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
Alik [6]
3 years ago
11

Q5: An ice skater moving at 12 m/s coasts

Physics
1 answer:
ExtremeBDS [4]3 years ago
8 0

u = 0.077

Explanation:

Work done by friction is

Wf = ∆KE + ∆PE

-umgx = ∆KE,. ∆PE =0 (level ice surface)

-umgx = KEf - KEi = -(1/2)mv^2

Solving for u,

u = v^2/2gx

= (12 m/s)^2/2(9.8 m/s^2)(95 m)

= 0.077

You might be interested in
A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s
Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

  • Time of flight = t = 20 s
  • Angle of the initial velocity of projectile with the horizontal = \theta = 30^\circ

<u>Assume:</u>

  • Initial velocity of the projectile = u
  • R = Range of the projectile during the time of flight
  • H = maximum height of the projectile
  • D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

  • Initial horizontal velocity of the projectile = u\cos \theta
  • Initial horizontal velocity of the projectile = u\sin \theta

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s

Hence, the initial speed  of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0
2 years ago
An inattentive driver is traveling 18.0 m/s when he notices a red light ahead. his car is capable of decelerating at a rate of 3
ryzh [129]

Speed of the car given initially

v = 18 m/s

deceleration of the car after applying brakes will be

a = 3.35 m/s^2

Reaction time of the driver = 0.200 s

Now when he see the red light distance covered by the till he start pressing the brakes

d_1 = v* t

d_1 = 18* 0.200 = 3.6 m

Now after applying brakes the distance covered by the car before it stops is given by kinematics equation

v_f^2 - v_i^2 = 2 a s

here

vi = 18 m/s

vf = 0

a = - 3.35

so now we will have

0^2 - 18^2 = 2*(-3.35)(s)

s = 48.35 m

So total distance after which car will stop is

d = d_1 + d_2

d = 48.35 + 3.6 = 51.95 m

So car will not stop before the intersection as it is at distance 20 m

3 0
3 years ago
Please please please please help
aivan3 [116]

Answer:

I am pretty sure its the second one but I could be wrong sorry if I am.

Explanation:

:D

4 0
3 years ago
Read 2 more answers
What is its rotational inertia about the meter stick in the
Sindrei [870]

Answer:

Honestly I cant understand that so please explain it someone

Explanation:

3 0
3 years ago
You leave the doctor's office after your annual checkup and recall that you weighed 688 N in her office. You then get into an el
Lapatulllka [165]

Answer:

a=0.5418\ m.s^{-2} upwards

a=1.283\ m.s^{-2} downwards

Explanation:

Given:

weight of the person, w=688\ N

So, the mass of the person:

m=\frac{w}{g}

m=\frac{688}{9.81}

m=70.132\ kg

  • Now if the apparent weight in the elevator, w_a= 726\ N

<u>Then the difference between the two weights is :</u>

\Delta w=w_a-w

\Delta w=726-688

\Delta w=38\ N is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

a=\frac{\Delta w}{m}

a=\frac{38}{70.132}

a=0.5418\ m.s^{-2} upwards, because the normal reaction that due to the weight of the body is increased here.

  • Now if the apparent weight in the elevator, w_a= 598\ N

<u>Then the difference between the two weights is :</u>

\Delta w=w-w_a

\Delta w=688-598

\Delta w=90\ N is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

a=\frac{\Delta w}{m}

a=\frac{90}{70.132}

a=1.283\ m.s^{-2} downwards, because the normal reaction that due to the weight of the body is decreased here.

7 0
3 years ago
Read 2 more answers
Other questions:
  • An automobile with a radio antenna 1.0 m long generates an emf, V1, since it is traveling at 100.0 km/h in a location where the
    14·2 answers
  • The slope of a distance vs. time graph is a measurement called
    10·1 answer
  • A force of 20 N acts on a rocket for 350 s, causing the rocket's velocity to increase. Calculate the impulse of the force and by
    7·1 answer
  • A bicycle wheel has an initial angular velocity of 1.10 rad/s . Part A If its angular acceleration is constant and equal to 0.20
    11·1 answer
  • I want to answer these questions about vectors
    13·1 answer
  • Which of the following sequences describes the path by which electrons travel downhill energetically in aerobic respiration
    9·1 answer
  • If my indicater shows a oh of 12 would it be acidic ,alkalinic or nuetral​
    7·1 answer
  • What is life gsjkdgkkceskjhvcskgjeghsvcjev
    14·1 answer
  • How many wavelengths are represented​
    9·1 answer
  • Jupiter has enough mass to make 318 earths. In contrast, uranus and neptune have only enough mass to make.
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!