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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

Tom [10]

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.

6 0

3 years ago

sing a rope that will snap if the tension in it exceeds 361 N, you need to lower a bundle of old roofing material weighing 455 N

yaroslaw [1]

Answer:

a)-2m/s^2

b)27.2m/s

Explanation:

Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)

W=455N=weight

W=mg

W=455N=weight

$m=\frac{W}{g} =\frac{455}{9.81}=46.38kg$

The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration

$a=\frac{T-W}{m} =\frac{361-455}{46.38kg} =-2m/s^2$

for point b we use the equations of motion with constant acceleration to find the velocity

$Vf=\sqrt{X(2)(a)+Vo^2}$

Where

Vf = final speed

Vo = Initial speed =0

A = acceleration =2m/s

X = displacement =6.8m

Solving

$Vf=\sqrt{(6.8)(2)(2)+0^2}=27.2m/s$

4 0

3 years ago

A person drives a car around a circular road with a constant speed of 20 m/s. The

ale4655 [162]

Answer:

16 m/s^2

Explanation:

acceleration tangential = (v^2)/r

a=400/25

a=16 m/s^2

Side note: next time, be more specific when asking about acceleration in circular motion. There's more than one type! Example:

angular acceleration=acceleration tangential/r

angular acc.=16/25

angular acc.=0.64 rad/s^2

5 0

3 years ago

Read 2 more answers

A uniformly charged conducting sphere of 1.3 m diameter has a surface charge density of 8.1 µc/m2. (a) find the net charge on th

8_murik_8 [283]

The surface charge density = q/A So q = surface charge density x Area The surface area of a sphere of radius R is 4*Pi*R^2. R = d/2 where d is diameter. This leaves us with 1.3/2 = 0.65. Area = 4 * pie * (0.65)^2 = 5.30998. So the net charge q = 8.1 * 10^(-6) * 5.30998 = 42.47998 * 10^(-6) The Total electric flux = Q/e_0 where , 8.854 Ă— 10â’12, e_0 is permitivity of free space. So Flux = 42.47998 * 10^(-6) / 8.854 * 10(â’12) = 4.833 * 10^(-6 - (-12)) = 4.833 * 10^(6)

8 0

3 years ago

A particle is constrained to move round a circle radius 382400km and makes a single revolution in 27.3 days. (i). Find the veloc

andreyandreev [35.5K]

The velocity and acceleration of the particle moving round the circle is;

Velocity = 162.12 m/s

Velocity = 162.12 m/sAcceleration = 6.873 × 10^(-5) m/s²

We are given;

Radius of circle; 382400 km = 382400000 m

Time; t = 27.3 days = 27.3 × 86400 s = 2358720 s

Now, formula for velocity is;

Velocity = distance/time

Thus;

I) velocity = 382400000/2358720

Velocity = 162.12 m/s

II) Acceleration is centripetal acceleration and is given by the formula;

a = v²/r

a = 162.12²/382400000

a = 6.873 × 10^(-5) m/s²

Read more at; brainly.com/question/12199398

5 0

3 years ago