Answer:
At the highest point when you toss a ball into the air.
Explanation:
At the higest point of a trajectory of a ball, the velocity is zero for a split second and there is no speed and direction. However, there still is acceleration of -10 m/s^2 because the force of gravity is still acting upon it at that point.
Hi there!
An example of this could be when a ball is thrown vertically into the air and reaches the TOP of its trajectory.
When an object is thrown with a vertical velocity, the acceleration due to gravity results in a decrease in its positive (upward) velocity until it reaches its highest point, where the instantaneous velocity = 0 m/s and the object begins to fall back down (negative velocity).
Additionally, throughout its entire trajectory, the ball experiences an acceleration due to gravity of g = 9.8 m/s², even at its highest point where there is a velocity = 0 m/s.
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:
Acceleration of a Car A car traveling along a straight road at accelerated to a speed of over a distance of ft. What was the acceleration of the car, assuming that it was constant
Answer:
how many feet?
Explanation:
how do all organisms begin life
Answer:
All organisms begin their lives as single cells.Overtime,these organisms grow and take on the characteristics of their species...All organisms grow,and different parts of organisms may grow at different rates.Organisims made out of only one cell
may change little during their lives, but they do grow
Explanation:
brainlest me please
what memory are you using to remember who the president of the united states is
Answer:
The First 8 Presidents
For this exercise, we're going to use a silly story made of silly sentences. The letters that represent the last names of these presidents are W, A, J, M, M, A, J, V. One silly sentence to help you remember this sequence is: Wilma and John made merry and just vanished
working memory.
sensory memory.
short-term memory.
long-term memory.
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]
What is the direction of the torque produced on the crankset by the 2-kg mass attached to the pedal bar
A Torque is a twisting force, or turning moment, it is a vector quantity with both magnitude and direction e.g Turning the handle of a cork-screw clockwise and then counterclockwise will advance the screw first inward and then outward By convention, counterclockwise torques are positive and clockwise torques are negative.
The direction is perpendicular to both the radius from the axis and to the force. It is conventional to choose it in the right hand rule direction along the axis of rotation.
Counterclockwise is the positive rotation direction and clockwise is the negative direction.
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3) A force of magnitude Fx acting in the x-direction on a 2.00 kg particle varies in time as shown
in FIGURE 2. Find
a) The impulse of the force
b) The final velocity of the particle if it is initially at rest
c) The final velocity of the particle if it is initially moving along the x-axis with velocity
of -2.00 ms -1
Answer:
a) Impuise of force =F∗?(t) = area of F-T graph area= impulse =triangle + rectangle + triangle = 0.5*4*2 + 4*1 + 0.5*4*2 = 12 N-s (b) impulse = change in momentum \(= mExplanation:
How is the wavelength of a sound affected when (a) a sound source moves toward a stationary observer and (b) the observer moves away from a stationary sound source
Answer:
If the observer is stationary but the source moves toward the observer at a speed vs, the observer still intercepts more waves per second and the frequency goes up. This time it is the wavelength of the wave received by the observer that is effectively shifted by the motion, rather than the speed.
This is two or more elements chemically combined in a fixed ratio.
Example: water, carbon dioxide, sodium chloride
I need help been struggling on this question
Answer:
440 m
Explanation:
S=(u+v) t / 2
S = (11+33) × 20/2
S= 44× 20/2
S=440 m
electron and proton are projected with same velocity normal to the magnetic field which one will suffer greater deflection? why
Answer:
Explanation:
The deflection of a charged particle by a magnetic field is proportional to its electric charge and to its velocity. The deflection is also inversely proportional to its mass. So given a proton and an electron going at the same velocity in a magnetic field and having equal (but opposite) electric charge the electron will deflect much more since the ratio of the masses is 1836.
Tectonic plate movement is the reason why northern California has a very different landscape than southern California. Two different tectonic plates, each moving in different directions, border the western side of the North American Plate. Use the map to identify the two tectonic plates that border the North American Plate to the west.
Answer:
Remember, NORTH ^, EAST >, SOUTH v, WEST <
Explanation:
It doesn't have to be a super complex answer. All you have to do is look to the left (west) of the North American plate. What are the 2 plates that you see? The Pacific and the Juan de Fuca, yeah? To the South, there is the Cocos amongst a few others.
I am not going to share the answer for sure as I haven't completed the test yet but that's how I'm solving it. You should write the answer in your own words anyways. Hope this helps! Have a good day :)
Answer:
The Juan de Fuca Plate and the Pacific Plate both border the west side of the North American Plate.
Explanation:
Edmentum
plz help me on this question thank you
Answer:
D
Explanation:
A cyclist on a training ride records the distance she travels away from home. The data only shows the first150 minutes of the ride before her cycling computer ran out of battery.
Answer:
A) 58 km
B) 30 mins
Explanation:
In pic details
graph in pic
Children in a tree house lift a small dog in a basket 3.85 m up to their house. If it takes 201 J of work to do this, what is the combined mass of the dog and basket
Answer:
Explanation:
The work will equal the increase in potential energy.
PE = mgh
m = PE/gh = W/gh = 201/(9.81(3.85)) = 5.32 kg
Find the time it takes for an object dropped from a building and reaches a final velocity of 20 m/s downward?
I need the formula
Answer:
Explanation:
v = at
t = v/a
t = 20 m/s / 9.8 m/s²
t = 2.0408163...
t = 2.0 s
A ball is dropped from an 80.0 m building. What is the ball's velocity after 3.00 s? Use an order-of-magnitude estimation to identify the correct choice.
A. -2.9 m/s
B. -29.4 m/s
C -8.8 m/s
D. -88.3 m/s
Answer:b
Explanation:
-29.4 m/s
The velocity of the ball dropped from 80 m if it reaches the ground within 3 seconds is 26.6 m/s. If it is in midway within this time, then the velocity will be 29.4 m/s.
What is velocity ?Velocity of a moving object is the measure of its distance travelled per unit time. Velocity is a vector quantity having both magnitude and direction. Acceleration is the rate of change in velocity.
Given that, height of the building = 80 m
the ball is moving downwards by acceleration due to gravity g = 9.8 m/s².
Then after 3 seconds, the velocity of the ball is calculated as follows:
velocity = acceleration × time
v = 9.8 m/s² × 3 s = -29.4 m/s
If the ball reaches the ground within the time of 3 s, then, the velocity is:
v = 80 m/3s = 26.6 m/s.
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PLEASE HELP!
A 9kg particle is initially at rest at x=0. It is subject to a single force Fx (N) which varies with x (m) as shown in the
diagram
F
2
1
0
ББ by
x
- 1
1
-2
The kinetic energy of the particle when it is at x = 3 m is:
Hi there!
With a Force/Displacement curve, we must take the integral (area underneath the curve) to calculate the work done.
We know that:
W = ΔKE
Calculate the work by finding the area underneath the force curve from
x = 0 to 3 m:
We can use a trapezoid:
A = 1/2(3 + 2)(3) = 7.5 J
This is the amount of work done, and since the object starts from rest:
7.5J = KEf - KEi (0 J)
7.5J = KEf = 7.5 J
Accelerations are produced by
A. Masses
B.accelerations
C. Velocities
D.unbalanced, net forces
An empty cylindrical barrel is open at one end and rolls without slipping straight down a hill. The barrel has a mass of 15.0 kg, a radius of 0.400 m, and a length of 0.800 m. The mass of the end of the barrel equals a fourth of the mass of its side, and the thickness of the barrel is negligible. The acceleration due to gravity is =9.80 m/s2.
What is the translational speed f of the barrel at the bottom of the hill if released from rest at a height of 33.0 m above the bottom?
Hi there!
We can use work and energy to solve this problem.
We know that:
Ei = Ef
Ei = Potential energy = mgh
Ef = Rotational kinetic + Translational kinetic = 1/2Iω² + 1/2mv²
The barrel is comprised of a hollow cylinder and disk-shaped bottom, so:
I (hollow cylinder) = mr²
I (disk) = 1/2mr²
Calculate the moment of inertias of each.
Since the mass on the base is one-fourth of its side:
x = mass of side
x + x/4 = 15
4x + x = 60
5x = 60
x = 12 kg
end mass = 3 kg
Solve for each moment of inertia:
Side: (12)(0.4²) = 1.92 Kgm²
Bottom: 1/2(3)(0.4²) = 0.24 Kgm²
Side + bottom = 2.16 Kgm²
We can now solve:
mgh = 1/2mv² + 1/2(2.16)v²/r²
(15)(9.8)(33) = 1/2(15)v² + 1/2(13.5)v²
4851 = 14.25v²
v = 18.45 m/s
An object is dropped from a vertical height of 1.89 m above the balcony level. What is the object’s speed when it is 2.20 m below the balcony level if 10.0% energy is lost due to the air resistance? Does it matter when to apply 10% loss before V calculations or after? [8.49m/s] [yes it does, 0.9Energy result in √0.9Velocity]
a.
The object's speed at 2.20 m below balcony level is 8.74 m/s
Let the balcony level be 0 m and the height above the balcony level be positive and height below the balcony level negative.
Using the principle of conservation of energy, the total energy at a vertical height of 1.89 m above the balcony level equals the total mechanical energy when the object is 2.20 m below the balcony level and
So, E = E'
U + K + f = U' + K' + f'
where U = initial potential energy at 1.89 m = mgh, K = initial kinetic energy at 1.89 m = 0 J(since it is released from rest), f = energy loss at 1.89 m = 0 J, U' = final potential energy at 2.20 m below balcony level = mgh', K = final kinetic energy at 2.20 m = 1/2mv², f' = energy loss at 1.89 m = 10%U = 0.10mgh(since 10% of the initial energy is lost).
So,
U + K + f = U' + K' + f'
mgh + 0 + 0 = mgh' + 1/2mv² + 0.10mgh
mgh = mgh' + 1/2mv² + 0.10mgh
Dividing through by m, we have
gh = gh' + 1/2v² + 0.10gh
So, gh - 0.10gh = gh' + 1/2v²
0.90gh = gh' + 1/2v²
1/2v² = 0.90gh - gh'
1/2v² = g(0.90h - h')
v² = 2g(0.90h - h')
Taking square-root of both sides, we have
v = √[2g(0.90h - h')]
where v = velocity of object at 2.20 m below balcony level, h = height above the balcony level = 1.89 m, h' = height below the balcony level = -2.20 m and g = acceleration due to gravity = 9.8 m/s²
Substituting the values of the variables into the equation, we have
v = √[2g(0.90h - h')]
v = √[2 × 9.8 m/s²{0.90 × 1.89 m - (-2.20 m)}]
v = √[2 × 9.8 m/s²(1.701 m + 2.20 m)]
v = √[2 × 9.8 m/s²(3.901 m)]
v = √[76.4596 m²/s²]
v = 8.74 m/s
So, the object's speed at 2.20 m below balcony level is 8.74 m/s
b.
Yes it does matter when we apply 10% loss before V calculations
We need to apply the 10 % loss before V calculations because this would give us a proper value for V since the energy is lost before V is obtained.
So, yes it does matter when we apply 10% loss before V calculations
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14 The radius of gyration of a body about an axis &ta
distance 6 cm from its centre of mass is 10 cm.
Then, its radius of gyration about a parallel axis
through its centre of mass will be
(a) 80 cm (b) 8 cm (c) 0.8 cm (d) 0.08 cm
Correct option is B 8 cm.
Let radius of gyration for the axis not passing through center of mass be r and that for the axis passing through the center of mass be k and the distance between the two parallel axes be a.
Parallel axes theorem gives:
[tex] {mr}^{2} = m( {k}^{2} + {a}^{2} ) \\ ⇒ {r}^{2} = {k}^{2} + a {}^{2} [/tex][tex]⇒k = \sqrt{ {10}^{2} - {6}^{2} } = 8cm.[/tex]Thus, option B is the correct answer.
Match the sport/physical activity in column B with the primary physical fitness component needed to perform it in column A . Write the letters of your answer in your activity notebook.
1. Power
A. Patintero
2. Speed
B. Marathon
3. Balance
C. dodgeball / tamaan bata
4. Coordination
D. 100m sprint
5. Flexibility
E. badminton and table tennis
6. Muscular Strength
F. exercise and proper diet
7. Agility
G. hopscotch/piko
8. Cardiorespiratory Endurance
H. Shotput
9. Reaction time
I. Archery
J. Leg Splits and yoga poseso
sagutan po plss
. Power
Patintero
, also known as harangang-taga or tubigan, (Intl. Translate: Escape from the hell or Block the runner) is a traditional Filipino children's game. Along with tumbang preso, it is one of the most popular outdoor games played by children in the Philippines.[1]
2. Speed The Barkley Marathonsis an ultramarathon trail race held in Frozen Head State Park near Wartburg, Tennessee. If runners complete 60 miles (97 km) this is known as a "fun run." The full course is about 100 miles (160 km). The race is limited to a 60-hour period and takes place in late March or early April of each year.
Convert 6 picoseconds into seconds.
Answer:
6e-12
Explanation:
divide the time value by 1e+12
What causes an astigmatism?
A. damaged lens
B. retina not focusing the image
C. cornea being wavy or not spherical
D. sclera not refracting light properly
Answer:
c) cornea being wavy or not spherical
100 J of work was done to lift a 10-N rock and set it at Position A near the edge of a cliff.
1. If the 100 Joules of work lifted the rock to the top of the cliff, how much potential energy did the rock gain?
2. At point C, the rock's potential energy will be
3. The rock's kinetic energy at point A is
4. At point B, some of the rock's potential energy will be changed to Kinetic energy
5. What is the mass of the rock?
6. What is the rock's velocity just before it hits the ground?
The rock to the right is sitting at the top of a ramp. I wonder how much work it required to get that rock up there.
Answer:
lol
Explanation:
Check Pic please, need help immediately
1.25 is the closest to 1.04 or not I want to answer please. I think it's true, but I want to prove it scientifically, please.
Answer:
in general context yes it is closest to 1.04
Explanation:
theres no right or wrong way to scientifically prove this though.
Overall in scale its closest to 1.04 hope that helped
Nikolai is using a hand-operated grain mill to grind wheatberries into flour. The mill is operated by spinning a fly-wheel with radiusR= 23 cm, which has a handle attachedto the outer edge. After grinding for a few minutes at a con-stant angular speedωi, he lets go of the handle and allows themechanism to come to rest as it undergoes constant angularacceleration. This happens over the course oft= 0.50 s, andthe flywheel undergoes a quarter of a rotation during this time.What is the linear tangential accelerationaof the handle as itcomes to rest? For the limits check, investigate what happenstoaas the time required to stop the flywheel becomes small(t→0).
Answer:
Explanation:
α = Δω/t = (0 - ωi)/0.50 = -2ωi rad/s²
ωf² = ωi² + 2αθ
θ = (ωf² - ωi²) / 2α
2π/4 = (0² - ωi²) / (2(-2ωi))
2π/4 = ωi / 4
ωi = 2π rad/s
α = -2(2π) = -4π rad/s²
a = rα = 0.23(-4π) = 0.92π m/s² ≈ -2.89 m/s²
as the time to stop approaches zero, acceleration goes toward infinity.
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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