A 0.60-kgkg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.3 mm on a frictionless horizontal surface. Part A If the cord will break when the tension in it exceeds 60 NN , what is the maximum speed the ball can have

Correct option is B 8 cm.

Let radius of gyration for the axis not passing through center of mass be r and that for the axis passing through the center of mass be k and the distance between the two parallel axes be a.

Parallel axes theorem gives:

Thus, option B is the correct answer.

9514 1404 393

Answer:

  about 5.89 seconds

Explanation:

The penny will hit the ground at the same time it would if it were simply dropped. The equation for the vertical motion is ...

  h(t) = -4.9t^2 +170 . . . . . where 170 is the initial height in meters

h(t) = 0 when ...

  4.9t^2 = 170

  t = √(170/4.9) ≈ 5.89

The penny will hit the ground in about 5.89 seconds.

Answer:

5 m/s

Explanation:

5000/1000=5 m/s

:))

Answer:

Approximately [tex]2\; \rm s[/tex], assuming that the floor of this parking lot is level, [tex]\mu_{\rm k} = 0.60[/tex], and [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].

Explanation:

Let [tex]m[/tex] denote the mass of this vehicle. Weight of this vehicle: [tex]m\, g[/tex].

If the floor of this parking lot is level, the normal force on this vehicle would be equal to its weight: [tex]N = m \, g[/tex].

Given that [tex]\mu_{\rm k}[/tex], the kinetic friction between this vehicle and the ground would be consistently [tex]\mu_{\rm k} \, N = \mu_{\rm k} \, m \, g[/tex] until the vehicle comes to a stop.

Assuming that all forces on this vehicle other than friction are balanced. The net force of this vehicle during braking would be [tex](-\mu_{\rm k} \, m \, g)[/tex] (negative because this force is opposite to the direction of the motion.)

By Newton's second law of motion, the acceleration of this vehicle would be:

[tex]\begin{aligned}a &= \frac{F_\text{net}}{m} \\ &= \frac{-\mu_{\rm k} \, m \, g}{m} \\ &= -\mu_{\rm k}\, g \\ &= -0.60 \times 9.81\; \rm m\cdot s^{-2} \\ &= -5.886\; \rm m\cdot s^{-2}\end{aligned}[/tex].

In other words, braking would reduce the velocity of this vehicle by a constant [tex]5.886\; \rm m\cdot s^{-1}[/tex] every second until the vehicle comes to a stop. Calculate the time it would take to reduce the velocity of this vehicle from [tex]v_{0} = 12\; \rm m\cdot s^{-1}[/tex] to [tex]v_{1} = 0\; \rm m\cdot s^{-1}[/tex]:

[tex]\begin{aligned}t &= \frac{v_{1} - v_{0}}{a} \\ &= \frac{0\; \rm m\cdot s^{-1} - 12\; \rm m\cdot s^{-1}}{-5.886\; \rm m \cdot s^{-2}} \\ &\approx 2.0\; \rm s \end{aligned}[/tex].

Acceleration is the slope of the velocity-time graph. Since the acceleration here is constant, the velocity-time graph of this vehicle would be a line with a negative slope.

Find the distance the ball travels:

3.4 meters - 1.5 meters = 1.9 meters

Now divide the distance the ball travels by the speed:

1.9 meters / 4.4 m/s = 0.43 seconds

Answer:

Explanation:

s = s₀ + v₀t + ½at²

There are an infinite number of solutions to this question as posed because we are not told the direction of the initial velocity.

Assuming ground is level and origin and UP the positive direction

The shortest amount of time possible is when the initial velocity is straight down

1.5 = 3.4 - 4.4t + ½(-9.8)t²

0 = -4.9t² - 4.4t + 1.9

t = (4.4 ±√(4.4² - 4(-4.9)(1.9))) / (2(-4.9))

positive answer is

t = 0.32 s

The longest amount of time possible is when the initial velocity is straight up.

1.5 = 3.4 + 4.4t + ½(-9.8)t²

0 = -4.9t² + 4.4t + 1.9

t = (-4.4 ±√(4.4² - 4(-4.9)(1.9))) / (2(-4.9))

positive answer

t = 1.22 s

If the initial velocity is horizontal, meaning no vertical velocity

1.5 = 3.4 + 0t + ½(-9.8)t²

-4.9t² = -1.9

t² = 0.38775...

t = 0.62 s

Any angle between UP and Down will have a different initial vertical velocity and result in a different time to catch height.

It appears from the comments on the other answer, that I have shown you how to arrive at three of the four possible solutions.  The initial direction is very important.

Hi there!

For this problem, we must use the conservation of angular momentum. This is an example of an inelastic "collision", so:

I₁w₁ + I₂w₂ = (I₁ + I₂)wf

We know that the moment of inertia of a disk is 1/2mR², so we can calculate the moments of inertia for both disks:

Disk 1: 1/2(2)(0.40²) = .16 kgm²/s

Disk 2: 1/2(2)(0.20²) = .04 kgm²/s

Plug in the values. Let counterclockwise be positive.

.16(-50) + .04(50) = (.16 + .04)wf

Solve:

wf = -30 rev/s

I think force can accelerate

Explanation:

i think force of an object

a. Assuming all energy involved is conserved, at the lowest point of the swing (which includes the moment the student grabs the rope), the student only has kinetic energy,

K = 1/2 m (3.5 m/s)²

and at the highest point of the swing, the student only has potential energy

P = mgh

The energies at the bottom and top of the swing must be equal, so

1/2 m (3.5 m/s)² = mgh

h = (3.5 m/s)² / (2g)

h = 0.625 m ≈ 63 cm

b. In part (a), we found the relationship

h = v²/(2g)

If we cut the speed v in half, we replace v in the equation above with v/2 :

h = (v/2)²/(2g)

and simplifying this gives

h = (v²/4)/(2g) = 1/4 • v²/(2g)

The factor of 1/4 tells you that reducing the speed by a factor of 1/2 reduces the height by a factor of 1/4. So he can swing as high as

1/4 (3.5 m/s)²/(2g) = 0.15625 m ≈ 16 cm

Answer:

Explanation:

If we assume the rod and sphere are of uniform construction so that their individual centers of mass are at their geometric centers, and that the rod end is attached to the surface of the sphere.

Balance moments about the rod free end of the assembly with its parts

(M + m)Cx = M(L/2) + m(L + R)

Cx = (M(L/2) + m(L + R)) / (M + m)

Answer:

192

Explanation:

The sun emits electromagnetic waves with a power of  

4.0 ∗ 10  (26)  W.

Answer:

the one in which the fluid has the lowest density

The impulse the floor exert on the box is 116 kgm/s.

The given parameters;

The final velocity of the box when it dropped to the floor is calculated as follows;

[tex]\frac{1}{2} mv^2 = mgh\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 1.1} \\\\v = 4.64 \ m/s[/tex]

The impulse the floor exert on the box is calculated as follows;

the impulse the floor exert on the box is equal to change in momentum of the book

[tex]J = \Delta P\\\\J = \Delta Mv\\\\J = M(v_f - v_0)\\\\J = 25(4.64 - 0)\\\\J = 116 \ kgm/s[/tex]

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Answer:

a) Impuise of force =F∗?(t) = area of F-T graph area= impulse =triangle + rectangle + triangle = 0.5*4*2 + 4*1 + 0.5*4*2 = 12 N-s (b) impulse = change in momentum \(= mExplanation:

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, C represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

[tex]tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}[/tex]

The steepness of the slope is therefore;

[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100[/tex]

Where;

[tex]\overline{AM}[/tex] = Half the width of the truck = [tex]\dfrac{2.4 \, m}{2}[/tex] = 1.2 m

[tex]\overline{CM}[/tex] = The elevation of the center of gravity above the ground = 2.2 m

[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%[/tex]

[tex]tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}[/tex]

[tex]Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}[/tex]

The maximum steepness of the slope where the truck can be parked is 54.55 %.

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Answer:

The angle of incidence = The angle of reflection

Explanation:

For instance, the formula is <i = <r which means if the surface is smooth then the reflacted anglw will be equal according to the normal (dotted 90°) line.

Upon being reflected backwards the angle of incidence is simply the same, except from the other side

Answer:

By applying the definition of torques ( [tex]\vec \tau = \vec r \times \vec F[/tex] ) and them remembering a few tricks.

Namely: if you wrap your RIGHT hand fingers around something and stick your thumb out, the direction your finger wraps gives you the verse of rotation and the thumb the orientation of the torque. Bottom force (4N) will give a counterclockwise rotation, torque is pointing up; top force (3N) will give a clockwise rotation and its torque its pointing down (read up and down as if the sheet the image is printed on is on your table).

In terms of magnitude the trick is easy: You want to multiply the intensity of the force (3N and 4N) by the distance between the point and the line the force it is applied to (that is, you don't care about the length of r itself, but the distance at a right angle, which is 0.9 and 0.8m respectively.

At this point, assuming "upwards" (relative to the plane of the sheet that is) torques positive, the 3N force gives you a torque of [tex]- 3N \times 0.9m = - 2.7N\cdot m[/tex] and the 4N force provides [tex]+4N\times 0.8 m = +3.2 N\cdot m[/tex]

Answer:

Light can travel in three ways from a source to another location: (1) directly from the source through empty space; (2) through various media; (3) after being reflected from a mirror.

Explanation:

The energy in the system is given by the initial potential energy at the point 1.

The linear velocity at point 3, is approximately 33.59 m/s.

Reasons:

The parameters are;

Height at point 1, h₁ = 83 m

Radius of the ring = 8 cm

Mass of the ring, M = 8 kg

Height at point 2, h₂ = 32 m

At point 2, we have;

Change in potential energy = Kinetic energy

Which gives;

(83 - 32) × 9.81 × 8 = 0.5 × 8 × v² + 0.5 × 8 × 0.08² × (v/0.08)²

Which gives;

v ≈ 22.37 m/s

At point 3, the rotational kinetic energy remains constant while the

translational kinetic energy increases as follows;

K.E. at point 3 = Initial kinetic energy + Change in potential energy

Which gives;

K.E. at point 3 = 0.5 × 8 × v₃³ ≈ 0.5×8×22.37² + 32×9.81×8

[tex]v_3^2 = \dfrac{0.5 \times 8 \times 22.37^2 + 32 \times 9.81 \times 8}{0.5 \times 8} = 1128.15[/tex]

v₃ ≈ √(1128.15) ≈ 33.59

The linear velocity at point 3, v₃ ≈ 33.59 m/s

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The probable question parameters as obtained from a similar question online are;

Height at point 1, h₁ = 83 m

Radius of the ring = 8 cm

Mass of the ring, M = 8 kg

Height at point 2, h₂ = 32 m

Answer:

An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.

Explanation:

Answer:

An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.

Explanation:

[tex]r = 1.29×10^8\:\text{m}[/tex]

Explanation:

According to Newton's law of universal gravitation, the gravitational force between Uranus and Miranda is

[tex]F_G = G\dfrac{M_UM_M}{r^2}[/tex]

where [tex]M_U[/tex] is the mass of planet Uranus, [tex]M_M[/tex] is the mass of its satellite Miranda, r is the distance between their centers and G is the universal gravitational constant. Moving the variable r to the left side, we get

[tex]r^2 = G\dfrac{M_UM_M}{F_G}[/tex]

Taking the square root of the equation above, we get

[tex]r = \sqrt{G\dfrac{M_UM_M}{F_G}}[/tex]

Plugging in the values, we get

[tex]r = \sqrt{(6.67×10^{-11}\:\text{N-m}^2{\text{/kg}}^2)\dfrac{(8.68×10^{25}\:\text{kg})(6.59×10^{19}\:\text{kg})}{2.28×10^{19}\:\text{N}}}[/tex]

[tex]\:\:\:\:\:=1.29×10^8\:\text{m}[/tex]

Answer:

4600kg

Explanation:

Density=mass÷volume

920=m/5

m=920×5=4600kg

Answer:

Explanation:

1)  average velocity is

v = (21 + 39)/2 = 30m m/s

d = vt 30(4.1) = 123 = 120 m

2)  d = ½gt²

d = ½(9.8)(4.1²)

d = 82.369 = 82 m

when rounding to the two significant digits of the question numerals.

Answer:

two of the same instrument playing notes at slightly different frequencies

Explanation:

The correct answer is two of the same instrument playing notes at slightly different frequencies.

Frequency is the number of occurrences of a periodic event per unit of time. In the context of sound waves, frequency refers to the number of complete cycles of a sound wave that occur in one second and is measured in units of Hertz (Hz).

In this question,

A beat frequency occurs when two sound waves with slightly different frequencies interfere with each other, resulting in a periodic variation in the amplitude of the resulting wave. The beat frequency is equal to the difference in frequency between the two original sound waves.

When two of the same instrument play notes at slightly different frequencies, the sound waves produced by each instrument will have slightly different frequencies due to differences in tuning or other factors. When the two sound waves interfere with each other, they will produce a beat frequency equal to the difference between their frequencies. This beat frequency will be audible as a periodic variation in the loudness or intensity of the sound.

In the other statements, There will not be any beat frequency because the sound waves being produced have either the same frequency or very different frequencies, which do not interfere with each other in a way that produces beats.

Therefore, The statement two of the same instrument playing notes at slightly different frequencies is correct.

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Explanation:

the speed of sound is affected by temperature and humidity

This question involves the concepts of the law of conservation of momentum, velocity, and mass.

The two quantities, the students should multiply before and after the collision are "A. mass and velocity".

According to the law of conservation of momentum, In an isolated system, the total momentum of the system before the collision is always equal to the total momentum of the system after the collision.

To prove the law of conservation of momentum, consider two balls of masses ‘m₁’ and ‘m₂’, moving with velocities ‘u₁’ and ‘u₂’, respectively, such that u₁ is greater than u₂. After some time, these balls collide with each other and their velocities become ‘v₁’ and ‘v₂’, respectively.

This situation is illustrated in the attached picture.

So, according to the law of conservation of momentum:

Total Momentum Before Collision = Total Momentum After Collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

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Sorry this is a year late, but here it is for those of you who are stuck on the same thing.

NOTE: Please Check and Confirm That You Are On The Same Assignment with The Same Questions and Number of Questions. Thank You and Good Luck!

1. Mass & velocity

2. The total momentum after the collision is the same as the total momentum before the collision.

3. 0.54 kg⋅m/s

4. The system has external forces, such as friction and air resistance, acting on it.

5. 3.0 m/s