All categories

3 years ago

A rock from the top of a hill is falling from rest.

Physics

1 answer:

dsp733 years ago

3 0

Object name:- Rock

Starting energy:- Kinetic energy

Conversion:- Potential to kinetic energy

Final energy form:- Potential energy

Non usable form if energy:- Nil [ As it has potential & kinetic both]

You might be interested in

What is the name of a very porous igneous rock that is so light that it floats?

Sholpan [36]

The answer is pumice rock

3 0

3 years ago

Cuales son los fallos en la teoría del big bang clasico?

IrinaK [193]

Answer:

english bro..

Explanation:

8 0

3 years ago

When an object is placed 110 cm from a diverging thin lens, its image is found to be 55 cm from the lens. The lens is removed, a

Kazeer [188]

Explanation:

Given that,

Object distance u= -110 cm

Image distance v= 55 cm

We need to calculate the focal length for diverging lens

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-f}=\dfrac{1}{55}-\dfrac{1}{-110}$

$\dfrac{1}{f}=-\dfrac{3}{110}$

$f=-36.6\ cm$

The focal length of the diverging lens is 36.6 cm.

Now given a thin lens with same magnitude of focal length 36.6 cm is replaced.

Here, The object distance is again the same.

We need to calculate the image distance for converging lens

Using formula of lens

$\dfrac{1}{36.6}=\dfrac{1}{v}-\dfrac{1}{-110}$

Here, focal length is positive for converging lens

$\dfrac{1}{v}=\dfrac{1}{36.6}-\dfrac{1}{110}$

$\dfrac{1}{v}=\dfrac{367}{20130}$

$v=54.85\ cm$

The distance of the image is 54.85 cm from converging lens.

Hence, This is the required solution.

5 0

3 years ago

A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?

Nostrana [21]

Answer:

The work done by friction was $-4.5\times10^{5}\ J$

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

$W=\Delta KE$

$W=K.E_{f}-K.E_{i}$

$W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2$

Put the value of m and v

$W=0-\dfrac{1}{2}\times1000\times(30)^2$

$W=-450000
\ J$

$W=-4.5\times10^{5}\ J$

Hence, The work done by friction was $-4.5\times10^{5}\ J$

3 0

3 years ago

Considerando que você comece a caminhar em velocidade constante, inicialmente a 350 m de um ponto referencial escolhido. Você ca

ExtremeBDS [4]

Answer:

a. S(t)=350−1t

Explanation:

To determine the equation of motion you take into account the general form of motion with constant velocity:

$S(t)=S_o+vt$ ( 1 )

So is the initial position from a specific reference frame. In this case is 350 m.

v is the speed of the motion, in this case is 1m/s. However, the motion is forward the zero point of the reference frame, hence, the speed is - 1m/s.

You replace the values of So and v in the equation ( 1 ) and you obtain:

$S(t)=350-(1m/s)t$

Hence, the answer is:

a. S(t)=350−1t

- - - - - - - - - - - - - - - - - - - - -

Para determinar a equação do movimento, você leva em consideração a forma geral do movimento com velocidade constante:

(1)

Assim é a posição inicial de um quadro de referência específico. Neste caso, é de 350 m.

v é a velocidade do movimento, neste caso é de 1m / s. No entanto, o movimento é avançar o ponto zero do quadro de referência, portanto, a velocidade é de - 1m / s.

Você substitui os valores de So ev na equação (1) e obtém:

Portanto, a resposta é:

uma. S (t) = 350-1t, movimento retrógrado

4 0

3 years ago