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lidiya [134]
3 years ago
6

A person holding a 30 kg box 1 m above the floor puts it down. How much work was done ? Remember , the force of an objects weigh

t = mass times gravity
A. 0.0j

B. 30j

C. 147 j

D. 294 j
Physics
2 answers:
IrinaK [193]3 years ago
4 0
Hey friend! The answer is D. 294 j
frosja888 [35]3 years ago
3 0
As long as the box was 1m off the floor. it's potential energy was m•g•h = 30•9.8 = 294 joules. When the person stopped restraining the box and allowed it to move, gravity did 294 joules of work on the box, and would have done more if the floor hadn't been in the way. The person did no work on the box.
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Cuando una persona sube y baja una escalera, Cuanto vale su desplazamiento y cual es la medida de su trayectoria.
adoni [48]

Answer:

Primero, definimos el desplazamiento como la distancia entre la posición final y la posición inicial.

Así, si comenzamos abajo, luego subimos la escalera, y luego bajamos, la posición final y la posición inicial serán la misma

por lo que el desplazamiento es igual a cero.

La medida recorrida es el espacio total recorrido.

Es decir, si entre el principio y el final de la escalera hay una distancia D.

La persona que sube y baja, recorre esta distancia dos veces.

Entonces cuando una persona sube y baja la escalera, la medida de su trayectoria será 2*D.

8 0
3 years ago
At the second maxima on either side of the central bright spot in a double-slit experiment, light from
Alex

Answer:

Option 3 is the correct option to the following question.

Explanation:

Double slit experiment:

The double-slit experiment in modern physics reveals that light and matter can exhibit characteristics of both classically described waves and particles; it also shows the inherently probabilistic existence of quantum mechanical phenomena.  

When two wavelength meets ,if the resultant amplitude is maximum then this is known as constructive interference and the resultant amplitude is maximum then this is known as destructive interference.

Therefore the answer is "3".

5 0
4 years ago
Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 5
LekaFEV [45]

Answer:

  • |\vec{F}_3| = 102.92 \ N
  • \theta = 57 \° 24 ' 48''

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

\vec{a} = \vec{0}.

As the net force equals acceleration multiplied by mass , this must mean:

\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}.

So, the sum of the three forces must be zero:

\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0},

this implies:

\vec{F}_3  = - \vec{F}_1 - \vec{F}_2.

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector \hat{i} pointing east,  with the y axis unit vector \hat{j} pointing south, so the positive angle is south of east. For this, we got for the first force:

\vec{F}_1 = 83.7 \ N \ (-\hat{j}),

as is pointing north, and for the second force:

\vec{F}_2 = 59.9 \ N \ (-\hat{i}),

as is pointing west.

Now, our third force will be:

\vec{F}_3  = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})

\vec{F}_3  =  83.7 \ N \ \hat{j}  + 59.9 \ N \ \hat{i}

\vec{F}_3  =  (59.9 \ N , 83.7 \ N )

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

|\vec{R}| = \sqrt{R_x^2 + R_y^2}

|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}

|\vec{F}_3| = 102.92 \ N

this is the magnitude.

To find the direction, we can use:

\theta = arctan(\frac{F_{3_y}}{F_{3_x}})

\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })

\theta = 57 \° 24 ' 48''

and this is the angle south of east.

7 0
3 years ago
Can someone please Help me with this? It’s Due today
kirza4 [7]

helium group no val elect

mg is reactive when activated. when burned, very intense

pot\asssium 1 valence elect ... KCl eg

theone with H and sodium in it

http://perendis.webs .com

6 0
4 years ago
A fish in an aquarium with flat sides looks out at a hungry cat. To the fish, does the distance to the cat appear to be less tha
lakkis [162]

Answer:

p = -q  

he distance is equal to the current distance, so the distance does not change

Explanation:

For this exercise we can solve it using the equation of the constructor

            1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

For a flat surface the radius is at infinity, therefore 1 / f = 0, which implies

          1 / p = - 1 / q

           p = -q

Therefore the distance is equal to the current distance, so the distance does not change

7 0
3 years ago
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