a 150 kg roller coaster is released from rest at the top of a 50 m hill. how fast will it be going if the second hill is 10 m high?

Answers

Answer 1

Final answer:

To find the final speed of the roller coaster at the second hill, we can use the conservation of energy equation. Substituting the given values into the equation, we find that the final speed of the roller coaster at the bottom of the first hill is approximately 14.14 m/s. Assuming negligible friction, this will be the speed of the roller coaster at the top of the second hill.

Explanation:

To find the final speed of the roller coaster at the second hill, we can use conservation of energy. At the top of the first hill, the roller coaster has gravitational potential energy which is converted into kinetic energy at the bottom of the first hill.

Using the conservation of energy equation: mgh = (1/2)mv², where m is the mass of the roller coaster, g is the acceleration due to gravity, h is the height of the hill, and v is the final velocity, we can solve for v. Plugging in the values, we have (150 kg)(9.8 m/s²)(50 m) = (1/2)(150 kg)v².

Solving this equation, we find that the final speed of the roller coaster at the bottom of the first hill is approximately 14.14 m/s. Since the work done by frictional forces is negligible, we can assume that the roller coaster will maintain this speed as it goes up the second hill.


Related Questions

A 1.6-m wire is wound into a coil with a radius of 3.2cm.

-If this coil is rotated at 93 rpm in a (7.1x10^-2)-T magnetic field, what is it’s maximum emf? (In 2 significant figures in mV)

Answers

Answer:

18 mV

Explanation:

The maximum emf is:

ε = N B A ω

where N is the number of turns in the coil,

B is the strength of the magnetic field,

A is the area of the coil,

and ω is the angular velocity.

The number of turns N is equal to the length of the wire divided by the circumference of the coil.

N = L / (2πr)

Area of the coil is:

A = πr²

The angular velocity in rad/s is:

ω = rpm × 2π / 60 s

Therefore:

ε = (L / (2πr)) B (πr²) (rpm × 2π / 60 s)

ε = L B r (rpm × π / 60 s)

Plugging in values:

ε = (1.6 m) (0.071 T) (0.032 m) (93 × π / 60 s)

ε = 0.018 Tm²/s

ε = 18 mV

Final answer:

To find the maximum emf for a rotating coil in a magnetic field, use Faraday's law. The given wire length, rotation speed, and magnetic field strength can be used to calculate the emf after determining the number of turns and the coil area. The calculation results in an approximate maximum emf of 18 mV.

Explanation:

To find the maximum emf induced in a rotating coil within a magnetic field, we can use Faraday's law of electromagnetic induction which states that the emf (ε) induced in a coil is directly proportional to the rate of change of magnetic flux through the coil. Mathematically, ε = NABωsin(ωt+φ), where N is the number of turns, A is the area of the coil, B is the magnetic field strength, ω is the angular velocity, and ωt+φ is the phase angle. In this scenario, the coil is rotating at 93 RPM (revolutions per minute), which we can convert to radians per second (rad/s) since 1 revolution is equal to 2π radians: 93 RPM * 2π rad/rev * 1 min/60 s = 9.735 rad/s. The area (A) of the coil with radius (r = 3.2 cm = 0.032 m) can be found using the formula for the area of a circle: A = πr².

As the wire length is 1.6 m, we need to calculate the number of turns (N) by dividing the length of the wire (L = 1.6 m) by the circumference of the coil (C = 2πr), yielding N = L/C. Finally, to find the peak or maximum emf, we consider that sin(ωt+φ) is 1 at maximum. Plugging in the values, we get the maximum emf:

N = 1.6 m / (2π * 0.032 m) ≈ 7.958 turns (approx. 8 turns since a coil cannot have a fraction of a turn).
A = π * (0.032 m)² ≈ 0.0032 m².
ε(max) ≈ 8 turns * 0.0032 m² * 9.735 rad/s * 7.1 x 10⁻² T.

Calculating this givesapproximately 18 mV as the maximum emf, which should be provided in millivolts (mV) and rounded to two significant figures according to the question, resulting in 18 mV.

Which statement about dormant volcanoes must be true?
AThey can never erupt again. incorrect answer
BThey have not erupted for a very long time. incorrect answer
CThey will erupt within the next 12 months. incorrect answer
DThey were formerly stratovolcanoes.

Answers

Answer:

They have not erupted for a long time

Explanation:

They have not erupted for a long time is correct because dormant volcanoes are volcanoes that are active and have erupted once in the last 10,000 years. Because they are active, they will erupt again in the future.

Which of the following is a result of gravitational forces in the Solar System?
A.
the radiation given off by Jupiter
B.
Saturn is further away from the Sun than Earth
C.
the difference in surface temperature on each of the planets
D.
the orbit of moons around their planets in the Solar system

Answers

Answer:

D

Explanation:

the answer is d because gravitational force is what allows them to rotate

hope this was helpful

Which processes cause rocks to be exposed
at Earth's surface? SC.7..6.2

Answers

Answer:

weathering

Explanation:

If the mass of an object is 44 kilograms and its velocity is 10 meters per second east, how much Kinetic Energy does it have?

Answers

Answer: 2200J

Explanation:

M = 44kg

V = 10m/s

K.E =?

K.E = 1/2MV2 = 1/2 x 44 x (10)^2

K.E = 22 x 100

K.E = 2200J

A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second calculate, the original length of the pendulum, the value of accelerations due to gravity​

Answers

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, [tex]L_1=L-1[/tex]

New time period of the pendulum is, [tex]T_1=2.81\ s[/tex]

We know that, the time period of a simple pendulum of length 'L' is given as:

[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex]-------------- (1)

So, for the new length, the time period is given as:

[tex]T_1=2\pi\sqrt{\frac{L_1}{g}}[/tex]------------ (2)

Squaring both the equations and then dividing them, we get:

[tex]\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1[/tex]

Now, plug in the given values and calculate 'L'. This gives,

[tex]L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m[/tex]

Therefore, the original length of the simple pendulum is 2.97 m

Final answer:

The period of a simple pendulum can be calculated with the formula T = 2π√(L/g). When the length of the pendulum is shortened by 1.0m, the period becomes 2.81 seconds. By solving a system of equations related to the period and the length of the pendulum, we can get the original length and the acceleration.

Explanation:

The subject matter relates to physics, specifically kinematics, and the topic is the period of a simple pendulum. The formula to calculate the period of a simple pendulum is T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. We know that when the length of the pendulum was shortened by 1.0m, its period became 2.81 seconds. We can plug these values into our formula and solve for the original length (L).

To find the original length, we first let L0 be the original length and L1 be the shortened length. Therefore, L1 = L0 - 1m. Now we have two equations, T0 = 2π√(L0/g) and T1 = 2π√(L1/g). Plug in the known values and solve the system of equations, you can get the original length and the acceleration. Note, that the typical value for acceleration due to gravity on Earth is g = 9.8 m/s².

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What is formed from nuclear decay?
A) A bond between two atoms
B) A radioactive particle
C) A new neutron
D) A solution of two or more elements

Answers

Answer

B

Explanation:

i took that same test i'm pretty sure it's B

Nuclear decay forms new B) A radioactive particle

What is nuclear decay?Nuclear decay occurs when the nucleus of an atom is unstable and emits energy in the form of radiation. As a result, the nucleus changes into the nucleus of one or more other elements. The daughter nuclei have a lower mass and are more stable than the parent nucleus.

There are many types of nuclear decay:-

Alpha decay

beta decay

gamma decay

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. A car moves forward up a hill at 12 m/s with a uni-
form backward acceleration of 1.6 m/s2.
a. What is its displacement after 6.0 s?
b. What is its displacement after 9.0 s?

Answers

A) Displacement after 6.0 s 43.2 m uphill.

B) Displacement after 9.0 s 43.2 m uphill.

Explanation:

A car moving upwards in a hill is [tex]12 ms^{-1}[/tex].

Its uniform backward acceleration is [tex]-1.6ms^{-2}[/tex]. (since backward acceleration is a negative acceleration, it is mentioned in negative)

We need to find the displacement of the car after some time.

Using the equation of the motion formula, we know can identify the displacement.

D=[tex]vt+\frac{1}{2} at^2[/tex].

a) Displacement after 6.0 seconds,

D = [tex]12(6.0)+\frac{1}{2}(-1.6)(6.0)^2[/tex].

=[tex]72+\frac{1}{2} (36)(-1.6).[/tex]

=[tex]72+\frac{1}{2}(-57.6).[/tex]

=72-28.8.

D=43.2 m.

b) Displacement after 9.0 seconds,

D= [tex]12(9.0)+\frac{1}{2}(-1.6)(9.0)^2[/tex].

=[tex]108+\frac{1}{2} (81)(-1.6).[/tex]

=[tex]108+\frac{1}{2}(-129.6).[/tex]

= 108-64.8.

D=43.2 m.

The car's displacement after 6.0 s is 14.4 m and after 9.0 s is 43.2 m.

To find the displacement of the car after a given time, we can use the equation:
Displacement (d) = Initial velocity (v) * time (t) + (1/2) * acceleration (a) * time^2

a. After 6.0 s:
Initial velocity (v) = 12 m/s
Acceleration (a) = -1.6 m/s^2 (negative because it's a backward acceleration)
Substituting the values into the equation:
d = (12 m/s) * (6.0 s) + (1/2) * (-1.6 m/s^2) * (6.0 s)^2 = 72 m - 57.6 m = 14.4 m

b. After 9.0 s:
Using the same equation and substituting the new time, we can calculate the displacement:
d = (12 m/s) * (9.0 s) + (1/2) * (-1.6 m/s^2) * (9.0 s)^2 = 108 m - 64.8 m = 43.2 m

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a child pulls on a string that is attached to a car. if the child does 80.2 J of work while pulling the car 25.0 m, with what force is the child pulling?

Answers

Answer:

F = 3.20 N

Explanation:

Given:

Work done by child = 80.2 j

Distance that the car moves = 25.0 m

We need to find the force acting on the car.

Solution:

Using work done formula as.

[tex]W = F\times d[/tex]

Where:

W = Work done by any object.

F = Force (push or pull)

d = distance that the object moves.

Substitute [tex]W = 80.2\ J\ and\ d =25.0\ m[/tex] in work done formula.

[tex]80.2 = F\times 25[/tex]

[tex]F=\frac{80.2}{25}[/tex]

F = 3.20 N

Therefore, force acting on the car F = 3.20 N

Can somebody please help me with sig figs for this question

Answers

Answer: 760 cW/microgram

2 significant figures

Explanation: solution attached:

Answer:

760 [tex]\frac{cW}{micrograms}[/tex]

Explanation:

You have to know that:

1 gigawatt (GW)=100000000000 centiwatt (cW) 1 kilogram (kg) = 1000000000 micrograms  (μg)

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

So:

[tex]x=\frac{c*b}{a}[/tex]

In this case, you have [tex]\frac{760 GW}{1 kg}[/tex]

In this case, the rule of three is applied as follows: if 1 gigawatt [GW] = 100000000000 centiwatt [cW], 7.600 GW how many cW are they?

[tex]\frac{7.600 GW}{kg} *\frac{100000000000 cW}{1 GW} *\frac{1 kg}{1000000000 micrograms}=760 \frac{cW}{micrograms}[/tex]

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