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2 years ago

What is the objects acceleration

Physics

2 answers:

alekssr [168]2 years ago

5 0

Answer:

Acceleration is a vector quantity which is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.

Morgarella [4.7K]2 years ago

4 0

It’s the movement’s of the object
Hehe

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A spring has a constant k= 10.32 N/m and is hung vertically. A 0.400kg mass is suspended from the spring. What is the displaceme

alukav5142 [94]

Answer:

displacement (x) = 0.003798 meters

Explanation:

from the fact that the string is hung vertically we can deduce that:

Total force acting on the mass = Fs (by spring) + Fg (by gravity)

where

Fs = k*x , x is the displacement..

Fg = m*g

then:

Ftot = m*a, but a = 0 m/(s^2) because the mass becames stationary.

Ftot = 0

Fs + Fg = 0

by direction, take down as negative.

Fs - Fg = 0

k*x = m*g

x = m*g/k = [(0.400)(9.8)]/(10.32)

= 0.3798 meters

4 0

3 years ago

A car travelled a distance of 200 m with initial velocity of 216 km/hr. Calculate the acceleration

pogonyaev

Answer:

$a = 16\ m/s^2$

Explanation:

When the velocity changes uniformly, the object has a constant acceleration. The acceleration, the velocities, and the distance are related by the equation:

$v_f^2=v_o^2+2ax$

Where:

vf = final velocity

vo = initial velocity

a = acceleration

x = distance

Solving for a:

$\displaystyle a=\frac{v_f^2-v_o^2}{2x}$

The car travels a distance of x=200 m and the velocities are:

vo = 216 Km/h

vf = 360 Km/h

Both velocities must be converted to meters by seconds.

vo = 216 Km/h *1000/3600 = 60 m/s

vf = 360 Km/h *1000/3600 = 100 m/s

The acceleration is:

$\displaystyle a=\frac{100^2-60^2}{2*200}$

$\displaystyle a=\frac{10000-3600}{400}$

$\displaystyle a=\frac{6400}{400}$

$\mathbf{a = 16\ m/s^2}$

7 0

3 years ago

Using the result of the preceding problem, (a) calculate the distance between fringes for 633-nm light falling on double slits s

vekshin1

To solve this problem it is necessary to apply the concepts related to the concept of superposition and the fringe separation for double slit experiment.

The equation can be written as

$\Delta y = \frac{x\lambda}{d}$

Where

$\Delta y$ = Distance between fringes

x = distance between slits and screen

d = Distance between slits

$\lambda$ = Wavelength

Our values are given as

d= 0.08mm

x =3m

$\lambda = 633nm$

In this way replacing in the equation,

$\Delta y = \frac{x\lambda}{d}$

$\Delta y = \frac{3m(633nm*(\frac{1*10^{-9}}{1nm}))}{0.08*(\frac{1*10^{-3}m}{1mm})}$

$\Delta y = 2.37*10^{-2}m$

$\Delta y = 2.37cm$

Therefore the distance between the fringes is 2.37cm

PART B) For the case in which it is submerged in water it is necessary to apply the relationship of the fringes with the index of refraction therefore

$\Delta Y_2 = \frac{\Delta Y_1}{n}$