A 40% pressure drop across an oil filter screen indicates that?

Which is affected more by an earthquake, : deep sea or less deep sea???

lisabon 2012 [21]

Answer:

deep sea will obviously be more affected because of the sea floor shaking

6 0

2 years ago

Read 2 more answers

A spherical bowling ball with mass m = 3.6 kg and radius R = 0.101 m is thrown down the lane with an initial speed of v = 8.7 m/

luda_lava [24]

Answer:

1) The magnitude of the angular acceleration = 67.92 rad/ $s^{2}$

2) Magnitude of the linear acceleration = 2.744 m/ $s^{2}$

3) How long does it take the bowling ball to begin rolling without slipping = 0.906 s

4) How long does it take the bowling ball to begin rolling without slipping = 6.75 m

5) the final velocity is 6.21 m/s

Explanation:

the given information :

Bowling mass m = 3.6 kg

Radius = 0.101 m

Initial speed $v_{0}$ = 8.7 m/s

Coefficient of kinetic friction μ = 0.28

1) he magnitude of the angular acceleration of the bowling ball is

F = m a

$F_{g}$ = μ N , N = m g

$F_{g}$ = μ m g

1) The magnitude of the angular acceleration of the bowling ball as it slides down the lane:

momen inersia of Bowling ball I = (2/5) m $R^{2}$

torque τ = I α

τ = F R

I α = F R

(2/5) m $R^{2}$ α = μ m g R

α = (5 μ g / 2R) μ g R

= (5 x 0.28 x 9.8/ 2 x 0.101)

= 67.92 rad/ $s^{2}$

2) Magnitude of the linear acceleration of the bowling ball as it slides down the lane

F = - $F_{g}$ , $F_{k}$ is the force of kinetic friction

m a = - μ m g, remove m

the magnitude of linear accelaration is

a = μ g

= (0.28) (9.8)

= 2.744 m/ $s^{2}$

3) The bowling ball takes time to begin rolling without slipping:

The linear speed, $v_{t}$ = $v_{0}$ - a t

$v_{t}$ = $v_{0}$ - μ g t

the angular speed, ω = ω0 + α t

ω = ω0 + (5 μ g/2R ) t

$v_{t}$ = ω R

$v_{0}$ - μ g t = ω0 R + (5 μ g/2R ) t R

7 μ g t/2 = $v_{0}$ + ω0 R

hence,

t = (2 $v_{0}$ + ω0 R)/ 7 μ g

ω0 = 0 (no initial spin), therefore

t = 2 $v_{0}$ / 7 μ g

= 2 x 8.7 / 7 (0.28) (9.8)

= 0.906 s

4) How long it takes for the bowling ball to begin rolling without slipping, S

S = $v_{0}$ t - (1/2) a $t^{2}$

= (8.7) (0.906) - (1/2) (2.744) $0.906^{2}$

= 6.75 m

5) The final velocity

$v_{t}$ = $v_{0}$ - a t

$v_{t}$ = 8.7 - (2.744) (0.906)

$v_{t}$ = 6.21 m/s

4 0

3 years ago

A. Compute the torque developed by an industrial motor whose ouputis 150kW at an angular speed of 4000 rev/min.

Gnom [1K]

Answer:

A. τ = 358 N.m

B. m = 73kg

C. V = 209.5 m/s

Explanation:

Let's first convert angular speed to rad/s:

ω = 4000 rev/min * 2π / 60 = 419 rad/s

Since Power is P = τ * ω,

τ = P / ω = 358 N.m

For part B: On the hanging weight:

T - m*g = 0

T = m*g

On the drum:

τ = T*R Replacing the expression for Tension of the rope:

τ = m*g*R

Solving for m:

m = 73 kg

For part C:

V = ω * R = 209.5 m/s

5 0

2 years ago

What limits the amplitude of motion of a real vibrating system that is driven at one of its resonant frequencies?

andriy [413]

To find limits of the amplitude of motion of real vibrating system that is driven at one of its resonant frequencies, Here the damping of vibrations turns the energy of vibration into internal energy.

5 0

1 year ago

Help me with theses 2 questions please.

spin [16.1K]

1)

Answer: A inclined plane and staircase reaching the same height. Work depends on initial and final position only.

Explanation:

The amount of work done does not depend on path length. It depends on initial and final position. It is given by gravitational potential energy.

W = m g h

Where, m is the mass of the object, g is the acceleration due to gravity and h is the height.

Let the mass of the object be m. Two different paths ( one inclines plane and another staircase) are taken to same point. The object would gain same gravitational potential energy because it depends on the height through which the object is taken and not the actual path length.

W ∝ h.

2)

Spring constant can be defined as the ratio of amount of force acting spring with the displacement caused due to the force. A deformed elastic object has elastic potential energy stored in it which is equal to the work done to deform the object. It is proportional to the spring constant and amount of stretched distance.

Equilibrium is reached when the net force acting on the spring is zero. In equilibrium state, the object has natural extension. An object is said to possess elastic potential energy when force is applied and object deforms relative to original equilibrium shape or length.

4 0

3 years ago