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4 years ago

7

A puppy weighing 3 kilograms races through the dog park. If she slows from a speed of 2 meters/second to 1 meter/second, what wi

ll happen to her kinetic energy?

2 answers:

Tomtit [17]4 years ago

8 0

Answer: Kinetic energy will be decreases by 4.5 Joules.

Explanation:

Kinetic energy is the energy which is possessed by the virtue of objects motion. It is defined as the product of object's mass and square of the objects's velocity.

Mathematically,

$K=\frac{1}{2}mv^2$

Change in kinetic energy will be: $K=\frac{1}[2}m\Delta v^2$

where,

m = mass of the object = 3 kg

$\Delta v=v_2-v_1$ = Change in velocity

$v_1=2m/s\\v=1m/s$

Now, calculating the change in kinetic energy:

$K.E.=\frac{1}{2}\times 3kg\times (1^2-2^2)m^2/s^2=-4.5J$

Negative sign indicates that the kinetic energy is getting reduced.

Hence, kinetic energy will be decreases by 4.5 Joules.

aliya0001 [1]4 years ago

5 0

Ans: Thus, the kinetic energy change = 4.5 J

The mass of puppy = 3kg

Kinetic energy initial = $\frac{1}{2} mv^{2}$
= $\frac{1}{2}*3*4 = 6 kg-m/s2$

Kinetic energy final = $\frac{1}{2} mv^{2} = \frac{1}{2}*3*1 = 1.5 kg-m/s2$

Thus, the kinetic energy change = 4.5 J

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A 1,100 kg car is traveling initially 20 m/s when the brakes are applied. The brakes apply a constant force while bringing the c

Natalka [10]

Answer:

Work done = -220,000 Joules.

Explanation:

<u>Given the following data;</u>

Mass = 1100kg

Initial velocity = 20m/s

To find workdone, we would calculate the kinetic energy possessed by the car.

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

$K.E = \frac{1}{2}MV^{2}$

Where,

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

Substituting into the equation, we have;

$K.E = \frac{1}{2}*1100*20^{2}$

$K.E = 550*400$

K.E = 220,000J

Therefore, the workdone to bring the car to rest would be -220,000 Joules because the braking force is working to oppose the motion of the car.

4 0

3 years ago

What is the role of the nervous system?

zimovet [89]

The nervous system is responsible for sending, receiving, and interpreting information from all parts of the body. The nervous system monitors and coordinates internal organ function and responds to changes in the external environment. (The role) The central nervous system consists of the brain and the spinal cord. It is part of the overall nervous system that also includes a complex network of neurons, known as the peripheral nervous system. (Central nervous system)

4 0

3 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

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3 years ago

A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an ang

adell [148]

Answer:

A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.

how much work is done on the monitor by (a) friction, (b) gravity

work(friction) = 453.5J

work(gravity) = -453.5J

Explanation:

Given that,

mass = 14kg

displacement length = 5.50m

displacement angle = 36.9°

velocity = 2.30cm/s

F = ma

work(friction) = mgsinθ .displacement

= (14) (9.81) (5.5sin36.9°)

= 453.5J

work(gravity)

= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)

= 126.9°

work(gravity) = (14) (9.81) (5.5cos126.9°)

= -453.5J

8 0

3 years ago

A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

yaroslaw [1]

The electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.

To determine σ:

σ = Q/A

Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:

σ = Q/d²

Make this substitution in the equation for E:

E = Q/(2ε₀d²)

We see that E is inversely proportional to the square of d:

E ∝ 1/d²

The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:

$E_{new} = E/4$

4 0

3 years ago