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
Oliga [24]
3 years ago
14

At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan

t speed. At time t2 = 5.00 s, its acceleration is 4.00 i - 6.00 j m/s2 . What is the radius of the path taken by the particle if t2-t1 is less than one period?

Physics
1 answer:
Jet001 [13]3 years ago
8 0

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time t_1, the acceleration vector has direction \theta_1 such that

\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ

which indicates the particle is situated at a point on the lower left half of the circle, while at time t_2 the acceleration has direction \theta_2 such that

\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ

which indicates the particle lies on the upper left half of the circle.

Notice that \theta_1-\theta_2=90^\circ. That is, the measure of the major arc between the particle's positions at t_1 and t_2 is 270 degrees, which means that t_2-t_1 is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}

where R is the radius of the circle and T is the period. We have

t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s

and the magnitude of the particle's acceleration toward the center of the circle is

\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}

So we find that the path has a radius R of

7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}

You might be interested in
A thunderstorm loses energy in the _______ stage.
Irina-Kira [14]
 dissipation is the answer ;(
3 0
3 years ago
Read 2 more answers
A force is applied to the rim of a disk that can rotate like a merry-go-round, so as to change its angular velocity. Its initial
yawa3891 [41]

Answer:

A = B = C > D

Explanation:

Work done to stop the disc is given as

W = \Delta K

W = \frac{1}{2}I(\omega_f^2 - \omega_i^2)

so we have

(a) –2 rad/s, 5 rad/s;

(b) 2 rad/s, 5 rad/s;

(c) –2 rad/s, –5 rad/s; and

(d) 2 rad/s, –5 rad/s.

So we have

W_a = W_b = W_c = \frac{1}{2}I(5^2 - 2^2)

W_d = \frac{1}{2}I(2^2 - 5^2)

so we have

A = B = C > D

5 0
3 years ago
Which is true about total internal reflection?
Shkiper50 [21]

Answer:

It can occur only when light is incident on an interface where the index of refraction on the other side is less.

Explanation:

When the light passes from a denser medium, with refractive index n1, to another less dense medium, with refractive index n2, the incident light beam is refracted in such a way that it is not able to cross the surface between both media, the light beam is fully reflected and completely confining in the optically denser medium through which it propagates. For this phenomenon to occur, it is necessary that the angle of the incident light beam with respect to the normal be greater than or equal to the critical incidence angle θc. The critical angle can be calculated as :

\theta_c=arcsin(\frac{n_2}{n_1})

7 0
3 years ago
g An object with mass m=2 kg is completely submerged, and tethered, to the bottom of a large body of water. If the density of th
Anit [1.1K]

Answer:

The tension in the rope is 20 N

Solution:

As per the question:

Mass of the object, M = 2 kg

Density of water, \rho_{w} = 1000\ kg/m^{3}

Density of the object, \rho_{ob} = 500\kg/m^{3}

Acceleration due to gravity, g = 10\ m/s^{2}

Now,

From the fig.1:

'N' represents the Bouyant force and T represents tension in the rope.

Suppose, the volume of the block be V:

V = \frac{M}{\rho_{ob}}              (1)

Also, we know that Bouyant force is given by:

N = \rho_{w}Vg

Using eqn (1):

N = \rho_{w}\frac{M}{\rho_{ob}}g

N = 1000\frac{2}{500}\times 10 = 40\ N

From the fig.1:

N = Mg + T

40 = 2(10) + T

T = 40 - 20 = 20 N

N = \rho_{w}Vg

3 0
3 years ago
Calculate the horizontal projectile range with the cannon elevated 6m above the x-axis, with an initial speed of 14 m/s and an i
Paul [167]

Answer:

d = 20.25 m

Explanation:

h = 6m, V = 14 m/s, θ=14°

⇒ Vx=V × Cosθ=13.58 m/s and Vy = V × Sinθ=3.39 m/s

to find time t by h=-Vy t + 1/2 g t² (after putting value and simplifying) (-ve for downward motion)

4.9  t² - 3.39 t - 6 =0

⇒ t=1.5 sec (ignoring -ve root)

Now Vx = d/t

13.5 m/s = d / 1.5 s    ⇒ d = 20.25 m

5 0
3 years ago
Read 2 more answers
Other questions:
  • Which is a consequence of the third law of thermodynamics Energy is not always conserved. Engines cannot discharge waste heat. H
    13·2 answers
  • The gravitational force between the charged constituents of the atom is negligible compared with the electric force between them
    7·1 answer
  • What is the longest wavelength of radiation that possesses the necessary energy to break the o-o bond?
    10·1 answer
  • How many nanoseconds does it take light to travel 1.00 ft in vacuum? (This result is a useful quantity to remember.)
    9·1 answer
  • Convert 26.4 mi to km
    10·1 answer
  • Where is the magnetic south pole compared to the geographical north pole?
    8·1 answer
  • Ehren is trying to increase his mile run from eight to six minutes. He has decided to run some sandy hills near his home. Which
    9·1 answer
  • Hi! Can somebody please help?
    6·1 answer
  • 3. Riddle:
    13·2 answers
  • What happens to heat energy after an object is cooled down lolololol asking for my bff
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!