Which force is a contact force

What is the net force on an object<br> at rest?

Natasha2012 [34]

Answer:

Net force equals 0

Explanation:

Newton's first law states that an object at rest tends to remain at rest, and an object in motion tends to remain in motion with a constant velocity (constant speed and direction of motion), unless it is acted on by a nonzero net force. Note that the net force is the sum of all the forces acting on an object.

4 0

4 years ago

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A beam of yellow light is made to pass through two slits that are 3.0 x 10−3 meters apart. On a screen 2.0 meters away from the

lakkis [162]

Answer:

585 nm

Explanation:

The formula that gives the position of the m-th maximum (bright fringe) relative to the central maximum in the interference pattern produced by diffraction from double slit is:

$y=\frac{m\lambda D}{d}$ $\Delta y =\frac{m\lambda D}{d}$

where

m is the order of the maximum

$\lambda$ is the wavelength

D is the distance of the screen from the slits

d is the separation between the slits

The distance between two consecutive bright fringes therefore is given by:

$\Delta y = \frac{(m+1)\lambda D}{d}-\frac{m\lambda D}{d}=\frac{\lambda D}{d}$

In this problem we have:

$\Delta y = 3.9\cdot 10^{-4} m$ (distance between two bright fringes)

D = 2.0 m (distance of the screen)

d = 3.0 x 10−3 m (separation between the slits)

Solving for $\lambda$ , we find the wavelength:

$\lambda=\frac{\Delta y d}{D}=\frac{(3.9\cdot 10^{-4})(3.0\cdot 10^{-3})}{2.0}=5.85\cdot 10^{-7} m = 585 nm$

4 0

4 years ago

On a straight road (taken to be in the x direction) you drive for an hour at 60 km per hour, then quickly speed up to 120 km per

Luda [366]

Answer:

The average velocity is 180 km/hr

Explanation:

Given;

initial velocity, u = 60 km per hour

final velocity, v = 120 km per hour

initial time = 1 hour

final time = 2 hour

Initial position = 60 km/h x 1 hour = 60 km

final position = 120 km/h x 2 hour = 240 km

The average velocity is given by;

$V_{avg} = \frac{Final \ position\ - \ Initial \ position}{final \ time\ - \ initial \ time}\\\\V_{avg} = \frac{240km \ - \ 60km}{2hr\ - \ 1hr} \\\\V_{avg} = \frac{180 \ km}{1hr} \\\\V_{avg}= 180 \ km/hr$