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
Whitepunk [10]
3 years ago
10

What happens when a hockey puck slides on a perfectly frictionless surface?

Physics
1 answer:
oksian1 [2.3K]3 years ago
7 0
It will slide for ever in the same direction, unless it hits an object! Please give me the brainlest answer1
You might be interested in
a particle moves along the x axis with an acceleration of a=18t, where a has units if m/s2. if the particle at time t=0 is at th
ludmilkaskok [199]

Answer:

 Position at t= 4 seconds is 144 m

Explanation:

 It is given that acceleration, a = 18 t, where t is the time.

 We know that Velocity, v = \int { a} \, dt

  Substituting value of a,

           Velocity, v = \int {18t} \, dt=\frac{18t^2}{2} +c=9t^2+c

 We know that at t = 0, v = -12 m/s

         So, 9*0^2+c=-12\\ \\ c=-12m/s

So velocity, v = (9t^2-12)m/s

  We also know that displacement, x = \int { v} \, dt

     Substituting value of v,  

        Displacement, x=\int {(9t^2-12)} \, dt=\frac{9t^3}{3} -12t+c=3t^3-12t+c

  We know that at t = 0, particle is at origin, x =0.

               So,  0=3*0^3-12*0+c\\ \\ c=0

   Displacement, x = 3t^3-12t

At t = 4 seconds

   x = 3*4^3-12*4=192--48=144m

Position at t= 4 seconds is 144 m  

4 0
3 years ago
A pushing force acts on a toy car for 3 seconds causing it to accelerate at a rate A’. If the car is replaced with a similar, bu
Alla [95]

Answer:

one-half of A'

Explanation:

3 0
3 years ago
Read 2 more answers
A car accelerates from rest at 3.6 m/s 2 . How much time does it need to attain a speed of 5 m/s?
Olenka [21]

car starts from rest

v_i = 0

final speed attained by the car is

v_f = 5 m/s

acceleration of the car will be

a = 3.6 m/s^2

now the time to reach this final speed will be

t = \frac{v_f - v_i}{a}

t = \frac{5 - 0}{3.6}

t = 1.39 s

so it required 1.39 s to reach this final speed

6 0
3 years ago
A 75-kg sprinter accelerates from rest to a speed of 11.0 m/s in 5.0 s. (a) calculate the mechanical work done by the sprinter d
mezya [45]

The mechanical work done by the sprinter during this time will be 4537.5 J , the average power the sprinter must generate will be 907.5 W and if the sprinter converts food energy to mechanical energy with an efficiency of 25% then he will be burning calories at 54.20 calories per second.

Work in physics is the energy that is transferred to or from an item when a force is applied along a displacement. It is frequently described in its most basic form as the result of force and displacement.

The quantity of energy moved or transformed per unit of time is known as power in physics. The watt, or one joule per second, is the unit of power in the International System of Units.. A scalar quantity is power.

Given 75-kg sprinter accelerates from rest to a speed of 11.0 m/s in 5.0 s.

So let,

m = 75 kg

v = 11.0 m/s

t = 5.0 s

So the mechanical work done by the sprinter during this time will be as follow:

W = 0.5 mv²

W = 0.5 (75)(11)²

W = 4537.5 J

The average power the sprinter must generate will be as follow:

Power(P) = W / t

P =  4537.5/5

P = 907.5 W

Only 25% is absorbed. So, the sprinter only absorbed 226.875 J per second which is equal to 54.20 calories per second.

Hence   mechanical work done by the sprinter during this time will be 4537.5 J , the average power the sprinter must generate will be 907.5 W and if the sprinter converts food energy to mechanical energy with an efficiency of 25% then he will be burning calories at 54.20 calories per second.

Learn more about mechanical power here:

brainly.com/question/25573309

#SPJ10

8 0
1 year ago
A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).
Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c.  The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia  has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω²  + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh =  1/2Iω²  + 1/2mv² =  1/2Iω²  + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω²  + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω²  + 1/2mr²ω²

mgh = mr²ω²/5  + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0
2 years ago
Other questions:
  • Drivers a and b travel west from boston. driver a leaves three hours earlier traveling at a constant speed of 68mph. driver b fo
    15·1 answer
  • A 7.00 kg mass is being pulled by a 53.8 n force what is the acceleration
    6·1 answer
  • A cell converts What energy
    6·1 answer
  • The femur is a bone in the leg whose minimum cross-sectional area is about 3.70 10-4 m2. A compressional force in excess of 6.60
    10·1 answer
  • What is the importance of the ozone layer?
    15·1 answer
  • A d'Arsonal meter with an internal resistance of 1 kohm requires 10 mA to produce full-scale deflection. Calculate thew value of
    12·1 answer
  • a proton moves in a circle of radius 0.4 when it enters a region with a magnetic field of 1.0t which points into the plane the s
    15·1 answer
  • Anser the photo and i'll give brainlest
    15·1 answer
  • What force does it take to accelerate a 9.2 kg object 7.0 m/s^2?
    10·1 answer
  • As a train accelerates away from a station, it reaches a speed of 4.9 m/s in 5.4 s. if the train's acceleration remains constant
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!