All categories

3 years ago

Why is it difficult to use the law of conservation of energy to calculate the

Physics

2 answers:

Liula [17]3 years ago

7 0

Answer:

Inelastic collisions actually do conserve energy, but the loss of energy to heat and mechanical vibration is hard to calculate so the math equating energy before and energy after is hard to balance.

Dmitriy789 [7]3 years ago

4 0

<u>Answer:</u>

The law of conservation of energy is difficult to use for calculate the

effects of a collision because of energy loss

<u>Explanation:</u>

We know that Collisions are of two types which is elastic and inelastic. The Law of Conservation of energy can be defined as in isolated system the total energy is said to remain constant and stays conserved over time can be applied on elastic collisions only as they don't lose energy and can not be applied on inelastic collisions due to energy loss in them due to heat and mechanical vibration. Hence, law of conservation of energy can not be used to calculate effects of collision

You might be interested in

A projectile is launched from ground level at angle u and speed v0 into a headwind that causes a constant horizontal acceleratio

ale4655 [162]

Answer:

Explanation:

Given

Launch angle =u

Initial Speed is $v_0$

Horizontal acceleration is $a_x=a$

At maximum height velocity is zero therefore

$v_f=v_i-gt$

$0=v_0\sin u-gt$

$t=\frac{v_0\sin u}{g}$

Total time of flight $T=2t=\frac{2v_0\sin u}{g}$

During this time horizontal range is

$R=v_o\cos u\cdot 2t-\frac{a(2t)^2}{2}$

$R=\frac{2v_0^2\sin u\cos u}{g}-\frac{2av_0^2\sin ^u}{g^2}$

For maximum range $\frac{\mathrm{d} R}{\mathrm{d} u}=0$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2\cos 2u}{g}-\frac{4av_0^2\sin u\cos u}{g^2}$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0$

$\tan 2u=\frac{g}{a}$

$u=\frac{1}{2}tan ^{-1}\frac{g}{a}$

(b)If a =10% g

$a=0.1g$

thus $u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}$

$u=42.14^{\circ}$

7 0

3 years ago

What has to happen for a force to do work on an object?

dexar [7]

It has to move through a distance over time.

4 0

3 years ago

What color will a yellow banana appear when illuminated by yellow light

Phantasy [73]

Answer:

A yellow banana reflects only yellow light,

Explanation:

3 0

3 years ago

A time-dependent but otherwise uniform magnetic field of magnitude B0(t) is confined in a cylindrical region of radius 6.5 cm. I

vodka [1.7K]

Answer:

a = 603.59 m/s^2

Explanation:

from the data given . the rate of change in magnetic field is as follow

$\frac{dB}{dt} = 280 G/s = 280 \times 10^{-4} T/s$

from the faraday's law of induction , the expression for the induced emf in region of radius r as follow

$\epsilon = \frac{d \phi}{dt}$

$\int E.dl = \frac{d(BA)}{dt}$

$E(2\pi r)= \pi r^2 \frac{dB}{dt}$

$E = \frac{r}{2} \frac{dB}{dt}$

electric field at point P_1 as follow

$E = \frac{r}{2} \frac{dB}{dt}$

$E = \frac{1.5\times 10^{-2}}{2} 280 \times 10^{-4}$

$E = 6.3\times 10^{-6} V/m$

from newton 2nd law of motion, the acceleration of proton is

F = ma

qE = ma

$a = \frac{qE}{m}$

$a = \frac{1.6 \times 10^{-19} (6.3\times 10^{-6})}{1.67\times 10^{-27}}$

a = 603.59 m/s^2

5 0

3 years ago

It took 1700 J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.

Alenkasestr [34]

The work to stretch a spring from its rest position is

   (1/2) (spring constant) (distance of the stretch)²

   E = 1/2 k x² .

You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write

   1700 joules = 1/2 k (3m)²

1 joule = 1 newton-meter

   1700 N-m = 1/2 k (3m)²

Multiply each side by 2: 3400 N-m = k · 9m²

Divide each side by 9m²    k = 3400 N-m / 9m²

   = (377 and 7/9)   newton per meter

7 0

3 years ago

Read 2 more answers