The fastest time trial for the first quarter checkpoint is 2.02 s.
The slowest time trial for the first quarter checkpoint is 2.7 s.
The range of the times measured for the checkpoint is 0.68 s.
The given parameters;
Time for quarter checkpoint, = 2.15, 2.05, 2.02, 2.7The fastest time trial for the first quarter checkpoint is the least measured time value.
fastest time trial = least time measured
fastest time trial = 2.02 s
The slowest time trial for the first quarter checkpoint is the highest measured time value.
slowest time trial = 2.7 s
The range of the times measured for the checkpoint is difference between the fastest time and slowest time.
Range = fastest time - slowest time
Range = 2.7 - 2.02
Range = 0.68 s
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Answer:
2.02, 2.15, 0.13
Explanation:
A 2-column table with 5 rows. The first column has entries empty, time of trial number 1 (seconds), time of trial number 2 (seconds), time of trial number 3 (seconds), average time (seconds). The second column labeled one quarter checkpoint has entries 2.15, 2.05, 2.02, 2.07.
Use the table to answer the questions.
What is the fastest time trial for the first quarter checkpoint?
2.02
seconds
What is the slowest time trial for the first quarter checkpoint?
2.15 seconds
What is the range of times measured for this checkpoint?
0.13 seconds
bow long does it take earth to rotate once on its axis
Answer:
23 hours, 56 minutes
. A teacher pushes a desk across the floor for a distance of 5 meters. She exerted a horizontal force of 20N. How much work was done?
Answer:
100 J work was done.
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How do you find the final velocity of two objects colliding when you only have the mass and initial velocity of both objects
Answer:
We are required to analyze the conservation of kinetic energy in a perfectly inelastic collision.
kinetic energy:K=1/2mv²
momentum:p=mv
Explanation:
Using the equations for final velocity in terms of masses and initial velocity for a perfectly inelastic collision, work out the final kinetic energy. Calculate the initial kinetic energy and show that kinetic energy is not conserved. Use variables, no numbers
You need to follow law of conservation of energy
Energy is neither created nor destroyedWe know
Kinetic Energy=mass(Velocity)^2You may use momentum formula mass×velocity
In both you need to form equations then solve for final velocity
two forces of magnitude 12 n and 24 n act at the same point. which force cannot be the resultant of these forces?
The force that cannot be the resultant of these forces is 10N since it is less than both given forces
Given the forces with a magnitude of 12N and 24N, the resultant of these forces must not be less than any of the two forces.
From the given options, the only force that is less than both 12N and 24 N is 10N. Hence the force that cannot be the resultant of these forces is 10N since it is less than both given forcesLearn more on resultant force here: https://brainly.com/question/14626208
. a. Calculate the work done while lifting 300 kg of wate through a vertical height of 6 m. (Assume g = 10 m a =
Answer:
potential energy = mgh = 300 × 10 × 6m = 18000 joule or 18 kilo joule.
Explanation:
A car accelerates at a constant rate of 3 m/s2 for 5 seconds. If it reaches a velocity of 27 m/s, what was its initial velocity?
A
15 m/s
B
42 m/s
C
12 m/s
D
2.1 m/s
The initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.
CALCULATE INITIAL VELOCITY:
The initial velocity of the car can be calculated by using one of the equation of motion as follows:
V = u + at
Where;
V = final velocity (m/s)u = initial velocity (m/s)a = acceleration due to gravity (m/s²)t = time (s)According to this question, a car accelerates at a constant rate of 3 m/s² for 5 seconds. If it reaches a velocity of 27 m/s, its initial velocity is calculated as follows:
u = v - at
u = 27 - 3(5)
u = 27 - 15
u = 12m/s.
Therefore, the initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.
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Two carts, A and B, are connected by a spring and sitting at rest on a track. Cart A has a mass of 0.4 kg and Cart B has a mass of 0.8 kg. The spring is released and causes the two carts to push off from each other. Which of the following correctly compares the motion and forces acting on the two carts?
A. Cart A experiences a greater force but has the same speed as Cart
B. Cart A experiences a greater force and has a greater speed after the recoil.
C. Both carts experience the same force but Cart A has a greater speed after the recoil.
D. Both carts experience the same force and have the same speed after the recoil.
Both carts experience the same force but Cart A has a greater speed after the recoil.
The given parameters;
Mass of the cart A = 0.4 kgMass of the cart B = 0.8 kgApply the principle of conservation of linear momentum to determine the velocity of the carts after collision;
[tex]m_Av_0_A\ + m_Bv_0_B = m_Av_f_A \ + m_Bv_f_B\\\\the \ initial \ velocity \ of \ both \ carts = 0\\\\0.4(0) + 0.8(0) = 0.4v_f_A + 0.8v_f_B\\\\0 = 0.4v_f_A + 0.8v_f_B\\\\0.4v_f_A = -0.8v_f_B\\\\v_f_A= \frac{-0.8 v_f_B}{0.4} \\\\v_f_A = - 2 \ v_f_B \ \ m/s[/tex]
According to Newton's third law of motion, action and reaction are equal and opposite. The force exerted on cart A is equal to the force exerted on cart B but in opposite direction.
[tex]F_A = -F_B[/tex]
Thus, the correct statement that compares the motion and forces acting on the two carts is "Both carts experience the same force but Cart A has a greater speed after the recoil."
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given how the quantity work—which has units of energy—was defined, another unit for energy, equivalent to the joule (j), is
Answer:
I think you ask for another way of writing the joule (J). It can be expressed as the product between a force applied to a body and the deployment of the body (not necessarily caused by that force applied), so N*m (newton * meter).
The quantity work—which has units of energy—was defined, another unit for energy, equivalent to the joule (j), is Newton ·meter (N· m).
What is unit of physical quantities?The quantity with a constant magnitude that is used to measure the magnitudes of other quantities of the same type is referred to as a unit in physics.
A unit shouldn't alter as time or environmental factors like pressure and temperature change.It ought to be precisely defined.It need to be readily available and repeatable.Unit of work is that of energy, that is, Joule (J).
Again, work = force × displacement
Unit of work = Unit of force × Unit of displacement = Newton . meter (N.m).
Hence, the quantity work—which has units of energy—was defined, another unit for energy, equivalent to the joule (j), is Newton ·meter (N· m).
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Why does gravity on Earth have a stronger attraction with you than the sun has with you? Plsssssss help I really need this rn
Answer:
Although the Earth has a lesser mass than the Sun, it is far closer to you, allowing for a stronger pull.
Explanation:
soda is an example of which of these?
liquid–liquid solution
gas–liquid solution
solid–liquid solution
solid solution
who made the first game
A William Shakespeare
B Horace Alexander
C Doris Twitchell Allen
D William higginbotham
PLEASE HELP also i know this isnt physics but there was no HOPE/Gym subject for me to choose.
Answer:
Plan B.
Because flexibility is best improved by stretching.
Explanation:
Improving and increasing flexibility is done by having stretching sessions daily which maintains and widens the range of motion in the joints and stretches muscles.
Can someone please give me the (Answers) to this? ... please ...
Explanation:
1. By Newton's 3rd Law, both forces are equal and opposite. Hence both forces are of the same magnitude.
2. The reaction force is the force of arrow on bowstring.
3. Forward force of fish on water and vice versa?
4. The mass of the person is very small as compared to mass of earth. Hence the force exerted is very small amd hence the recoil is not noticed.
5. Reaction force is the force of ball against the bat. By Newton's Third Law, they form an action reaction pair. Hence, they are equal in magnitude and opposite in direction.
6. The reaction force will also decreases as both forces are equal in magnitude.
7 im nt sure sorry Ahahah
What happens to all light bulbs in a series circuit when one burns out? How would the situation change when the lights are hooked up in parallel?
The ratio between the force of friction and the normal force is called:
A.the drag
B.rolling friction
C.the coefficient of friction
D.the starting friction
Answer:
C.the coefficient of friction
Explanation:
Hope this helps. can i have brainliest thanks!
can someone explain it with steps?
A car was moving on a road at a constant speed of 15 m/s when suddenly the car driver saw some animal on the road at a distance of 21 m from the car, so he applied the brakes after a response time of 0.4 s and stopped before hitting the animal by 1 m. What was the deceleration of the car?
a-7.5 m/s^2
b-5.2 m/s^2
c-8.0 m/s^2
d-5.6 m/s^2
Answer:
Option C is the correct answer
Explanation:
Distance travelled by car during reaction time
[tex]=15\times0.4\\\\=6m[/tex]
The car stopped before hitting the animal by [tex]1 m[/tex]
Distance travelled during deceleration is [tex]21-6-1=14m[/tex]
Hence by [tex]v^2=u^2+2as[/tex]
We have
[tex]0^2=15^2+2 \cdot a \cdot 14\\\\a=\frac{-225}{28} \\\\=-8.03m/s^2[/tex]
Option C is the correct answer
Distance traveled during reaction time
15(0.4)=6mTotal distance
21-6-1=14m[tex]\\ \sf\longmapsto v^2-u^2=2as[/tex]
[tex]\\ \sf\longmapsto -(15)^2=2(14)a[/tex]
[tex]\\ \sf\longmapsto -225=28a[/tex]
[tex]\\ \sf\longmapsto a=-8.0m/s^2[/tex]
Tool
Scientists are searching for planets that orbit stars outside Earth's solar system by collecting data from different technology tools. A student lists a few tools and their cases that would provide clients
with information about these planets in a table, as shown
Use
spectroscope
gathers information about the
planets in any weather conditions
space probe
collects data from the planets that
would otherwise not be available
identifies different elements present
reflector telescope
around the planets
radio telescope produces images of the planets
Which tool is paired correctly with its intended use?
space probe
Opectroscope
radio telescope
reflector telescope
Answer:
9758 how many significant figures
I need help to build a mouse trap race car for my science class and this is my final and these are the materials I have.
So, your science teacher has given your class the classic "mousetrap car" assignment: to make, design, and build a small vehicle powered by the snapping action of a mousetrap to make your car travel as far as possible. If you want to come out ahead of all the other students in your class, you'll need to make your car as efficient as possible so you can squeeze every last inch out of your "car". With the right approach, it's possible to streamline your car's design for maximum distance using only common home materials. You could also buy a mousetrap car kit from any craft store and skip wondering if it will work.
Use large rear wheels. Large wheels have greater rotational inertia than small wheels. In practice, this means that once they start rolling, they're harder to stop rolling. This makes large wheels perfect for distance-based contests — theoretically, they'll accelerate less quickly than smaller wheels, but they'll roll much longer and they'll travel a greater distance overall. So, for maximum distance, make the wheels on the drive axle (the one the mousetrap is tied to, which is usually the rear one) very large. The front wheel is a little less important — it can be large or small. For a classic drag racer look, you'll want big wheels in the back and smaller ones in front.
Use thin, light wheels. Thinner wheels have less friction and may go farther if the distance is what you want or need with your mousetrap racer. It's also important to take the weight of the wheels themselves into account — any unneeded weight will ultimately slow your car down or lead to added friction. In addition, it's worth noting that wide wheels can even have a small negative effect on the car's drag due to air resistance. For these reasons, you'll want to use the thinnest, lightest wheels available for your car.
Old CDs or DVDs work fairly well for this purpose — they're large, thin, and extremely light. In this case, a plumbing washer may be used to reduce the hole size in the middle of the CD (to fit the axle better).
If you have access to old vinyl, these also work extremely well, though they may be too heavy for the smallest mousetraps.
Use a narrow rear axle. Assuming your car is a rear-wheel-drive car, each time your rear axle turns, the rear wheels turn. If your rear axle is extremely skinny, your mousetrap car will be able to turn it more times for the same length of string than it would if it were wider. This translates to turning your rear wheels more times, meaning greater distance! For this reason, it's a wise idea to make your axle out of the skinniest material available that can still support the weight of the frame and wheels.
Narrow wooden dowel rods are a great, easily-accessible choice here. If you have access to thin metal rods, these are even better — when lubricated, they usually have less friction.
Create traction by giving the edges of the friction of the wheels. If the wheels slip against the ground when the trap is sprung, energy is wasted — the mousetrap works to make the wheels turn, but you don't get any extra distance. If this happens with your car, adding a friction-inducing material to the rear wheels may reduce their slippage. To keep your weight requirements down, use only as much as is necessary to give the tips of the wheels some grip and no extra. Some suitable materials are:[1]
Electrical tape
Rubber bands
Additionally, placing a piece of sandpaper under the rear wheels at the start line can reduce slippage as the car begins to move (when it is most likely)
In the United States, radioactive waste is divided into three main categories based on their activity, their heat generation potential, and what they physically contain.
Low level radioactive waste is the least dangerous type of nuclear waste. It contains relatively small amounts of radioactive substances with short half-lives.
Which of the following methods is most commonly used to dispose of low level radioactive wastes?
A.
activating reactions so the material is no longer radioactive
B.
sealing the waste into steel tanks and burying the tanks deep into the ground
C.
compacting the waste and burying it in a shallow landfill
D.
sealing in a biohazard bag and disposing the bag with ordinary trash
Answer:
C.Compacting the waste and burying it in a shallow landfill.
Explanation:
It is buried in a shallow landfill because it is less reactive.
what is the wavelength of a sound wave that travels at 340m/s and has a frequency of 710Hz
Answer:
0.479 m
Explanation:
v = fl
340 = 710l
l = 0.479 m
classroom and draw an approxim object Three forces are acting on an object (Figure 1.32) which is in equilibrium. Determine force A 1200 N Force A 51 39 42.00 1400 N Figure 1.32 Three forces, acting on
i. la protagonista ama a su gata. escribimos, en nuestros cuadernos , las razones por las que pensamos que las personas aman de manera exagerada a sus mascotas. Proponemos un antídoto para la soledad. Lo escribrimos
Las mascotas son una buena fuente de compañía y socias en la soledad.
La protagonista ama a su gato porque las mascotas son una buena fuente de compañía y es un antídoto para la soledad. Estas mascotas como perros, gatos, loros, etc. dan placer y felicidad a las personas.
La gente tiene mascotas para pasar tiempo con ellas y jugar con ellas para refrescar sus mentes. Tener mascotas es una actividad maravillosa que hace que nuestra mente esté relajada y llena de felicidad, por lo que podemos concluir que las mascotas son una buena fuente de compañía y compañera en la soledad.
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Which source of evidence did Wegener use to support his theory of continental drift?
fossils found on Earth
magnetic fields of Earth
satellite mapping of the tropical islands
glaciers found near the poles
Answer:
fossils found on Earth
Explanation:
i just did the test 100%
Answer:
Fossils on Earth
Explanation:
Fossils can show that the continents were all together by putting them together on a map you can see the fit. and when you look where fossils were, they would be right next to each other.
what is the difference between an asteroid and a meteoroid
Answer:
Asteroids smaller than planets and meteroids are small piece of an asteroid burns up upon enetering Earths atmosphere
A 16.0 kg canoe moving to the left at 12.5 m/s makes an elastic head on collision with a 14.0 kg raft moving to the right at 16.0 m/s after the collision the raft moves to the left at 14.4 m/s assuming water simulates a frictionless surface, what is the final velocity of the canoe?
Answer:
The final velocity of the canoe is -14.1 m/s or 14.1 m/s to the right.
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Doubt clarification - use comment section.
If the forces acting on an object at rest are ______,the object will remain at rest.
Answer:
friction. or gravity both ofbthem .....................
..
A 1000 kg car and a 1500 kg car are driving in opposite directions at 20 m/s and 10 m/s respectively. If the cars collide head-on and stick together, determine their combined speed after the collision.
Hi there!
Recall the conservation of momentum for an inelastic collision:
[tex]\large\boxed{m_1v_1 + m_2v_2 = (m_1 + m_2)v_f}}[/tex]
Remember, velocity is a VECTOR and direction must be accounted for. Let the 1000 kg car have a positive velocity and the 1500 kg car have a negative velocity (opposite direction).
Plug in the given values:
[tex](1000)(20) + (1500)(-10) = (1000 + 1500)v_f}}[/tex]
Solve:
[tex]20000 -15000 = 2500v_f}}\\\\5000 = 2500v_f\\\\v_f =\boxed{ 2 m/s}[/tex]
what is the net force required to give an automobile with a mass of 1,600 kg an acceleration of 4.5m/s
Hi there!
Recall Newton's Second Law:
[tex]\large\boxed{\Sigma F = ma}}[/tex]
∑F = Net force (N)
m = mass (kg)
a = acceleration (m/s²)
We are already given the mass and acceleration, so we can plug these values into the equation:
∑F = 1600 · 4.5 = 7200 N
what does the slope of the curve on a velocity vs. time graph represent?
Answer:
the slope of velocity-time graph represent an object acceleration
A horizontal force is applied to a box with a mass equal to 5.5 kg, The graph shows the net force acting on a box as a function of its horizontal position x.
Part A
Find the net work done on the box as it moves from x = 0 m to x = 8.0 m. .
Part B
Apply concepts of work and energy to solve for the speed of the box when it reaches position x = 8.0 m. Assume the box started from rest.
From the graph, net work done on the box and the speed of the box when it reaches position x = 8.0 m are 28 Nm and 3.2 m/s respectively
From the question, these are the given parameters;
Mass M = 5.5 KgPosition X = 0 to X = 8mPart A
The net work done will be the area under the graph.
From position X = 2m to X = 8m gives us the shape of a trapezium.
A = 1/2( a + b )h
A = 1/2( 2 + 6 ) x 8
A = 8 x 4
A = 32 Nm
From X = 0 to X = 2 gives us the shape of a triangle.
A = 1/2bh
A = 1/2 x 2 x (-4)
A = -4 x 1
A = -4 Nm
Net Work done = 32 - 4
Net work done = 28 Nm
Part B
Applying the concepts of work and energy to solve for the speed of the box when it reaches position x = 8.0 m
Net Work done = 1/2m[tex]V^{2}[/tex]
Substitute all the necessary parameters
28 = 1/2 x 5.5 x [tex]V^{2}[/tex]
5.5[tex]V^{2}[/tex] = 56
[tex]V^{2}[/tex] = 56/5.5
[tex]V^{2}[/tex] = 10.18
V = [tex]\sqrt{10.1818}[/tex]
V = 3.19 m/s
Therefore, net work done on the box and the speed of the box when it reaches position x = 8.0 m are 28 Nm and 3.2 m/s respectively
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