Two water-slide riders, A and B, start from rest at the same time and same height, h but on differently shaped slides.

Required:
a. Which rider is traveling faster at the bottom?
b. Which rider makes it to the bottom first? Ignore friction and assume both slides have the same path length.

Answers

Answer 1

Answer:

a. None

b. Both

Explanation:

a. Which rider is traveling faster at the bottom?

Since both riders fall from the same height, h, their potential energy, U at the top equals their kinetic energy, K at the bottom.

U = mgh and K = 1/2mv²

Since U is he same for both water-slide riders, then K will be the same and thus their speed at the bottom will be the same. This is shown below.

K = U

1/2mv² = mgh

v² = 2gh

v =√(2gh) where v = speed of rider at the bottom, g = acceleration due to gravity and h = height of slide.

Since the height is the same, so their speed at the bottom is the same. So, none of the riders travels faster than the other since they have the same speed at the bottom.

b. Which rider makes it to the bottom first? Ignore friction and assume both slides have the same path length.

Since the path length of the water slides are the same and friction is neglected, both water-slide rider get to the bottom at the same time since the distance moved is the same and they both start from rest.

So, both riders make it to the bottom at the same time.

Answer 2

The acceleration due to gravity acts vertically downwards, and the component of gravity acceleration is larger when the slope is steeper.

a. Rider Bb. Rider B

Reasons:

The acceleration of the riders are due to gravity

The component acceleration due to gravity acting on a slope is a = g·sin(θ)

As the steepness of the slope increase, the angle, θ, and sin(θ) increases, therefore, the acceleration increases.

Rider A is on a slide with gentle slope, such that if the slide is flat, rider A will be stationary.

The shape of the water slide rider B is on is steeper, and therefore, rider B is accelerating more than rider A. The higher acceleration of rider B, gives rider B a higher speed than rider, such that rider B, is riding faster than rider A

Therefore;

a. The rider that is travelling faster at the bottom is rider B

b. Given that friction is ignored, and the path have the same length to the

bottom, the rider that makes to the bottom first is the rider that is moving

faster, which is rider B

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Related Questions

A marshmallow in a vacuum becomes

A) larger.
B) smaller.
C) does not change.

Answers

puffs up because of aur traping in marshmallow

therefore A is the answer

thank you

B. Complete the lists:
Things that I must do for my family
Things I must never do to my family
1.
2.
2.
3.
3.
4.
5.
5.​

Answers

Answer:

Things you should do for your family

help your parentstreat them kindlylisten and obey themappreciate them for anything they do for you talk softly

things you shouldn't

backanswering them Disobey And anything that's harsh or make it parents sad

A force of 3 newtons moves a 10 kilogram mass horizontally a distance of 3 meters. The mass does not slow down or speed up as it moves. Which of the following must be true?
a) 9 joules of kinetic energy were produced
b) 9 joules of gravitational potential energy were produced
c) 9 joules of heat energy were produced
d) 9 joules of kinetic energy and heat were produced

Answers

Answer:

9 joules of heat energy was produced

Explanation: there is no acceleration therefore its not a kinetic energy

Energy= force × distance

= 3×3

=9

A positively charged plastic ruler is brought close to a piece paper resting on the desk. The piece of paper was initially neutral. When the ruler was brought closer, the paper is attracted to the ruler. The surface of the paper became charged through:_________

Answers

Answer: static electricity

Explanation:

When the plastic ruler is rubbed, friction opposes the motion and causes the transfer of electron from one surface to another such that plastic becomes negatively charged. When ruler is brought nearer to the paper, it induces the   positive charge in the piece of paper.

A tennis ball of mass of 0.06 kg is initially traveling at an angle of 47o to the horizontal at a speed of 45 m/s. It then was shot by the tennis player and return horizontally at a speed of 35 m/s. Find the impulse delivered to the ball.

Answers

Answer:

The impulse delivered to the ball is [tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex].

Explanation:

By Impulse Theorem, the motion of the tennis ball is modelled after the following expression:

[tex]Imp = m\cdot (\vec v_{f} - \vec v_{o})[/tex] (1)

Where:

[tex]m[/tex] - Mass of the ball, in kilograms.

[tex]\vec v_{o}[/tex] - Vector of the initial velocity, in meters per second.

[tex]\vec v_{f}[/tex] - Vector of the final velocity, in meters per second.

[tex]Imp[/tex] - Impulse, in meters per second.

If we know that [tex]m = 0.06\,kg[/tex], [tex]\vec v_{o} = \left(45\,\frac{m}{s} \right)\cdot (\cos 47^{\circ}, \sin 47^{\circ})[/tex] and [tex]\vec v_{f} = \left(35\,\frac{m}{s} \right)\cdot (-1, 0)[/tex], then the impulse delivered to the ball is:

[tex]Imp = (0.06\,kg)\cdot \left[\left(35\,\frac{m}{s} \right)\cdot (-1,0) -\left(45\,\frac{m}{s} \right)\cdot (\cos 47^{\circ}, \sin 47^{\circ})\right][/tex]

[tex]Imp = (0.06\,kg)\cdot (-65.670, -32.911)\,\left[\frac{m}{s} \right][/tex]

[tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex]

The impulse delivered to the ball is [tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex].

sanaysay tungkol sa pangangalaga ng mga endangered animals​

Answers

What details do you need in this essay exactly?

According to Newton’s law of universal gravitation, which statements are true?

Answers

1,3,5 it should be right because i have took that thing before

The food calorie, equal to 4186 J, is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit-and-cereal bar contains 160 food calories per bar.
Part A
If a 67.0 kg hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential energy?
Express your answer in meters.
Part B
If, as is typical, only 20.0 % of the food calories go into mechanical energy, what would be the answer to Part A? (Note: In this and all other problems, we are assuming that 100% of the food calories that are eaten are absorbed and used by the body. This is actually not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the rest are eliminated by the body. Metabolic efficiency varies considerably from person to person.)
Express your answer in meters.

Answers

It’s definitely gonna be A

a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficient of static friction between the car and the road is 0.78. What is the magnitude of the force of static friction acting on the car

Answers

Answer:

The magnitude of the friction force is 8197.60 N

Explanation:

Using the definition of the centripetal force we have:

[tex]\Sigma F=ma_{c}=m\frac{v^{2}}{R}[/tex]

Where:

m is the mass of the carv is the speed R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

[tex]F_{f}=m\frac{v^{2}}{R}[/tex]

[tex]F_{f}=2100\frac{18^{2}}{83}[/tex]

[tex]F_{f}=8197.60 \: N[/tex]

Therefore the magnitude of the friction force is 8197.60 N

I hope it helps you!

A ball is thrown vertically upward at 24.0 ms can reach a height of 28.8m ( neglecting air resistance).The speed,in m/s,when it is halfway to its highest point is (using g= 10 ms ^2)

Answers

Answer:

The answer is "[tex]16.79\ \frac{m}{s}[/tex]"

Explanation:

In this question, the halfway indicates the height that is [tex]\frac{28.8}{2}=14.4 \ m[/tex]

Using formula:

[tex]v^2=u^2+2as\\\\v^2=24^2+2(-10)(14.4)\\\\[/tex]

[tex]v^2=576-288\\\\v^2=288\\\\v=\sqrt{288}\\\\v=16.97 \ \frac{m}{s}[/tex]

What is the main way in which heat transfer occurs in liquids and gases?

Answers

The answer is convection

A spring with a 10-kg mass and a damping constant 15 can be held stretched 2 meters beyond its natural length by a force of 6 newtons. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity.

Required:
Find the position of the mass at any time t.

Answers

Answer:

Explanation:

Given that:

mass = 10kg

damping constant C = 15 kg/s

length = 2 m

Force F = 6N

Using the Hooke's law:

F = kx

6 = 15x

k = 6 N /2 m

spring constant k = 3 N/m

For the critical damping

C² - 4k*m= 0

m = C²/4k

m = (15)²/4(3) kg

m = 225/12 kg

m = 18.75 kg

A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity (-21 m/s) and a second piece, also of mass m, moves with velocity (-21 m/s) . The third piece has mass 3m. Just after the explosion, what are the (a) magnitude and (b) direction (as an angle relative to the x axis) of the velocity of the third piece

Answers

Answer:

25

Explanation:

magnitude and (b) direction (as an angle relative to the x axis) of the velocity

At 20 ◦C a copper wire has a resistance of 4×10−3 Ω and a temperature coefficient of resistivity of 3.9×10−3 (C◦)−1, its resistance at 100 ◦C is

A.
52.5 × 10-3 Ω
B.
5.25 × 10-3 Ω
C.
5.25 × 10-4 Ω
D.
5.25 × 10-2 Ω


E.
25.5 × 10-3 Ω

Answers

Answer:

[tex]R _{t} = R _{0}( \alpha t + 1) \\ = 4 \times {10}^{ - 3} (3.9 \times {10}^{ - 3} \times 20 + 1) \\ = 4 \times {10}^{ - 3} (1.078) \\ = 4.312 \times {10}^{ - 3} \: Ω[/tex]

A body is dropped from a height H. In how much time will it reach the ground?

Answers

Answer:

[tex]s = ut + \frac{1}{2} g {t}^{2} \\ t(u + \frac{1}{2} gt) = H \\ u + \frac{1}{2} gt = H \\ t = 2(H - u) \div g \\ t = \frac{(H - u)}{5} \\ u \: is \: speed \: or \: velocity[/tex]

Calculate the current flowing when the voltage across is 35V and the resistance is 7ohms.​

Answers

Explanation:

V= IR

35=I×7

I=35/7

I=5amperes

pls give brainliest

What is the Voltage of a circuit that has a resistance of 50 and a current of 2 A?
A 25 V
B. 107
C. 2.5 V
D. 0.4V

Answers

Answer:

100 V

Explanation:

V = I * R

V = 2 * 50

V = 100 V

{ I think there is a mistake in the options. }

a car's acceleration is negative. This means the car is _____.
a. slowing down
b. speeding up
c. changing direction
d. traveling in a circle​

Answers

A
Acceleration is a forward motion. To be negative it would mean the opposite of accelerations functions
Not sure if this is correct but my best guess
The answer is “A” just to keep it simple

why does a spherometer have three legs?​

Answers

spherometer is a device used to measure curved in surface

it have 3 legs which form equivalent triangle.

geometry says that 3 point determine a plane that's why it have 3 legs


how to calculate sound of an echo ​

Answers

by an echo meter

please flw me and thank my answers

#Genius kudi

A hammer has a mass of 1 kg. What is its weight (i) on Earth (ii) on the
Moon (iii) in outer space

Answers

Given mass= 1kg

Weight on earth = mg(gravity of earth) = 9.8N

weight on moon = mg(gravity of moon)= 1.62N

weight on outer space mg(gravity outer space = 0) = 0N

URGENTT

Which statement best defines the term "superconductivity"?

Answers

Answer:

the ability of certain substances at very low temperatures to conduct electricity with no resistance

A lumberjack is trying to drag a small tree that he cut down. If the static
coefficient of friction of the tree on the ground is 0.5 and the tree weighs 430
N, what is the minimum amount of horizontal force that he will need to apply
so that the tree will start moving?
A. 215 N
B. 430 N
C. 365 N
D. 500 N

Answers

Answer:

A

Explanation:

weight of the tree =normal force

Horizontal force =coefficient of friction x Fnormal

0.5×430=215

What is the energy equivalent of an object with a mass of 1.05g?​

Answers

Answer:

The equivalent energy of an object given its mass is calculated through the equation,

                             E = mc²

where c is the speed of light (3 x 10^8 m/s)

Substituting the known values,

                            E = (1.05 g/ 1000) (3 x 10^8 m/s)²

                               E = 9.45x10^13 J

Explanation:

como calcular la velocidad un atleta en los 100 metros planos?

Answers

Answer:

Explanation:

9ooooo

The large scale structure of the universe has been carefully mapped using redshift surveys of a very large number of galaxies. Answer the following: Suppose that a large telescope with modern equipment can measure the redshift to a galaxy in just 10 minutes. And suppose we want to spend no more than a year mapping the distribution of galaxies. How many redshifts can be surveyed

Answers

Answer:

26280

Explanation:

In current time, good telescope can measure redshift to a galaxy in 10 minutes.  

Thus, in one year that has on an average 365 days, the total time taken to  measure redshifts is = ( 365 *12 *60) minute  

= 262800 minutes .

Hence, the number of redshifts observed in a year = (262800/10) = 26280

A gymnast of mass 70.0 kgkg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value 9.81m/s29.81m/s2 for the acceleration of gravity.
PART A Calculate the tension T in the rope if the gymnast climbs the rope at a constant rate.
PART B Calculate the tension TTT in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 1.00 m/s2
PART C Calculate the tension TTT in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 1.00 m/s2m/s2 .

Answers

Answer:

43994

Explanation:

Hope this helps!

A 56 kg pole vaulter falls from rest from a height of 5.1 m onto a foam rubber pad. The pole vaulter comes to rest 0.29 s after landing on the pad.

Required:
a. Calculate the athlete's velocity just before reaching the pad
b. Calculate the constant force exerted on the pole vaulter due to the collision

Answers

a. The athlete's velocity just before reaching the pad is [tex]35.21m/s[/tex]

b. The constant force exerted on the pole vaulter is 6799.52 N

a. We use Newton's equation of motion,

                    [tex]v=u+at\\\\S=ut+\frac{1}{2}at^{2}[/tex]

Where u is initial velocity, v is final velocity, a is acceleration , t is time and S represent distance.

Given that,  s = 5.1 m , t = 0.29s, u = 0

Substitute in above equation.

            [tex]5.1=\frac{1}{2}*a*(0.29)^{2} \\\\a=\frac{5.1*2}{0.084}=121.42m/s^{2}[/tex]

the athlete's velocity, [tex]v=0+121.42*(0.29)=35.21m/s[/tex]

b. The constant force exerted on the pole vaulter due to the collision is given as,             [tex]Force=mass*acceleration[/tex]

             [tex]Force=56*121.42=6799.52N[/tex]

Learn more:

https://brainly.com/question/13532462

what is the difference between VELOCITY and SPEED?​

Answers

Answer:

Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement. Put another way, speed is a scalar value, while velocity is a vector. ... In its simplest form, average velocity is calculated by dividing change in position (Δr) by change in time (Δt).

Explanation:

A 35 kg child slides down a playground slide at a constant speed. The slide has a height of 3.8 m and is 8.0 m long. Find the magnitude of the kinetic friction force acting on the child.

Answers

Answer:

The magnitude of the kinetic frictional force acting on the child is 162.93 N

Explanation:

Given;

mass of the child, m = 35 kg

height of the slide, h = 3.8 m

length of the slide, d = 8.0 m

The change in thermal energy associated with the kinetic frictional force is calculated as follows;

[tex]\Delta E_{th} + \Delta K.E + \Delta U = 0\\\\\Delta E_{th} + (\frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2) + (mgh_f - mgh_i) =0\\\\since \ the \ speed \ is \ constant, \ v_f = v_i \ and \ \Delta K.E = 0\\\\Also, \ final \ height \ , h _f= 0\\\\\Delta E_{th} - mgh_i = 0\\\\\Delta E_{th} = mgh_i\\\\\Delta E_{th} = 35 \times9.8 \times 3.8\\\\\Delta E_{th} = 1303.4 \ J[/tex]

The magnitude of the kinetic frictional force that produced this thermal energy is calculated from the work done by frictional force;

[tex]\Delta E_{th} = F \times d\\\\F = \frac{\Delta E_{th} }{d} \\\\F = \frac{1303.4}{8} \\\\F = 162.93 \ N[/tex]

Therefore, the magnitude of the kinetic frictional force acting on the child is 162.93 N

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