PHYSICS COLLEGE

Suppose that an object undergoes simple harmonic motion, and its displacement has an amplitude A = 15.0 cm and a frequency f = 11.0 cycles/s (Hz). What is the maximum speed ( v ) of the object?A. 165 m/s
B. 1.65 m/s
C. 10.4 m/s
D. 1040 m/s

Answers

Answer 1
Answer:

Answer:

Maximum speed ( v ) = 10.4 m/s (Approx)

Explanation:

Given:

Amplitude A = 15.0 cm = 0.15 m

Frequency f = 11.0 cycles/s (Hz)

Find:

Maximum speed ( v )

Computation:

Angular frequency = 2πf

Angular frequency = 2π(11)

Angular frequency = 69.14

Maximum speed ( v ) = WA

Maximum speed ( v ) = 69.14 x 0.15

Maximum speed ( v ) = 10.371

Maximum speed ( v ) = 10.4 m/s (Approx)

Related Questions

In general terms, the efficiency of a system can be thought of as the output per unit input. Which of the expressions is a good mathematical representation of efficiency e of any heat engine? Where, Qh: the absolute value (magnitude) of the heat absorbed from the hot reservoir during one cycle or during some time specified in the problem Qc: the absolute value (magnitude) of the heat delivered to the cold reservoir during one cycle or during some time specified in the problem W: the amount of work done by the engine during one cycle or during some time specified in the problem A) e=QhW B) e=QcQh C) e=QcW D) e=WQh E) e=WQc
Instantaneous speed is...a) A speed of 1000 km/hb) The speed attained at a particular instant in time.c) The speed that can be reached in a particular amount of time.PLEASE HURRY
Find the intensity III of the sound waves produced by four 60-WW speakers as heard by the driver. Assume that the driver is located 1.0 mm from each of the two front speakers and 1.5 mm from each of the two rear speakers.
When dots are further apart on a ticker-tape diagram, it indicates an object is moving
You have a small piece of iron at 75 °C and place it into a large container of water at 25 °C. Which of these best explains what will occur over time. A:The iron will cause the water to boil and turn to steam. B:The iron will take the heat from the water and get hotter. C:The water will cool significantly due to its colder temperature. D:Some heat from the iron will move to the water causing both to change temperatures.

What power lens is needed to correct for farsightednesswherethe uncorrrected near point is 75 cm?

Answers

To solve this problem we will apply the concept related to the lens power with which farsightedness can be corrected. Mathematically this value is given by the relationship,

P = (1)/(f)

Here,

f =focal length

In turn, said expression can be exposed in terms of the distance of the object and the image as:

P = (1)/(p)+(1)/(q)

Here,

p = Object Distance ( By convention is 25cm)

q = Image distance

Replacing we have,

P = (1)/(0.25)+(1)/(-0.75)

P = +2.67D

Therefore the power lens that is needed to correct for farsightedness is +2.67D

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 25 kg and the larger bottom crate has a mass of m2 = 91 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is µs = 0.79 and the coefficient of kinetic friction between the two crates is µk = 0.62. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).1)The rope is pulled with a tension T = 234 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?2)In the previous situation, what is the magnitude of the frictional force the lower crate exerts on the upper crate?3)What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?4)The tension is increased in the rope to 1187 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?m/s25)As the upper crate slides, what is the acceleration of the lower crate?

Answers

Answer:

Explanation:

Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.

1 ) Common acceleration a = force / total mass

= 234 / ( 25 +91 )

= 2.017 m s⁻².

2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂

= mass x acceleration

= 25 x 2.017

= 50.425 N

The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.

3 ) Maximum friction force that is possible between m₁ and m₂

= μ_s m₁g

= .79 x 25 x 9.8

= 193.55 N

Acceleration of m₁

= 193 .55 / 25

= 7.742 m s⁻²

This is the common acceleration in case of maximum tension required

So tension in rope

= ( 25 +91 ) x 7.742

= 898 N

4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁

=  μ_k m₁g

= .62 x 25 x 9.8

= 151.9 N

Acceleration of m₁

= 151.9 / 25

= 6.076 m s⁻².

A 0.060 ???????? tennis ball, moving with a speed of 5.28 m/???? , has a head-on collision with a 0.080 ???????? ball initially moving in the same direction at a speed of 3.00 m/ ???? . Assume that the collision is perfectly elastic. Determine the velocity (speed and direction) of both the balls after the collision.

Answers

Explanation:

It is given that,

Mass of the tennis ball, m_1=0.06\ kg

Initial speed of tennis ball, u_1=5.28\ m/s

Mass of ball, m_2=0.08\ kg

Initial speed of ball, u_2=3\ m/s

In case of elastic collision, the momentum remains conserved. The momentum equation is given by :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

v_1\ and\ v_2 are final speed of tennis ball and the ball respectively.

0.06* 5.28+0.08* 3=0.06v_1+0.08v_2

0.06v_1+0.08v_2=0.5568..............(1)

We know that the coefficient of restitution is equal to 1. It is given by :

(v_2-v_1)/(u_1-u_2)=1

(v_2-v_1)/(5.28-3)=1

{v_2-v_1}=2.28.................(2)

On solving equation (1) and (2) to find the values of velocities after collision.

v_1=5.28\ m/s

v_2=3\ m/s

So, the speed of both balls are 5.28 m/s and 3 m/s respectively. Hence, this is the required solution.

Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second particle is at (−7 + 2t, −6 + 2t, −1 + t). (a) (5 Points) Do the paths of the two particles cross? If so, where?

Answers

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the paths meet, not the point where the particles meet themselves.

So, we can name the time of the first particle T_F ,  and the time of the second particle T_S.

Setting the locations equal, we get the following equations to solve for T_F and T_S:

(-1 + T_F) = (-7 + 2T_S)                     Equation 1

(4 - T_F) = (-6 + 2T_S)                        Equation 2

(-1 + 2T_F) = (-1 + T_S)                     Equation 3

Solving these three equations simultaneously we get:

T_F = 2 seconds

T_S = 4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of T_For T_Sin the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

A 0.010 kg ball is shot from theplunger of a pinball machine.Because of a centripetal force of0.025 N, the ball follows a
circulararc whose radius is 0.29 m. What isthe speed of the
ball?

Answers

Answer:

v = 0.85 m/s

Explanation:

Given that,

Mass of the ball, m = 0.01 kg

Centripetal force on the ball, F = 0.025 N

Radius of the circular path, r = 0.29 m

Let v is the speed of the ball. The centripetal force of the ball is given by :

F=(mv^2)/(r)

v=\sqrt{(Fr)/(m)}

v=\sqrt{(0.025* 0.29)/(0.01)}

v = 0.85 m/s

So, the speed of the ball is 0.85 m/s. Hence, this is the required solution.

How does an increase in cold working effect Modulus of Elasticity and why?

Answers

Answer:

There is a decrease in modulus of elasticity

Explanation:

Young's Modulus of elasticity also known as elastic modulus is the deformation of a body along a particular axis under the action of opposing forces along that axis. at atomic levels, it depends on bond energy or strength.

In cold working processes, plastic deformation a metal occurs below its re-crystallization temperature due to which crystal structure of metal gets distorted and as a result of dislocations fractures also occur resulting in hardening of metal but bonds at atomic levels defining elasticity are temporarily affected.

Thus an increase in cold working results in a decrease in modulus of elasticity.
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