1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ganezh [65]
3 years ago
6

The children at the local park think the slide is too slow. There is too much fiction. What could you do to decrease the amount

of friction on the slide?
Physics
1 answer:
igor_vitrenko [27]3 years ago
7 0
In order to decrease the friction on the slide,
we could try some of these:

-- Install a drippy pipe across the top that keeps continuously
dripping olive oil on the top end of the slide.  The oil oozes
down the slide and keeps the whole slide greased.

-- Hire a man to spread a coat of butter on the whole slide,
every 30 minutes.

-- Spray the whole slide with soapy sudsy water, every 30 minutes.

-- Drill a million holes in the slide,and pump high-pressure air
through the holes.  Make the slide like an air hockey table.

-- Keep the slide very cold, and keep spraying it with a fine mist
of water.  The water freezes, and a thin coating of ice stays on
the slide.

-- Ask a local auto mechanic to please, every time he changes
the oil in somebody's car, to keep all the old oil, and once a week
to bring his old oil to the park, to spread on the slide.  If it keeps
the inside of a hot car engine slippery, it should do a great job
keeping a simple park slide slippery.

-- Keep a thousand pairs of teflon pants near the bottom of the ladder
at the beginning of the slide.  Anybody who wants to slide faster can
borrow a set of teflon pants, put them on before he uses the slide, and
return them when he's ready to go home from the park.
You might be interested in
You, a 70 kg person, leap from a 10 m tall building and land feet first on a trampoline. The center of the trampoline where you
Anon25 [30]

Answer:

1373.4 N/m

Explanation:

Hooke's law states that the extension of a spring and force are related by the expression, F=kx where k is spring constant, x is extension of spring and F is the applied force. Making k the subject of the formula then

k=\frac {F}{x}

Also, F=gm hence the above formula is modified as

k=\frac {gm}{x}

Taking g as 9.81 m/s2 , x as 0.5 m and m as 70 kg then

k=\frac {9.81\times 70 kg}{0.5m}=1373.4 N/m

4 0
3 years ago
A 0.01-kg object is initially sliding at 9.0 m/s. It goes up a ramp (increasing its elevation by 1.5 m), and then moves horizont
barxatty [35]

Answer:

During this motion, 0.133 J of heat energy was created

Explanation:

Hi there!

Let´s calculate the energy of the object in each phase of the motion.

At first, the object has only kinetic energy (KE):

KE = 1/2 · m · v²

Where:

m = mass of the object.

v = velocity.

KE = 1/2 · 0.01 kg · (9 m/s)²

KE = 0.405 J

When the object goes up the ramp, it gains some gravitational potential energy (PE). Due to the conservation of energy, the object must convert some of its kinetic energy to obtain potential energy. By calculating the potential energy that the object acquires, we can know the loss of kinetic energy:

PE = m · g · h

Where:

m = mass of the object.

g = acceleration due to gravity (9.81 m/s²)

h = height.

PE = 0.01 kg · 9.81 m/s² · 1.5 m

PE = 0.147 J

The object "gives up" 0.147 J of kinetic energy to be converted into potential energy.

Then, after going up the ramp, the kinetic energy of the object will be:

0.405 J - 0.147 J = 0.258 J

When the object reaches the spring, kinetic energy is used to compress the spring and the object obtains elastic potential energy (EPE). Let´s calculate the EPE obtained by the object:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compression of the spring

EPE = 1/2 · 100 N/m · (0.05 m)² = 0.125 J

Then, only 0.125 J of kinetic energy was converted into elastic potential energy. The object is at rest at the end of the motion, i.e., the object does not have kinetic energy when it compresses the spring by 5.0 cm. Since energy can´t be lost, the rest of the kinetic energy, that was not used to compress the spring, had to be converted into heat energy:

Heat energy = initial kinetic energy - obtained elastic potential energy

Heat energy = 0.258 J - 0.125 J = 0.133 J

During this motion, 0.133 J of heat energy was created.

7 0
3 years ago
A group of students left school at 8:00 am on a field trip to a science museum 90 miles away. Which best describes the average s
Aliun [14]

Answer:

45

Explanation:

because the equation for speed is distance divided by time! hope that helps gave a nice day!

7 0
2 years ago
Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected b
zmey [24]

Answer:

The maximum amount of work is  W = 1563.289 \ J

Explanation:

From  the question we are told that

   The temperature of the environment is  T = 280\ K

    The volume of container A is  V_A = 2 m^3

    Initially the number of moles  is  n = 1.2 \ moles

     The volume of container B is V_B = 3.5 \ m^3

     

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

             W =  P_A V_A  ln[ \frac{V_B}{V_A} ]

Now from the Ideal gas law

          P_A V_A =  nRT

So substituting for P_A V_A in the equation above

          W =  nRT ln [\frac{V_B}{V_A} ]

Where R is the gas constant with a values of  R =  8.314 \  J/mol

Substituting values we have that

            W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]

          W = 1563.289 \ J

8 0
3 years ago
A tow truck exerts a net horizontal force of 1050 N on a 760-kg car. What is the acceleration of the car during this time?
vovikov84 [41]
We can solve the problem by using Newton's second law of motion:
F=ma
where
F is the net force applied to the object
m is the object's mass
a is the acceleration of the object

In this problem, the force applied to the car is F=1050 N, while the mass of the car is m=760 kg. Therefore, we can rearrange the equation and put these numbers in, in order to find the acceleration of the car:
a= \frac{F}{m}= \frac{1050 N}{760 kg}=1.4 m/s^2

The equation also tells us that the acceleration and the force have same directions: therefore, since the force exerted on the car is horizontal, the correct answer is
<span>B) 1.4 m/s2 horizontally.</span>
5 0
3 years ago
Other questions:
  • Which statement would be the best evidence that heat transfer through radiation has occurred and why? A) A pan gets hot on a sto
    10·2 answers
  • According to Faraday's law, voltage can be changed by moving magnets away from the coil of wire. True False
    14·1 answer
  • You heat a mug of water to make hot chocolate. Which statement best describes the changes in the water?
    8·2 answers
  • Which of the following is true about ALL energy conversions?
    13·2 answers
  • Meg walks 2 m from her desk to the teacher's desk. From the teacher's desk, she then walks 4 m in the opposite direction to the
    12·2 answers
  • A stunt cyclist needs to make a calculation for an upcoming cycle jump. The cyclist is traveling 100 ft/sec toward an inclined r
    14·1 answer
  • A 9.5 kg object undergoes an acceleration of 3.3m/s^2. What is the magnitude of the next external force acting on it
    7·1 answer
  • You hit a hockey puck and it slides across the ice at nearly a constant speed.Is a force keeping it in motion?Explain.
    9·1 answer
  • A skipper on a boat notices wave crests passing the anchor chain every 6.0 seconds. The skipper estimates the distance between c
    7·1 answer
  • Which of the following best explains why energy transfers are never 100%
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!