An object is constrained by a cord to move in a circular path of radius 1-m on a frictionless, horizontal surface. The cord will break if the tension exceeds 26.9-N. The maximum kinetic energy that this object can have is _____ J. Round your answer to the nearest whole number.

Answers

Answer 1

Answer:

Answer:

Explanation:

Let m be the mass of the object and v be the maximum velocity . The tension will provide centripetal force for the circular motion .

T = mv² / R where R is radius of circular path . T is tension .

putting the values given in the equation above

26.9 = m v² / 1

m v² = 26.9

kinetic enrgy = 1/2 m v²

= 26.9 / 2

= 13.45 J

13 J .

Maximum kinetic energy = 13 J .

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The time taken by a mass projected verticallyupwards to reach the maximum height (with air
resistance not neglected) is 10 sec. The time of
descent of the mass from the same height will be

Answers

Answer:

10s

Explanation:

The time to get to the maximum would be the same as the time to get down to the maximum unless somehow gravity’s changes during the duration it goes up to and from maximum height.

Which state of matter is most similar to solids

Answers

Answer:

liquids

Explanation

You wish to buy a motor that will be used to lift a 10-kg bundle of shingles from the ground to the roof of a house. The shingles are to have a 1.5-m/s2 upward acceleration at the start of the lift. The very light pulley on the motor has a radius of 0.17 m . Part A Determine the minimum torque that the motor must be able to provide. Express your answer with

Answers

To lift a 10-kg bundle of shingles with an upward acceleration of 1.5 m/s², a motor with a pulley of radius 0.17 m must provide a torque of at least 19 N.m.

We wish to use a motor to lift a 10-kg (m) bund of shingles with an upward acceleration of 1.5 m/s² (a).

The resulting force (F = m.a) is the difference between the tension (T) of the pulley and the weight (w = m.g) of the shingles.

$T-m.g=m.a\nT = m.g+m.a = m(g+a) = 10 kg (9.8m/s^(2)+1.5m/s^(2) )=113 N$

where,

g: gravity

Then, we can calculate the minimum torque (τ) that the motor must apply using the following expression.

$\tau = r * T = 0.17m * 113N = 19N.m$

where,

r: radius of the pulley

To lift a 10-kg bundle of shingles with an upward acceleration of 1.5 m/s², a motor with a pulley of radius 0.17 m must provide a torque of at least 19 N.m.

Learn more: brainly.com/question/19247046

Answer:

$\tau=19.21\ N-m$

Explanation:

It is given that,

Mass of bundle of shingles, m = 10 kg

Upward acceleration of the shingles, $a=1.5\ m/s^2$

The radius of the motor of the pulley, r = 0.17 m

Let T is the tension acting on the shingles when it is lifted up. It can be calculated as :

$T-mg=ma$

$T=m(g+a)$

$T=10* (9.8+1.5)$

T = 113 N

Let $\tau$ is the minimum torque that the motor must be able to provide. It is given by :

$\tau=r* T$

$\tau=0.17* 113$

$\tau=19.21\ N-m$

So, the minimum value of torque is 19.21 N-m. Hence, this is the required solution.

A bag of cement of Weight 1000N hangs from ropes. Two of the ropes make angles of 1=60 and 2=30 with the horizontal.if the system is in equilibrium,find the tension T1,T2andT3 in the ropes

Answers

Answer:

T1 = 499.9N, T2 = 865.8N, T3 = 1000N

Explanation:

To find the tensions we need to find the vertical and horizontal components of T1 and T2

T1x = T1 cos60⁰, T1y = T1 sin60⁰

Also, T2x = T2 cos30⁰, T2y = T2 sin30⁰

For the forces to be in equilibrium,

the sum of vertical forces must be zero and the sum of horizontal forces must also be zero

Sum of Fx = 0

That is, T1x - T2x=0

NB: T2x is being subtracted because T1x and T2x are in opposite directions

T1 cos60⁰ - T2 cos30⁰ = 0

0.866T1 - 0.5T2 = 0 ............ (1)

Sum of Fy = 0

T1y + T2y - 1000 = 0

T1 sin60⁰ + T2 sin30⁰ - 1000 = 0

NB: The weight of the bag of cement is also being subtracted because it's in an opposite direction.

0.5T1 - 0.866T2 - 1000 = 0 ........(2)

From (1)

make T1 the subject

T1 = 0.5T2/0.866

Substitute T1 into (2)

0.5 (0.5T2/0.866) - 0.866T2 = 1000

(0.25/0.866)T2 - 0.866T2 = 1000

0.289T2 - 0.866T2 = 1000

1.155T2 = 1000

T2 = 865.8N

Then T1 = 0.5 x 865.8 / 0.866

T1 = 499.9N

T3 = 1000N

NB: The weight of the bag is the Tension above the rope, which is T3

A 84-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in 0.73 as. The elevator travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.4 s, and then comes to rest.What does the spring scale register During the first 0.80s of the elevator’s ascent?

Answers

Answer:

$SR=949.2N$

Explanation:

From the question we are told that:

Mass $M=84kg$

Speed $V=1.2m/s$

Acceleration Time $t_a=0.73$

Constant speed Time $t_s=5.0s$

Deceleration time $t_d=1.4s$

Generally the equation for Acceleration is mathematically given by

$a=(v)/(t)$

Therefore acceleration for the first 0.80 sec is

$a=(1.2)/(0.80)$

$a=1.5m/s^2$

Therefore

Spring Reading=Normal force -Reaction

$SR=m(g+a)$

$SR=84(9.8+1.5)$

$SR=949.2N$

A 2.4 kg toy oscillates on a spring completes a cycle every 0.56 s. What is the frequency of this oscillation?

Answers

Answer:

$f=1.79Hz$

Explanation:

The period is defined as the time taken by an object to complete a cycle in a simple harmonic motion. As the frequency of the motion increases, the period decreases. Therefore, they are inversely proportional. The frequency does not depend on the mass of the object.

$f=(1)/(T)\nf=(1)/(0.56 s)\nf=1.79Hz$

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