A car on the roller coaster begins with 0j of kinetic energy and 12,928j of potential energy and finishes the track with 3,715j of potential energy. How much kinetic energy dose the car finish with????

Explanation:

Energy input = F×d = (150 N)(5.5 m) = 825 J

Energy output = mgh = (66 kg)(9.8 m/s^2)(1.10 m) = 711 J

efficiency = [tex]\dfrac{\text{output}}{\text{input}}[/tex]×100% = 86.2%

Answer:

30

Explanation:

since  [tex]\frac{30}{5}[/tex]=6

         [tex]\frac{30}{6}[/tex]=5

then both 5 and 6 are factors of 30

Have a nice day

Answer:

C. both a speed and a direction

Complete Question

A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s.

(a) What is the moment of inertia of the system

(b) What is its rotational kinetic energy? axis 50 cm 30 cm 20 cm

Answer:

a) [tex]M.I=0.32kg.m^2[/tex]

b) [tex]K.E=621.8J[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=2.0kg[/tex]

Disk Radius [tex]r_d=50cm=0.5m[/tex]

Mass of annular cylinder [tex]M_c=1.0kg[/tex]

Inner Radius of cylinder [tex]R_i=20cm=0.2m[/tex]

Outer Radius of cylinder [tex]R_o=0.3m[/tex]

Angular  Velocity [tex]\omega=10rev/s=2 \pi*10=62.83[/tex]

Generally the equation for moment of inertia is mathematically given by

[tex]M.I=0.5M r_d^2+0.5M_c(R_i^2+R_o^2)[/tex]

[tex]M.I=0.5(2*0.50)^2)+0.5*1*(0.2^2+0.30^)[/tex]

[tex]M.I=0.32kg.m^2[/tex]

Generally the equation for Rotational Kinetic Energy  is mathematically given by

[tex]K.E=0.5M.I \omega^2[/tex]

[tex]K.E=0.5 *0.32*62.83[/tex]

[tex]K.E=621.8J[/tex]

Answer:

Answer:

An Object is positively charged if it has more Positive Electrons in that object

Answer:

[tex] = \frac{30 \times {10}^{ - 3} }{1} \\ = 0.03 \: km \: per \: second[/tex]

Answer:

108 km/hr or 0.03 km/s

Explanation:

conversion factor for m/s to km/hr is 5/18

conversion factor for m/s to km/s is 1/1000

Answer:

Distance = speed * time

55*5

275 meters.

The train would have covered a distance of 275 m

We can define distance as to how much ground an object has covered despite its starting or ending point.

Distance = speed * time

given

speed= 55 m/s

time = 5 sec

Distance = 55 * 5 = 275 m

The train would have covered a distance of 275 m

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Answer:

By measure the effective length and the time period of the pendulum.

Explanation:

Let the student take the oscillating pendulum at the planet.

He measure the time period of the pendulum  by using the stop watch or the ordinary watch.

Then measure the effective length of the pendulum which is the distance between the center of gravity of the bob and the point of suspension of the pendulum.

Now, use the formula of the time period of the pendulum,

[tex]T =2\pi\sqrt\frac{L}{g}[/tex]

Here, L is the effective length of the pendulum, g is the acceleration due to gravity at the planet and T is time period of the pendulum.  

By rearranging the terms, we get

[tex]T =2\pi\sqrt\frac{L}{g}\\\\T^{2}=4\pi^2\times\frac{L}{g}\\\\g =\frac{4\pi^2L}{T^2}[/tex]

Here, by substituting the values of L and T, the student get the value of acceleration due to gravity at that planet.

Answer:

Answer:

A). The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity.

Explanation:

Edge.

Answer:

The motion of the paper airplane  is best explained by horizontal inertia and vertical pull of gravity.

Explanation:

Inertia is the property by which the body wants to remain in its position unless any external for is applied. Here horizontal inertia is inertia of motion which is acting horizontally .

While vertical pull is due to the earth .

Hence horizontal inertia and vertical pull best explain the projectile motion of paper airplane.

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Answer:

[tex]v=6.65m/sec[/tex]

Explanation:

From the Question we are told that:

Mass [tex]m=97.6[/tex]

Coefficient of kinetic friction  [tex]\mu k=0.555[/tex]

Generally the equation for Frictional force is mathematically given by

 [tex]F=\mu mg[/tex]

 [tex]F=0.555*97.6*9.8[/tex]

 [tex]F=531.388N[/tex]

Generally the  Newton's equation for Acceleration due to Friction force is mathematically given by

 [tex]a_f=-\mu g[/tex]

 [tex]a_f=-0.555 *9.81[/tex]

 [tex]a_f=-54455m/sec^2[/tex]

Therefore

 [tex]v=u-at[/tex]

 [tex]v=0+5.45*1.22[/tex]

 [tex]v=6.65m/sec[/tex]

Answer:

if 1 light year was one millimeter then 105,700 light years = 105,700 mm, (or 105.7 meters in case you needed to simplify or something)

Answer:

B

Explanation:

The seasons are measured in how far or close the earth is to the sun.

Answer:

So the answer is B. A is wrong because negative answer = deceleration

Answer:

The correct answer is "4000 J".

Explanation:

Given that,

Force,

= 200 N

Displacement,

= 20 m

Now,

The work done will be:

⇒ [tex]Work=Force\times displacement[/tex]

By putting the values, we get

              [tex]=200\times 20[/tex]

              [tex]=4000 \ J[/tex]

Please delete my answer. I made a mistake

Answer:

225,000 kilometers per second

Explanation:

Have a nice day

Answer:

M = 1.409 10¹⁴ kg

Explanation:

In this exercise we have that the prioress with a minimum speed can escape from the asteroid, therefore we can use the conservation of energy relation.

Starting point. When you drop the stone

         Em₀ = K + U

         Em₀ = ½ m v² - G m M / r

where M and r are the mass and radius of the asteroid

Final point. When the stone is too far from the asteroid

          Em_f = U = - G m M / R_f

as there is no friction, the energy is conserved

          Em₀ = Em_f

          ½ m v² - G m M / r = - G m M / R_f

          ½ v² = G M (1 / r - 1 /R_f)

indicate that for the speed of v = 45.7 m /s, the stone does not return to the asteroid so R_f = ∞

          ½ v² = G M (1 /r)

          M = [tex]\frac{v^2 r}{2G}[/tex]

let's calculate

          M = [tex]\frac{45.7^2 \ 9}{ 2 \ 6.67 \ 10^{-11}}[/tex]

          M = 1.409 10¹⁴ kg

Complete question is;

Use a variation model to solve for the unknown value.

The stopping distance of a car is directly proportional to the square of the speed of the car.

a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.

b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?

Answer:

A) d = 333.2 ft

B) 60 mph

Explanation:

Let the stopping distance be d

Let the speed of the car be v

We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;

d ∝ v²

Therefore, d = kv²

Where k is constant of variation.

A) Speed is 50 mph and stopping distance of 170 ft.

v = 50 mph

d = 170 ft = 0.032197 miles

Thus,from d = kv², we have;

0.032197 = k(50²)

0.032197 = 2500k

k = 0.032197/2500

k = 0.0000128788

If the car is now travelling at 70 mph, then;

d = 0.0000128788 × 70²

d = 0.06310612 miles

Converting to ft gives;

d = 333.2 ft

B) stopping distance is now 244.8 ft

Converting to miles = 0.046363636 miles

Thus from d = kv², we have;

0.046363636 = 0.0000128788(v²)

v² = 0.046363636/0.0000128788

v² = 3599.99658

v = √3599.99658

v ≈ 60 mph

Complete question is;

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:

(a) the work done by the force of gravity on the projectile,

(b) the change in kinetic energy of the projectile since it was fired, and

(c) the final kinetic energy of the projectile.

(d) Are any of the answers changed if the initial angle is changed?

Answer:

A) W = mgh

B) ΔKE = mgh

C) K2 = mgh + ½mv_o²

D) No they wouldn't change

Explanation:

We are expressing in terms of m, v0​, h, and g. They are;

m is mass

v0 is initial velocity

h is height of projectile fired

g is acceleration due to gravity

A) Now, the formula for workdone by force of gravity on projectile is;

W = F × h

Now, Force(F) can be expressed as mg since it is force of gravity.

Thus; W = mgh

Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.

Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.

B) Change in kinetic energy is simply;

ΔKE = K2 - K1

Where K2 is final kinetic energy and K1 is initial kinetic energy.

However, from conservation of energy, we now that change in kinetic energy = change in potential energy.

Thus;

ΔKE = ΔPE

ΔPE = U2 - U1

U2 is final potential energy = mgh

U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.

Thus;

ΔKE = ΔPE = mgh

Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.

C) As seen in B above,

ΔKE = ΔPE

Thus;

½mv² - ½mv_o² = mgh

Where final kinetic energy, K2 = ½mv²

And initial kinetic energy = ½mv_o²

Thus;

K2 = mgh + ½mv_o²

Similar to a and B above, this will not change even if initial angle is changed

D) All of the answers wouldn't change because their equations don't depend on the angle.

Answer:

B.

Explanation:

sana makatulong sayo

Answer:

The output electric power is 1338876 W.

Explanation:

Volume, V = 690 cubic meter

height, h = 220 mm = 0.22 m

efficiency = 90 %

time , t = 1 s

Let the mass is m.

m = volume x density  

m = 690 x 1000 = 690000 kg

The input power is

P = m g h = 690000 x 9.8 x 0.22 = 1497640 W

The electric power out put is

[tex]P' = 90 % of 1487640\\\\\\P' = 1338876 W[/tex]

Answer:

22.15 N/m

Explanation:

As we know potential energy = m*g*h

Potential energy of spring = (1/2)kx^2

m*g*h = (1/2)kx^2

Substituting the given values, we get -  

(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)

k = 39200/2.645

k = 19600 N/m

For safety reasons, this spring constant is increased by 13 % So the new spring constant is  

 k = 19600 * 1.13 = 22148 N/m = 22.15 N/m

Answer:

less than the weight of the block.

Explanation:

From the free body diagram, we get.

The normal force is N = Mg cosθ

The tension in the string is T = Mg sinθ

Wight of the block when the block is static, W = Mg

Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,

therefore, the normal force is less than the weight of the static block.

Answer:

The ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of  second coil.

And

The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of  fourth coil.

Explanation:

The resistance of the coil is directly proportional to the length of the coil and inversely proportional to the area of coil and hence inversely proportional to the square of radius of the coil.

So, the ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of  second coil.

And

The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of  fourth coil.

Answer:

c. kilometers and kilograms

Explanation:

If two units represent different things, like for example grams and joules, where the first one is a unit for mass, and the second one a unit for energy we can not add them together. (

Let's analyze each one of the given options:

a: grams and milligrams

both grams and milligrams are units of mass, so after the appropriate unit conversions, we can add them.

b: miles and kilometers

Both are units for distance, after the appropriate unit conversions, we can add them.

c: kilometers and kilograms

"Kilometer" is a unit for distance, and "kilogram" a unit for mass, we can not add distance and mass, so these two can not be added together.

d: centimeters and yards

Both are units of distance,  after the appropriate unit conversions, we can add them.

Then the only one that we can not add together is option c:  kilometers and kilograms

Answer: The correct option is C (kilometers and kilograms).

Explanation:

UNITS is the quantity of a constant magnitude which is used to measure the magnitudes of other quantities of the same nature. The magnitude of a physical quantity is expressed as, Physical quantity = (numerical value) × (unit). There are six basic units of measurement which includes:

--> metre (m) - unit of length.

--> kilograms (kg) - unit of mass.

--> second (s) - unit of time.

--> ampere (A) - unit of electrical current.

--> kelvin (K) - unit of temperature and

--> mole (mol) - unit of the amount of substance.

Mass is the quantity of matter present in a body which is constant and is measured in KILOGRAMMES or GRAMS. While position is the distance and direction of an object from a particular reference point which is known to vary and is measured in METERS OR KILOMETRES.

From the above definition, KILOGRAMMES is used to measure the magnitude of a different quantity from KILOMETERS and therefore should not be added up after conversation.

Answer:

Option B. 3.25×10¯¹⁰ mm.

Explanation:

Measurement (cm) = 0.00325×10⁻⁸ cm

Measurement (mm) =?

The measurement in mm can be obtained as follow:

1 cm = 10 mm

Therefore,

0.00325×10⁻⁸ cm = 0.00325×10⁻⁸ cm × 10 mm / 1 cm

0.00325×10⁻⁸ cm = 3.25×10¯¹⁰ mm

Thus, 0.00325×10⁻⁸ cm is equivalent to 3.25×10¯¹⁰ mm.

The conversion from centimeter to millimeter of the number 0.00325*10^-8cm is 3.25*10^-10mm

The number given is in standard form and can be written as 3.25*10^-11 cm.

To convert this from centimeter to millimeter, we have to multiply this value by 10.

So, let's 3.25*10^-11 by 10 and get our value in mm

[tex]3.25*10^-^1^1 * 10 = 3.25*10^-^1^0[/tex]

From the calculation above, we can see that option B is the right answer since it carries [tex]3.25*10^-^1^0mm[/tex]

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Answer:

the law of conservation of energy.

Explanation:

An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.

Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.

In Physics, a point where at least three circuit paths (wires) meet is referred to as a junction.

The Kirchhoff’s circuit laws are two(2) equations first published by Gustav Kirchhoff in 1845.

Fundamentally, they address the conservation of energy and charge in the context of electrical circuits.

One of the laws known as Kirchoff's Current Law (KCL) deals with the principle of application of conserved energy in electrical circuits.

Kirchoff's Current Law (KCL) states that the sum of all currents entering a junction must equal the sum of all currents leaving the junction.

This simply means that the algebraic sum of currents in a network of conductors(wires) meeting at a point is equal to zero.

The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.

This ultimately implies that, Kirchhoff's junction rule is a statement of the law of conservation of energy.