse of rotation and the thumb the orientation of the torque. Bottom force (4N) will give a counterclockwise rotation, torque is pointing up; top force (3N) will give a clockwise rotation and its torque its pointing down (read up and down as if the sheet the image is printed on is on your table).
In terms of magnitude the trick is easy: You want to multiply the intensity of the force (3N and 4N) by the distance between the point and the line the force it is applied to (that is, you don't care about the length of r itself, but the distance at a right angle, which is 0.9 and 0.8m respectively.
At this point, assuming "upwards" (relative to the plane of the sheet that is) torques positive, the 3N force gives you a torque of [tex]- 3N \times 0.9m = - 2.7N\cdot m[/tex] and the 4N force provides [tex]+4N\times 0.8 m = +3.2 N\cdot m[/tex]
If an electron moves in a direction perpendicular to the same magnetic field with this same linear speed,
what is the radius of its circular orbit?
Answer:
An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.
Explanation:
Answer:
An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.
Explanation:
A pendulum is made from a long rod of mass M and length L with a solid sphere (ball) of mass m and radius R attached to one end. As measured from the top of the pendulum (the end of the rod without the sphere), how far down the rod is the center of mass of the pendulum located
Answer:
Explanation:
If we assume the rod and sphere are of uniform construction so that their individual centers of mass are at their geometric centers, and that the rod end is attached to the surface of the sphere.
Balance moments about the rod free end of the assembly with its parts
(M + m)Cx = M(L/2) + m(L + R)
Cx = (M(L/2) + m(L + R)) / (M + m)
A student that is running in a gym at a speed of 3.5m/s grabs the rope hanging from the ceiling and swings on it.
a. how high will he swing? [63cm]
b. How high will he be when his speed reduced to half of its initial value? [16cm, ¼ of the initial value]
Can someone explain the logic behind the second part of the question (why is it 1/4 the initial value)?
a. Assuming all energy involved is conserved, at the lowest point of the swing (which includes the moment the student grabs the rope), the student only has kinetic energy,
K = 1/2 m (3.5 m/s)²
and at the highest point of the swing, the student only has potential energy
P = mgh
The energies at the bottom and top of the swing must be equal, so
1/2 m (3.5 m/s)² = mgh
h = (3.5 m/s)² / (2g)
h = 0.625 m ≈ 63 cm
b. In part (a), we found the relationship
h = v²/(2g)
If we cut the speed v in half, we replace v in the equation above with v/2 :
h = (v/2)²/(2g)
and simplifying this gives
h = (v²/4)/(2g) = 1/4 • v²/(2g)
The factor of 1/4 tells you that reducing the speed by a factor of 1/2 reduces the height by a factor of 1/4. So he can swing as high as
1/4 (3.5 m/s)²/(2g) = 0.15625 m ≈ 16 cm
How do light travels
Answer:
Light can travel in three ways from a source to another location: (1) directly from the source through empty space; (2) through various media; (3) after being reflected from a mirror.
Explanation:
Wrte down the effect of humidity and temperature in the speed of sound....
Explanation:
the speed of sound is affected by temperature and humidity
The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.2 m above the ground, how steep a slope can the truck be parked on without tipping over
The slope of the road can be given as the ratio of the change in vertical
distance per unit change in horizontal distance.
The maximum steepness of the slope where the truck can be parked without tipping over is approximately 54.55 %.Reasons:
Width of the truck = 2.4 meters
Height of the truck = 4.0 meters
Height of the center of gravity = 2.2 meters
Required:
The allowable steepness of the slope the truck can be parked without tipping over.
Solution:
Let, C represent the Center of Gravity, CG
At the tipping point, the angle of elevation of the slope = θ
Where;
[tex]tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}[/tex]
The steepness of the slope is therefore;
[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100[/tex]
Where;
[tex]\overline{AM}[/tex] = Half the width of the truck = [tex]\dfrac{2.4 \, m}{2}[/tex] = 1.2 m
[tex]\overline{CM}[/tex] = The elevation of the center of gravity above the ground = 2.2 m
[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%[/tex]
[tex]tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}[/tex]
[tex]Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}[/tex]
The maximum steepness of the slope where the truck can be parked is 54.55 %.
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A baseball player notices the ball when it is 3.4 m above the
ground, traveling at 4.4 m/s. He wants to make the catch when
the ball is 1.5 m above the ground, how long does it take to reach
his glove?
Find the distance the ball travels:
3.4 meters - 1.5 meters = 1.9 meters
Now divide the distance the ball travels by the speed:
1.9 meters / 4.4 m/s = 0.43 seconds
Answer:
Explanation:
s = s₀ + v₀t + ½at²
There are an infinite number of solutions to this question as posed because we are not told the direction of the initial velocity.
Assuming ground is level and origin and UP the positive direction
The shortest amount of time possible is when the initial velocity is straight down
1.5 = 3.4 - 4.4t + ½(-9.8)t²
0 = -4.9t² - 4.4t + 1.9
t = (4.4 ±√(4.4² - 4(-4.9)(1.9))) / (2(-4.9))
positive answer is
t = 0.32 s
The longest amount of time possible is when the initial velocity is straight up.
1.5 = 3.4 + 4.4t + ½(-9.8)t²
0 = -4.9t² + 4.4t + 1.9
t = (-4.4 ±√(4.4² - 4(-4.9)(1.9))) / (2(-4.9))
positive answer
t = 1.22 s
If the initial velocity is horizontal, meaning no vertical velocity
1.5 = 3.4 + 0t + ½(-9.8)t²
-4.9t² = -1.9
t² = 0.38775...
t = 0.62 s
Any angle between UP and Down will have a different initial vertical velocity and result in a different time to catch height.
It appears from the comments on the other answer, that I have shown you how to arrive at three of the four possible solutions. The initial direction is very important.
an observer sees two spaceships flying apart with speed .99c. What is the speed of one spaceship as viewed by the other
Answer:
V2 = (V1 - u) / (1 - V1 u / c^2)
V1 = speed of ship in observer frame = .99 c to right
u = speed of frame 2 = -.99 c to left relative to observer
V2 = speed of V1 relative to V2
V2 = (.99 - (-.99 ) / (1 - .99 (-.99)) c
V2 = 1.98 / (1 + .99^2) c = .99995 c
I need help with this equation. 4 tutors so far on the math side are unable to help me solve the problem.
Which of the following can cause an object to accelerate?
Select one:
a. Force
b. Inertia
c. Mass
d. Kinetic Energy
I think force can accelerate
Explanation:
i think force of an object
What is the intensity of the electromagnetic light waves coming from the Sun just outside of the atmosphere of Venus, Earth and Mars
The sun emits electromagnetic waves with a power of
4.0 ∗ 10 (26) W.
please help me
please help me
please help me
Answer:
do it got a picture
on the edge
Explanation:
The control center of human body is * 1/1 heart
brain
liver
We have ________ sense organs * 1/1
6
4
5
Which energy we get by burning of fuels * 1/1
Solar
Heat
Light
Sources of energy that can be used again and again will never run out * 1/1 Non renewable energy Renewable energy Running water What is called structure that build on rivers to get electricity * 1/1 Dam Check dam Tunnel Which nutrients protect us from diseases * 1/1 Carbohydrate Vitamin and Mineral Protein When is the best time of the day to water your lawn * 1/1 Early morning and late evening Afternoon Night Which of these ways to wash the car saves the most water? * 1/1 Wash it in the driveway with the garden hose Drive it into the lake Take it through a car wash that recycles water Why do we keep food and vegetables in refrigerator? * 1/1
After passing point 2 the hill becomes frictionless and the ring's rotational velocity remains constant. What is the linear velocity of the ring at point 3 in m/s
The energy in the system is given by the initial potential energy at the point 1.
The linear velocity at point 3, is approximately 33.59 m/s.
Reasons:
The parameters are;
Height at point 1, h₁ = 83 m
Radius of the ring = 8 cm
Mass of the ring, M = 8 kg
Height at point 2, h₂ = 32 m
At point 2, we have;
Change in potential energy = Kinetic energy
Which gives;
(83 - 32) × 9.81 × 8 = 0.5 × 8 × v² + 0.5 × 8 × 0.08² × (v/0.08)²
Which gives;
v ≈ 22.37 m/s
At point 3, the rotational kinetic energy remains constant while the
translational kinetic energy increases as follows;
K.E. at point 3 = Initial kinetic energy + Change in potential energy
Which gives;
K.E. at point 3 = 0.5 × 8 × v₃³ ≈ 0.5×8×22.37² + 32×9.81×8
[tex]v_3^2 = \dfrac{0.5 \times 8 \times 22.37^2 + 32 \times 9.81 \times 8}{0.5 \times 8} = 1128.15[/tex]
v₃ ≈ √(1128.15) ≈ 33.59
The linear velocity at point 3, v₃ ≈ 33.59 m/s
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The probable question parameters as obtained from a similar question online are;
Height at point 1, h₁ = 83 m
Radius of the ring = 8 cm
Mass of the ring, M = 8 kg
Height at point 2, h₂ = 32 m
A 25-kg box of books is dropped on the floor from a height of 1.1 m and comes to rest. What impulse did the floor exert on the box
The impulse the floor exert on the box is 116 kgm/s.
The given parameters;
mass of the books, m = 25 kgheight of the books, h = 1.1 mThe final velocity of the box when it dropped to the floor is calculated as follows;
[tex]\frac{1}{2} mv^2 = mgh\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 1.1} \\\\v = 4.64 \ m/s[/tex]
The impulse the floor exert on the box is calculated as follows;
the impulse the floor exert on the box is equal to change in momentum of the book
[tex]J = \Delta P\\\\J = \Delta Mv\\\\J = M(v_f - v_0)\\\\J = 25(4.64 - 0)\\\\J = 116 \ kgm/s[/tex]
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A 64 kg student is standing atop a spring in an
elevator that is accelerating upward at 3.0 m/s2
The spring constant is 3000 N/m.
A) by how much is the spring compressed?
Answer:
192
Explanation: