How long will it take a car, starting from rest, accelerating at 2 meters per second square to travel the same distance that another car traveling at a constant rate of 20m/s will travel?

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

Answer:

I feel like its the second one but I'm not completely sure..

Explanation:

Answer:

D. Tungsten

Explanation:

W - Wolfram

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)

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

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]

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.

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]

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

Answer:

she will eventually slow down and come to a stop

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

Answer:

c) cornea being wavy or not spherical

Answer:

D = 25g/cm³

Explanation:

1ml = 1cm³

D = m/V

D = 500g/20cm³

D = 25g/cm³

Answer:

Explanation:

v = at

t = v/a

t = 20 m/s / 9.8 m/s²

t = 2.0408163...

t = 2.0 s

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.

Answer: increases

Explanation:

Temperature is a measure of the average velocity of the molecular particles. The faster they go, the higher the temperature.

Answer:

ma*XsinB

option 2 is correct

Answer:

Explanation:

If 2×2 is 4 so 1 kg can be 1 gram if it belive on it self some people change

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!

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

Answer:

Summertime

Explanation:

the sun never sets south of the Antarctic circle in the summertime.

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.

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

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.

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.

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.

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]