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 car was traveling at 25 m/s when it slammed on the brakes and came to a complete stop in 3 seconds. What is the cars INITIAL/FINAL VELOCITY?
Answer:
Explanation:
Initial velocity 25 m/s
final velocity 0 m/s
The ratio of the two is undefined as dividing by zero is wonky.
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
PLEASE HELP ME WITH THISSSS
Answer:
she will move in the same direction at the same speed forever.
Explanation:
If there are no outside forces like gravity the net force will never change, she will just keep flying for forever and ever! poor lady
what two things make up an ionic bond?
Describe the concept of energy quanta of EM radiation which was explained by Planck.
Answer:
Planck postulated that the energy of light is proportional to the frequency, and the constant that relates them is known as Planck's constant (h). His work led to Albert Einstein determining that light exists in discrete quanta of energy, or photons.
Explanation:
Answer:
Energy does not occur in continuous amounts but in discrete amounts described by:
E = N h ∨ where N is the number of quanta (energy units), ∨ the frequency of the energy, and h Planck's constant (6.63E-34 J-sec)
Give an example of intense aerobics activity. Prompt must be accurate.
Answer:
Explanation:
An example of an intense aerobic activity would be running/ sprinting sprinting targets six specific muscle groups: hamstrings, quadriceps, glutes, hips, abdominals and calves. Sprinting is a total body workout featuring short, high-intensity repetitions and long, easy recoveries.
what would happen if gravity were to stop everywhere?
Answer:
everything will float up and go up to space and die
Explanation:
gravity keeps us down and once it stops everything will float up. And if it were to stop everywhere everything and everyone will die and everything will be destroyed.
The volume of a toy car was calculated by displacing water. The water
rose by 20ml when the object was placed into the graduated cylinder. The balance showed the toy car had a
mass of 500grams. Calculate the density of the toy car
Answer:
D = 25g/cm³
Explanation:
1ml = 1cm³
D = m/V
D = 500g/20cm³
D = 25g/cm³
At the molecular level, as the kinetic energy increases, what happens to the temperature?
decreases
increases
stays the same
Answer: increases
Explanation:
Temperature is a measure of the average velocity of the molecular particles. The faster they go, the higher the temperature.
A spherical ball of lead (density 11.3 g/cm 3) is placed in a tub of mercury (density 13.6 g/cm 3). Which answer best describes the result
The lead ball will float with about 17% of its volume above the surface of the mercury.
We know that density is defined as mass per unit volume of a substance. The density of a substance is an intrinsic property which can be used to identify a substance.
Given that Lead is less dense that mercury, we know that lead will float on mercury. Since the density of mercury is 13.6 g/cm3 and that of lead is 11.3 g/cm3, lead ball will float with about 17% of its volume above the surface of the mercury.
Learn more: https://brainly.com/question/12108425
Missing parts;
A spherical ball of lead (density 11.3 g/cm3) is placed in a tub of mercury (density 13.6 g/cm3). Which answer best describes the result?
A.The lead ball will float with about 83% of its volume above the surface of the mercury.
B.The lead ball will float with about 17% of its volume above the surface of the mercury.
C.The lead ball will float with its top exactly even with the surface of the mercury.
D.The lead will sink to the bottom of the mercury.
E.none of the above
Clothes stick together when you pull them out of the dryer because
clothing is a conductor.
clothing is an inductor.
they are not charged.
of static electricity.
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
Pendulum makes 12 complete swings in 8 seconds, what are its frequency and period on earth
Hi there!
We can begin by finding the period of the pendulum.
[tex]T = \text{ # of complete swings / seconds} = 12 / 8 = \boxed{\text{1.5 sec}}[/tex]
The frequency is simply the reciprocal of the period, so:
[tex]f = \frac{1}{T} = \frac{1}{1.5} = \frac{2}{3}Hz \text{ or } \boxed{0.67 Hz}[/tex]
Please help me with this problem
Answer:
Summertime
Explanation:
the sun never sets south of the Antarctic circle in the summertime.
What happens to the iron in the coilgun if the electricity in the coil was turned on
PLEASE HELP I DONT GET THISS
Answer:
I feel like its the second one but I'm not completely sure..
Explanation:
MCQ
A body of mass 5kg is pushed for distance x with accleration a. Then workdone against static friction is
1.ma*X cosB
2.ma*X sinB
3.zero
4.ma/X
Answer:
ma*XsinB
option 2 is correct
g What is the CD's moment of inertia for rotation about a perpendicular axis through the edge of the disk
Answer:
Explanation:
A CD has an OD of 120 mm and an ID of 15 mm and has a mass between 14 and 33 grams. Let's call it m
Lets call the outer and inner radii R and r respectively
Find the moment of inertia about a line perpendicular to the surface of the disc through its center. We can integrate or look up the result from standard tables
I = ½m(R² + r²)
then use the parallel axis theorem to shift the position of the axis
I = ½m(R² + r²) + md²
where d is the distance of the shift. In this case d = R
I = ½m(R² + r²) + mR²
I = m(1.5R² + 0.5r²)
If we select a mass of say 20 grams
I = 0.020(1.5(0.060²) + 0.5(0.0075²))
I = 0.0001085625 kg•m²
A rollercoaster car passes the hill which is 5.5m above the ground at speed 9.3m/s, and rolls over the second hill which is 2.5m above the ground, and heads toward the third hill which is 4.0 m higher than the first one. If the track is frictionless,
a. What maximum height will the car climb on the third hill? [h max = 9.9m, so car will climb the entire 9.5m hill]
b. Will the speed of the car on top of the hill 3 be lower or higher than its speed on the top of the hill one? [lower]
c. Calculate the speed of the car when it is 1m lower than the top of the third hill. [5.3m/s]
Would somebody kindly go over the questions :D
Answer:
Explanation:
Without friction, a roller coaster continuously converts potential energy to kinetic energy and back again. Total energy will be constant.
Let m be the mass of the car and ground level is the origin.
on the 5.5 m hill, total energy is
E = PE + KE
E = mgh + ½mv²
E = m(9.8)(5.5) + ½m(9.3)² = 97m J
a) The maximum height will occur when the total energy is all potential energy.
E = mgh
h = E/mg
h = 97m/m(9.8) = 9.9 m
As this value is greater than the height of the third hill at 5.5 + 4.0 = 9.5 m The car will cross the last hill with some remaining velocity in kinetic energy.
b) As 9.5 m is greater than 9.3 m, the 9.5 m hill will have more of the total energy of the system as potential energy, This mean there is less kinetic energy and therefore less velocity (and speed) on top of the 9.5 m hill.
c) KE = E - PE
KE = 97m - m(9.8)(9.5 - 1.0)
KE = 97m = 83.3m
KE = 13.7m = ½mv²
v² = √(2(13.7)
v = 5.2345...
v = 5.2 m/s
Objects 1 and 2 attract each other with a gravitational force of 178 units. If the mass of object 1 is one-fourth the original value AND the mass of object 2 is tripled AND the distance separating objects 1 and 2 is halved, then the new gravitational force will be _____ units.
Explanation:
Fgravity = G*(mass1*mass2)/D²
G is the gravitational constant, which has the same value throughout our universe.
D is the distance between the objects.
now, several numbers change.
Fgravitynew = G*((1/4)*mass1*3*mass2)/(1/2 * D)² =
= G*((3/4)*mass1*mass2)/(D²/4) =
= (3/4)* (G*(mass1*mass2)/D²) *4 =
= 4*(3/4)* (G*(mass1*mass2)/D²) =
= 3* (G*(mass1*mass2)/D²) = 3* Fgravity
the new gravitational force will be 3×178 = 534 units.
Question: A NEO distance from the Sun is 1.18 AU. What is its relative speed compared to Earth (round your answer to 3 decimal places)
Its relative speed compared to Earth is 0.921
The speed of the object v = 2πr/T where r = radius of orbit and T = period of orbit.
Let v = speed of earth, r = radius of earth orbit = 1 AU and T = period of earth orbit.
So, v = 2πr/T
Also, v' = speed of NEO, r' = radius of NEO orbit = distance of NEO from sun = 1.18 AU and T' = period of NEO orbit.
So, v' = 2πr'/T'
v'/v = 2πr'/T' ÷ 2πr/T
v'/v = r'/r × T/T'
From Kepler's law, T² ∝ r³
So, T'²/T² = r'³/r³
(T'/T)² = (r'/r)³
T'/T = √[(r'/r)]³
T/T' = √[(r'/r)]⁻³
So, substituting this into the equation, we have
v'/v = r'/r × T/T'
v'/v = r'/r × √[(r'/r)]⁻³
v'/v = √[(r'/r)]⁻¹
Since r' = 1.18 AU and r = 1 AU, r'/r = 1.18
So, v'/v = √[(r'/r)]⁻¹
v'/v = √[(1.18)]⁻¹
v'/v = [1.0863]⁻¹
v'/v = 0.921
So, its relative speed compared to Earth is 0.921
Learn more about NEO here:
https://brainly.com/question/24157038
calculate the surface area of a box whose mass is 200 kg and exerts a pressure of 100 Pascal on the floor.
Answer:
Explanation:
If 2×2 is 4 so 1 kg can be 1 gram if it belive on it self some people change
An object is moving with an initial velocity of 3.3 m/s. It is then subject to a constant acceleration of 3.7 m/s2 for 10 s. How far will it have traveled during the time of its acceleration?
I also need the complete Formula (Nothing left out)
Answer:
Explanation:
s = s₀ + v₀t + ½at²
ASSUMING the acceleration is in the direction of initial motion.
s = 0 + 3.3(10) = ½(3.7)(10²)
s = 218 m
Which region of electromagnetic spectrum will provide photons of the least energy
Answer:
Explanation:
Radio waves
Radio waves have photons with the lowest energies. Microwaves have a little more energy than radio waves. Infrared has still more, followed by visible, ultraviolet, X-rays and gamma rays.
A 0,9 -kg object attached to the end of a string swings in a vertical circle (radius = 75 cm). At the top of the circle the speed of the object is 6,5 m/s. What is the magnitude of the tension in the string at this position?
An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.44 m/s2. Determine the orbital period of the satellite.
Explanation:
The artificial satellite experiences a centripetal force [tex]F_c[/tex] as it moves around the earth and it is defined as
[tex]F_c = m\dfrac{v^2}{r} = m\left(\dfrac{2\pi r}{T}\right)^2\left(\dfrac{1}{r}\right) = \dfrac{4\pi^2mr}{T^2}[/tex]
where m is the mass of the satellite, r is its orbital radius and T is its orbital period. But we need to find the radius first.
Recall that the satellite is orbiting at a height where its acceleration due to gravity is 6.44 m/s^2. Since we know that the weight mg of the satellite is equal to the gravitational force [tex]F_G[/tex] between the earth and the satellite, we can write
[tex]mg = F_G = G\dfrac{mM}{r^2}[/tex]
[tex]\Rightarrow g = G\dfrac{M}{r^2}[/tex]
where M is the mass of the earth (=[tex]5.972×10^{24}\:\text{kg}[/tex]) and G is the universal gravitational constant (=[tex]6.674×10^{-11}\:\text{N-m}^2\text{/kg}[/tex]). Plugging in the values, we find that the radius of the satellite's orbit is
[tex]r = \sqrt{\dfrac{GM}{g}} = \sqrt{\dfrac{(6.674×10^{-11}\:\text{N-m}^2\text{/kg})(5.972×10^{24}\:\text{kg})}{6.44\:\text{m/s}^2}}[/tex]
[tex]\:\:\:\:\:= 7.87×10^6\:\text{m}[/tex]
Now that we have the value for the radius, we can now calculate the orbital period T. Recall that the centripetal force is equal to the weight of the satellite at its orbital radius. Therefore,
[tex]F_c = mg \Rightarrow \dfrac{4\pi^2mr}{T^2} = mg[/tex]
or
[tex]4\pi^2r = gT^2[/tex]
Solving for T, we get
[tex]T^2 = \dfrac{4\pi^2r}{g} \Rightarrow T = \sqrt{\dfrac{4\pi^2r}{g}}[/tex]
We can further simplify the above expression into
[tex]T = 2\pi\sqrt{\dfrac{r}{g}}[/tex]
Plugging in the values for r and g, we get
[tex]T = 2\pi\sqrt{\dfrac{(7.87×10^6\:\text{m})}{(6.44\:\text{m/s}^2)}}[/tex]
[tex]\:\:\:\:\:= 6945\:\text{s} = 1.93\:\text{hrs}[/tex]
CAN SOMEONE PLEASE HELP ME
Answer:
she will eventually slow down and come to a stop
Uranus (mass = 8.68 x 1025 kg) and its moon Miranda (mass = 6.59 x 1019 kg) exert a gravitational force of 2.28 x 1019 N on each other. How far apart are they? cs [?] x 10?'m Coefficient (green) Exponent (yellow) Enter
Answer:
Explanation:
F = GMm/d²
d = √(GMm/F)
d = √(6.674e-11(8.68e25)(6.59e19) / 2.28e19)
d = 1.29398e8 = 1.29 x 10^8 m center to center
Answer:
1.29 x 10^8 m apart
Explanation:
Works in Acellus!
Help pls!
A 3 kg mass is raised a distance of 14 m above the earth by a vertical force of 93 N.
The final kinetic energy of the mass, to 3 significant figures, if it was originally at rest is:
[tex] \large★·.·´¯`·.·★ {Answer}★·.·´¯`·.·★[/tex]
As we know that Kinetic Energy is the Energy that is possessed by a moving object. and if the object is at rest then it doesn't have velocity therefore there is no kinetic Energy.
In the numerical terms we can express it as : -
[tex] \sf0.00 \: \: joules[/tex][tex]꧁ \: \large \frak{Eternal \: Being } \: ꧂[/tex]
this is electricity in physics please help
Explanation:
a. (i) When the variable resistor is set at zero, the only resistance in the circuit is due to the lamp. So the current flowing through the circuit is
[tex]I = \dfrac{V}{R} = \dfrac{220\:\text{V}}{440\:Ω} = 0.5\:\text{A}[/tex]
(ii) The power output P of the lamp is given by
[tex]P = I^2R = (0.5\:\text{A})^2(440\:Ω) = 110\:\text{W}[/tex]
b. (i) The variable resistor is in a series connection to the lamp so when its value is set to its maximum value of 660 Ω, the total resistance of the circuit is simply the sum of the two resistances:
[tex]R_T = R_{vr} + R_L = 660\:Ω + 440\:Ω = 1100\:Ω[/tex]
Therefore, the current through the circuit is
[tex]I = \dfrac{V}{R_T} = \dfrac{220\:\text{V}}{1100\:Ω} = 0.20\:\text{A}[/tex]
(ii) Using the result in Part (ii), we can solve for the potential difference across the lamp as follows:
[tex]V_L = IR_L = (0.20\:\text{A})(440\:Ω) = 88\:\text{V}[/tex]
(iii) The power output of the lamp is
[tex]P = I^2R_L = (0.20\:\text{A})^2(440\:Ω) = 17.6\:\text{W}[/tex]
(iv) The rate at which electrical energy is supplied, i.e., the power output of the circuit is equal to the square of the current multiplied by the total resistance of the circuit:
[tex]P = I^2R_T = (0.20\:\text{A})^2(1100\:Ω) = 44\:\text{W}[/tex]