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
vichka [17]
2 years ago
7

A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg⋅m2 about a vertical axle through its center, and it

turns with negligible friction.
(a) A child applies an 18.0 N force tangentially to the edge of the merry-go-round for 15.0 s. If the merry-go-round is initially at rest, what is its angular speed after this 15.0 s interval?
(b) How much work did the child do on the merry-go-round?
(c) What is the average power supplied by the child?
Physics
1 answer:
daser333 [38]2 years ago
7 0

Answer:

a) 0.31 rad/s

b) 100 J

c) 6.67 W

Explanation:

(a) the force would generate a torque of:

T = FR = 18 * 2.4 = 43.2 Nm

According to Newton 2nd law, the angular acceleration would be

\alpha = \frac{T}{I} = \frac{43.2}{2100} = 0.021 rad/s^2

It starts from rest, then after 15s it would achieve a speed of

\omega = \alpha t = 0.021 * 15 = 0.31 rad/s

(b) The distance angle swept by it is:

\theta = \frac{\alpha t^2}{2} = \frac{0.021 * 15^2}{2} = 2.314 rad

Hence the work by the child

W = T\theta = 43.2 *2.314  \approx 100 J

c) Average power to work per time unit

P = \frac{W}{t} = \frac{100}{15} = 6.67 W

You might be interested in
Write the ground-state electron configurations of the following ions. (a) Li+ (b) N3− (c) In3+ (Use the noble gas core electron
maria [59]

Li+  [He]

N³-  [Ne]

In³+  [Kr] 4d10

Tl+  [Xe] 4f14 5d10 6S2

6 0
3 years ago
Read 2 more answers
On what principle does a bottle opener work
Vsevolod [243]

Answer:

Bottle opener works on a fulcrum.

Explanation:

A bottle opener is a second-class lever because the pivot point is at one end of the opener and the load is in the middle.

8 0
3 years ago
How much force is needed to accelerate a 4kg mass at 12m/s/s?
zloy xaker [14]
I guess 48 , but I’m not sure
7 0
2 years ago
Read 2 more answers
A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

\mathbf{v_{stone}= 14.6 \ m/s}

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

5 0
2 years ago
What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?
Daniel [21]

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

F=\frac{kq_{1}q_{2}  }{r^{2} }

Here, q_{1} andq_{2} are the charges on the pith balls, r is the separation between the charges and k is constant and its value is  8.99\times 10^9 N m^2/C^2.

Given  q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C and  r=8 cm=8\times10^{-2} m.

Substituting these values in above formula we get,

F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9}  N\\\\F=8.8\times10^{-4}N

Thus, the repulsive force between two pith balls is 8.8\times10^{-4}N.

7 0
3 years ago
Read 3 more answers
Other questions:
  • A surface encloses the charges q1 = 3.2 μC , q2 = 6.7 μC , and q3 = -3.8 μC. Find the electric flux through this surface.
    10·1 answer
  • The diagram shows that currently about 100 gigatons/year of carbon are moved from the atmosphere to vegetation on land. Which pr
    15·2 answers
  • A train travelling at a speed of 108km/h.how much distance is it traveling every second ?
    6·1 answer
  • A gas is in a sealed container.Part ABy what factor does the gas temperature change if the volume is doubled and the pressure is
    14·1 answer
  • Star formation begins when
    6·2 answers
  • Using the standard convention "positive is up", what is the sign of the acceleration of a rock you throw up (a) as it is going u
    10·1 answer
  • Light travels fastest when moving through
    15·1 answer
  • A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes
    10·1 answer
  • A car goes from a speed of 10m/s to 30m/s over 4s. If the car has a mass of 1000kg, How strong is the force pushing the car forw
    5·1 answer
  • Which item is a pure substance?
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!