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antoniya [11.8K]
3 years ago
10

A constant torque is applied to a rigid wheel whose moment of inertia is 2.0 kg · m2 around the axis of rotation. If the wheel s

tarts from rest and attains an angular velocity of 25.0 rad/s in 13.0 s, what is the applied torque (in N · m)? (Indicate the direction with the sign of your answer.)
Physics
1 answer:
Jlenok [28]3 years ago
4 0

Answer:

The applied torque is 3.84 N-m.      

Explanation:

Given that,

Moment of inertia of the wheel is 2\ kg-m^2

Initial speed of the wheel is 0 (at rest)

Final angular speed is 25 rad/s

Time, t = 13 s

The relation between moment of inertia and torque is given by :

\tau=I\alpha \\\\\tau=I\times \dfrac{\omega_f}{t}\\\\\tau=2\times \dfrac{25}{13}\\\\\tau=+3.84\ N-m

So, the applied torque is 3.84 N-m.

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A bowler throws a bowling ball of radius R = 11.0 cm down the lane with initial speed = 8.50 m/s. The ball is thrown in such a w
Ad libitum [116K]

Answer:

a) 1.18 seconds

b) 8.6 m

c) 5.19 revolutions

d) 6.07 m/s

Explanation:

<u>Step 1: </u>Data given

radius of the ball = 11.0 cm

Initial speed of the ball = 8.50 m/s

The coefficient of kinetic friction between the ball and the lane is 0.210.

<em></em>

<em>(a) For what length of time does the ball skid?</em>

The velocity at time t can be written as v(t) = v0 + at

 ⇒ with v(t) = the velocity at time t

⇒ with v0 : the initial velocity = 8.50 m/s

⇒ with a = the acceleration (in m/s²)

   ⇒The acceleration (negative) due to friction: a = -µg

           ⇒ with µ = 0.210

          ⇒ with g = 9.81 m/s²

v(t) =8.5m/s - 0.21*9.81m/s² * t = 8.5 - 2.06t

Torque τ = Iα = (2m(0.11m)²/5)α = 0.00484m*α

τ = F * r = µm*g*R = 0.21 * M * 9.81m/s² * 0.11m = 0.227m

so α = 0.227m / 0.00484m = 46.9 rad/s²

angular velocity ω(t) = ωo + αt = 0 + 46.9 rad/s² * t

The ball stops sliding when v(t) = ω(t) * r

8.5 - 2.06t  = 46.9*0.11*t = 5.159t

7.219t = 8.5

<u>t = 1.18 seconds</u>

<em>b) How far down the lane does it skid?</em>

s = Vo*t + ½at² = 8.5m/s * 1.18s - ½* 2.06 m/s² * (1.18s)² = <u>8.6 m</u>

<em>c) How many revolutions does it make before it starts to roll?</em>

The angular acceleration of the ball is:

α =  τ/I

 ⇒ with  τ = the torque experienced by the ball due the frictional force

   ⇒  τ = fk*R

α = fk*R /I

 ⇒ I = 2/5 m*R²

 ⇒ fk = µk*m*g

α = (µk*m*g*R)/(2/5mR²)

α = 5µk*g /2R

The angular displacement of the ball is:

∅ = 1/2αt²

⇒ The ball does not have an initial angular velocity

∅ =1/2*(5µk*g/2)*t²

∅ = 5µkgt²/4R

∅ = (5*0.21*9.81*1.18²)/(4*11.0 *10^-2)

∅ = 32.6 rad

Number of revolutions = 32.6 rad /2π

<u>Number of revolutions = 5.19</u>

<em>(d) How fast is it moving when it starts to roll?</em>

v = Vo + at = 8.5m/s - 2.06m/s² * 1.18s = <u>6.07 m/s</u>

7 0
3 years ago
Derive velocity-time relation from velocity-time graph ​
Brrunno [24]

Explanation:

hope this helps you

.....,.......

7 0
3 years ago
"An alpha particle (He2+) with a velocity of 2.6x106m/scrosses a uniform magnetic field at an angle of 37.0°to the field lines a
sergey [27]

Answer:

B = 5.59x10⁹ T

Explanation:

The magnetic force (F), on a the alpha particle with charge (q) that is moving at velocity (v) as the cross product of the velocity and magnetic field (B) is:

F = qvBsin(\theta)

<u>We have:</u>

F = 1.4x10⁻³ N

v = 2.6x10⁶ m/s

θ = 37.0°

q = 2*p = 2*1.6x10⁻¹⁹ C

Hence, the strength of the magnetic field is:

B = \frac{F}{qvsin(\theta)} = \frac{1.4 \cdot 10^{-3}}{1.6 \cdot 10^{-19} C*2.6 \cdot 10^{6}*sin(37)} = 5.59 \cdot 10^{9} T

Therefore, the strength of the magnetic field is 5.59x10⁹ T.

I hope it helps you!

7 0
3 years ago
Earth has a radius of 6370 km what is earth's mass?
Oliga [24]

Answer:

approx 6 × 10^24 kg is earth's mass

3 0
3 years ago
A truck accelerating at 0.0083 meters/second2 covers a distance of 5.8 × 104 meters. If the truck's mass is 7,000 kilograms, wha
anygoal [31]
<span>F = m*a = 7000kg * 9.8N/kg = 68,600 N. 

68,600 N is your answer
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3 0
3 years ago
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