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Hunter-Best [27]
1 year ago
6

the initial kinetic energy of an object moving on a horizontal surface is K. Friction between the object and the surface causes

the velocity of the object to decrease uniformly to zero in time t. What is the kinetic energy of the object at time = t/2?​
Physics
2 answers:
Ugo [173]1 year ago
8 0

Answer:

........................

baherus [9]1 year ago
4 0

Answer:

Kinetic energy is the energy an object has because of its motion.

If we want to accelerate an object, then we must apply a force. Applying a force requires us to do work. After work has been done, energy has been transferred to the object, and the object will be moving with a new constant speed. The energy transferred is known as kinetic energy, and it depends on the mass and speed achieved.

Kinetic energy can be transferred between objects and transformed into other kinds of energy. For example, a flying squirrel might collide with a stationary chipmunk. Following the collision, some of the initial kinetic energy of the squirrel might have been transferred into the chipmunk or transformed to some other form of energy.

How can we calculate kinetic energy?

To calculate kinetic energy, we follow the reasoning outlined above and begin by finding the work done, WWW, by a force, FFF, in a simple example. Consider a box of mass mmm being pushed through a distance ddd along a surface by a force parallel to that surface. As we learned earlier

\begin{aligned} W &= F \cdot d \\ &= m · a · d\end{aligned}

W

=F⋅d

=m⋅a⋅d

Huh? I'm lost already.

If we recall our kinematic equations of motion, we know that we can substitute the acceleration if we know the initial and final velocity—v_\mathrm{i}v

i

v, start subscript, i, end subscript and v_\mathrm{f}v

f

v, start subscript, f, end subscript—as well as the distance. What kinematic formula is this?

\begin{aligned} W &= m\cdot d\cdot \frac{v_\mathrm{f}^2-v_\mathrm{i}^2}{2d} \\ &= m\cdot \frac{v_\mathrm{f}^2-v_\mathrm{i}^2}{2} \\ &= \frac{1}{2}\cdot m \cdot v_\mathrm{f}^2 - \frac{1}{2}\cdot m \cdot v_\mathrm{i}^2 \end{aligned}

W

=m⋅d⋅

2d

v

f

2

−v

i

2

=m⋅

2

v

f

2

−v

i

2

=

2

1

⋅m⋅v

f

2

−

2

1

⋅m⋅v

i

2

So, when a net amount of work is done on an object, the quantity \dfrac{1}{2}mv^2

2

1

mv

2

start fraction, 1, divided by, 2, end fraction, m, v, squared—which we call kinetic energy KKK—changes.

\text{Kinetic Energy: } K=\frac{1}{2}\cdot m\cdot v^2Kinetic Energy: K=

2

1

⋅m⋅v

2

start text, K, i, n, e, t, i, c, space, E, n, e, r, g, y, colon, space, end text, K, equals, start fraction, 1, divided by, 2, end fraction, dot, m, dot, v, squared

Alternatively, one can say that the change in kinetic energy is equal to the net work done on an object or system.

W_{net}=\Delta KW

net

=ΔKW, start subscript, n, e, t, end subscript, equals, delta, K

This result is known as the work-energy theorem and applies quite generally, even with forces that vary in direction and magnitude. It is important in the study of conservation of energy and conservative forces.

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A closed and elevated vertical cylindrical tank with diameter 2.00 m contains water to a depth of 0.800 m. A worker accidentally
vazorg [7]

Answer:

a. v1 = 5.06 m/s,  v2 = 3.96 m/s ,  R = 1.27

b. t = 1 hr, 11 min, 26 sec  

Explanation:

Using the Bernoulli's laws to use the conserved energy

a. Solve the speed and the radio of this speed of the tank is open to the air

p₀ + ρgh₀ + ½ρv₀² = p₁ + ρgh₁ + ½ρv₁²

5000Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²

v² = 25.68  m²/s²

v1 = 5.06 m/s

Because it is open the tank so P=0 pa so:

0 Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²

v² = 15.68  m²/s²

v2 = 3.96 m/s

The ratio on the air is solve using both velocities so:

R = v1/v2 = 5.06 m/s / 3.96 m/s

R = 1.27

b. Now to find the time it takes for the tank to drain if the tank is open to the air

dh/dt = -u

dh/dt = -v * A/A'

dh/dt = v*(.02m)²/(2.0m)² = -v / 10000

and we can further substitute for v:

dh/dt = -(1/1e4)*√[(p+9800h)/500]

Solve replacing

-(1000/49)*√(49000h) = t + C

-(1000/49)*√(49000*0.8) = 0 + C

C = - 4040.6

Then when h = 0,

t = 4286 s

t = 1 hr, 11 min, 26 sec  

6 0
2 years ago
A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8  + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

    T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

 T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0
2 years ago
The human ear canal is, on average, 2.5cm long and aids in hearing by acting like a resonant cavity that is closed on one end an
Troyanec [42]

Answer:

3400 Hz

Explanation:

We know that

1 cm = 0.01 m

L = Length of the human ear canal = 2.5 cm = 0.025 m

V = Speed of sound = 340 ms⁻¹

f = First resonant frequency

The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as

f = \frac{(2n - 1)V}{4L}

for first resonant frequency, we have n = 1

Inserting the values

f = \frac{(2(1) - 1) 340}{4(0.025)}

f = \frac{340}{4(0.025)}

f = 3400 Hz

4 0
2 years ago
Calculate the amount of work done to draw a current of 8A from a point at 100V to a point at 120V in 2 seconds?
Morgarella [4.7K]
Given:
I=8A
t=2second
Potential difference,V=120-100=20volt
Workdone=V×i×t
=20×8×2
=320 joule.
3 0
2 years ago
A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 30. wha
Alecsey [184]

Answer:

<h2>9.375Nm</h2>

Explanation:

The formula for calculating torque τ = Frsin∅ where;

F = applied force (in newton)

r = radius (in metres)

∅ = angle that the force made with the bar.

Given  F= 25N, r = 0.75m and ∅ = 30°

torque on the bar τ  = 25*0.75*sin30°

τ = 25*0.75*0.5

τ = 9.375Nm

The torque on the bar is 9.375Nm

6 0
2 years ago
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