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marshall27 [118]
3 years ago
11

A boulder of mass 4 kg rolls over a cliff and reaches the beach below with a velocity of 20 metres per second minus 1.

Physics
1 answer:
Ksenya-84 [330]3 years ago
5 0

Given:-

Mass of boulder = 10kg

Final Velocity = 15m/s

Acceleration due to gravity = +10m/s

To Find:-

The kinetic energy of the boulder as it lands

The potential energy of the boulder when it was at the top of the cliff

The height of the cliff

Formulae used:-

Kinetic energy = mv²/2

Potential energy = mgh

Now,

We will first find the Kinetic enrgy,

→ K. E = ½mv²

→ K. E = ½ × 10 × (15)²

→ K. E = 5 × 225

→ K. E =1125J

But we know that when the body starts moving the Potential enrgy of the body starts Converting into Kinetic enrgy. Before reaching to the last spot it has equal kinetic energy and Potential energy.

Therefore,

→ mgh = 1125J

The Potential enrgy of the boulder when it was on top is 1125J

Now,

→ mgh = 1125J

→ 10 × 10 × h = 1125J

→ 100h = 1125J

→ h = 1125/100

→ h = 11.25m

Hence, The Height of the Cliff is 11.25m

Thanks!

Mark me brainliest!

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a 4lb ball is travelling in a circle radius 3ft speed 6ft/s. the attached cord pulled through the center has constant speed 2ft/
lora16 [44]

Answer:

Explanation:

Circle motion

Weight=4lb

Radius=3ft

Velocity Vb= 6ft/s

Velocity Vr=2ft/s

Velocity Vo=12

V2=√Vo²+Vr²

12=√Vo²+2²

Square both side

144=Vo²+4

Vo²=140

Vo=11.83ft/s

Applying conservation of angular momentum

Ha1=Hb2

MbVbr1=MbVor2

r2=Vbr1/Vo

r2=6×3/11.83

r2=1.52ft

The require time is written as.

∆r=Vrt

t=∆r/Vr

t=r1-r2/Vr

t=3-1.52/2

t=0.74sec

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4 years ago
40 POINTS <br> Can someone please answer me questions 16 and 17
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3 years ago
Which sentence correctly describes a friction force? A. It acts in the same direction as the motion of an object. B. It acts in
KatRina [158]

Answer:

B

Explanation:

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5 0
4 years ago
Describe the mechanical energy of a roller coaster car immediately before it begins traveling down a long track
GrogVix [38]
At the top of the hill, the cars possess a large quantity of potential energy. Potential energy - the energy of vertical position - is dependent upon the mass of the object and the height of the object. The car's large quantity of potential energy is due to the fact that they are elevated to a large height above the ground. As the cars descend the first drop they lose much of this potential energy in accord with their loss of height. The cars subsequently gain kinetic energy. Kinetic energy - the energy of motion - is dependent upon the mass of the object and the speed of the object. The train of coaster cars speeds up as they lose height. Thus, their original potential energy (due to their large height) is transformed into kinetic energy (revealed by their high speeds). As the ride continues, the train of cars are continuously losing and gaining height. Each gain in height corresponds to the loss of speed as kinetic energy (due to speed) is transformed into potential energy (due to height). Each loss in height corresponds to a gain of speed as potential energy (due to height) is transformed into kinetic energy (due to speed). A roller coaster ride also illustrates the work and energy relationship. The work done by external forces is capable of changing the total amount of mechanical energy from an initial value to some final value. The amount of work done by the external forces upon the object is equal to the amount of change in the total mechanical energy of the object. The relationship is often stated in the form of the following mathematical equation.

KEinitial + PEinitial + Wexternal = KEfinal + PEfinal

The left side of the equation includes the total mechanical energy (KEinitial + PEinitial) for the initial state of the object plus the work done on the object by external forces (Wexternal) while the right side of the equation includes the total mechanical energy (KEfinal + PEfinal) for the final state of the object.

Once a roller coaster has reached its initial summit and begins its descent through loops, turns and smaller hills, the only forces acting upon the coaster cars are the force of gravity, the normal force and dissipative forces such as air resistance. The force of gravity is an internal force and thus any work done by it does not change the total mechanical energy of the train of cars. The normal force of the track pushing up on the cars is an external force. However, it is at all times directed perpendicular to the motion of the cars and thus is incapable of doing any work upon the train of cars. Finally, the air resistance force is capable of doing work upon the cars and thus draining a small amount of energy from the total mechanical energy which the cars possess. However, due to the complexity of this force and its small contribution to the large quantity of energy possessed by the cars, it is often neglected. By neglecting the influence of air resistance, it can be said that the total mechanical energy of the train of cars is conserved during the ride. That is to say, the total amount of mechanical energy (kinetic plus potential) possessed by the cars is the same throughout the ride. Energy is neither gained nor lost, only transformed from kinetic energy to potential energy and vice versa.

The conservation of mechanical energy by the coaster car in the above animation can be studied using a calculator. At each point in the ride, the kinetic and potential energies can be calculated using the following equations.

<span> KE = 0.5 * mass * (speed)^2 PE = mass * g * height</span>

If the acceleration of gravity value of 9.8 m/s/s is used along with an estimated mass of the coaster car (say 500 kg), the kinetic energy and potential energy and total mechanical energy can be determined

5 0
3 years ago
While chatting with a friend you place your book bag on a nearby slide in the playground at school. The bag remains stationary.
alukav5142 [94]

Answer:

3. fs < μmg

4. fs = mg sinθ

Explanation:

For any object placed on a slide, there are 3 external forces acting on it:

  • Fg = m*g (always downward)
  • N (normal force, always perpendicular to the surface of the slide. going upward)
  • Fs (Friction Force, always opposite to the movement of the object, parallel to the slide)

As we have only one force with components along the normal and parallel to the slide directions (gravity force), it is advisable to find the components of  this force, along these directions.

If θ is the angle of the slide above the horizontal, we have the following components of Fg:

Fgn = m*g*cosθ

Fgp = m*g*sin θ

We can apply Newton's 2nd Law to these perpendicular directions:

Fp = m*g*sin θ - Fs

Fn = N -m*g*cosθ = 0 (as the object has no movement in the direction perpendicular to the slide) (1)

Looking at the equation for the parallel direction, we have two forces, the component of Fg along the slide (which tries to accelerate the object towards the bottom of the slide), and the friction force.

While the object remains stationary, the equation for Newton's 2nd Law along this direction is as follows:

m*g*sin θ - fs =0 ⇒ fs = m*g*sinθ  (4.)

This force can take any value (depending on the angle θ) to equilibrate the component of Fg along the slide, up to a limit value, which  is given by the following expression:

fsmax = μN (2)

From (1), N= m*g*cos θ

Replacing in (2):

fsmax = μ*m*g*cos θ

While the bag remains at rest, we can say:

fs < μ*m*g*cosθ < μ*m*g (as in the limit cosθ =1)

So, the following is always true:

fs < μmg (3.)

6 0
3 years ago
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