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
vitfil [10]
3 years ago
11

A playground merry-go-round of radius R = 1.40 m has a moment of inertia I = 265 kg · m2 and is rotating at 11.0 rev/min about a

frictionless vertical axle. Facing the axle, a 27.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round?
Physics
1 answer:
Tom [10]3 years ago
6 0

Answer:

The value of new value of angular speed of merry go round.\omega_{2} = 0.96 \frac{rad}{sec}

Explanation:

Given data

r = 1.4 m

Moment of inertia I_{1} = 265 kg - m^{2}

N_{1} = 11 RPM

\omega_{1} = \frac{2 \pi N}{60}

\omega_{1} = \frac{2 \pi (11)}{60}

\omega_{1} = 1.15 \frac{rad}{sec}

From conservation of momentum principal

I_{1} \omega_{1}  = I_{2} \omega_{2} ------- (1)

I_{2} = m r^{2} + 265

I_{2} = 27 (1.4)^{2} + 265

I_{2} = 317.92 \ kg m^{2}

Put all the values in equation  (1)

265 × 1.15 = 317.92 × \omega_{2}

\omega_{2} = 0.96 \frac{rad}{sec}

This is the value of new value of angular speed of merry go round.

You might be interested in
Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and
djverab [1.8K]

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>

To calculate the magnitude of the average friction force exerted on the collar we are using the formula,

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Here we are given,

k = The spring has a spring constant.

= 25.5 N/m.

x_f = Final length of the spring .

= \sqrt{1.25^2+1.8^2}  -0.60

= 1.591 m

x_i= The initial length of the spring.

= 1.25−0.60

=0.65 m

y=The collar then travels downward a distance.

=  1.80 m.

m= The mass of the collar.

=3.55 kg

v_c = the velocity of the collar.

= 3.39 m/s.

g = The acceleration due to gravity.

= 9.81 m/s²

We have to calculate the magnitude of the average friction force exerted on the collar = F

Now we put the known values in the above equation, we get;

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Or, \frac{1}{2} \times 25.5 \times((1.591)^2 - (0.65)^2 ) + F\times 1.80 + \frac{1}{2}\times 3.55\times (3.39)^{2}  = 3.55\times 9.81\times 1.80

Or, F= 8.641 N

From the above calculation we can conclude that,

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

Learn more about friction:

brainly.com/question/24338873

#SPJ4

Disclaimer: This question is incomplete in the portal. Here is the complete question.

Question:

The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .

6 0
2 years ago
Which of the following is an example of the conclusion phase of the scientific method?
Burka [1]

Answer:

a scientist examines the results and answers the lab question- last choice

5 0
3 years ago
A piece of glass has a temperature of 72.0 degrees Celsius. The specific heat capacity of the glass is 840 J/kg/deg C. A liquid
Nezavi [6.7K]

Answer:

741 J/kg°C

Explanation:

Given that

Initial temperature of glass, T(g) = 72° C

Specific heat capacity of glass, c(g) = 840 J/kg°C

Temperature of liquid, T(l)= 40° C

Final temperature, T(2) = 57° C

Specific heat capacity of the liquid, c(l) = ?

Using the relation

Heat gained by the liquid = Heat lost by the glass

m(l).C(l).ΔT(l) = m(g).C(g).ΔT(g)

Since their mass are the same, then

C(l)ΔT(l) = C(g)ΔT(g)

C(l) = C(g)ΔT(g) / ΔT(l)

C(l) = 840 * (72 - 57) / (57 - 40)

C(l) = 12600 / 17

C(l) = 741 J/kg°C

5 0
3 years ago
How are musical instruments grouped?
rosijanka [135]

A and B are equivalent.  That's one way instruments are often grouped. (the "sopranos", the "altos", the "bass")

C is another way instruments are often grouped; (the "woods", the "brass")

D is another way instruments are often grouped; (the "strings", the "percussions")

6 0
3 years ago
Read 2 more answers
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
3 years ago
Other questions:
  • 13. An aircraft heads North at 320 km/h rel:
    5·1 answer
  • The distance between two cities is 144km,it takes me 3hours to travel between these cities.What speed did I travel at?​
    15·1 answer
  • Use the law of conservation of energy to explain why the work output of a machine can never exceed the work input
    7·2 answers
  • Pcollisin physics :)
    7·1 answer
  • Which of the following oceans lies between North America, South America, Europe and Africa?
    5·2 answers
  • Name and draw the circuit symbols for the
    10·1 answer
  • Need some help please answer please
    11·1 answer
  • Q3. A large block of stone is held at a height of 15m having gained 4500J
    8·1 answer
  • Which is the most common direction of motion in the solar system, both for orbital revolution and axial rotation?.
    6·1 answer
  • Rearrange the equation = KE<br> -1<br> 2<br> - my? to solve for v. Show your work.
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!