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3 years ago

Calculate the coefficient of friction of a 12kg object that has a force of friction of 3N

Physics

1 answer:

postnew [5]3 years ago

3 0

Answer:

0.026

Explanation:

The force of friction acts in the direction perpendicular to the norm force of the surface on which the object rests, induced by gravity. The magnitude of the friction force is

(Friction) = (mass) x (gravitational acceleration g) x (coefficient of friction)

from which the coefficient of friction can be determined:

(coefficient of friction) = (Friction) / ((mass)x(g)) = 3 N / (12 kg * 9.8 m/s^2) = 0.026

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56 kg girl slides down the frictionless slip-n-slide at a constant speed of 35 m/s. What is the kinetic energy in Joules of the

melisa1 [442]

Answer:

93 joules

Explanation:

7 0

3 years ago

A diffraction pattern is formed on a screen 130 cm away from a 0.420-mm-wide slit. Monochromatic 546.1-nm light is used. Calcula

notka56 [123]

Answer:

The fractional Intensity $\frac{I}{I_{max}}$ = 0.0146

Given:

wavelength of the light, $\lambda = 546.1 nm = 546.1\times 10^{-9} m$

slit and screen separation difference, D = 130 cm = 1.3 m

distance of the point from the center of the principal maximum, y = 4.10 mm = 0.041 m

slit width, d = 0.420 mm = $0.420\times 10^{-3}$

Solution:

To calculate the fractional intensity, we use the given formula:

$\frac{I}{I_{max}} = \frac{sin^{2}\delta }{\frac({\delta}{2})^{2}}$ (1)

$\delta = \frac{\pi }{\lambda}dsin\theta$

For very small angle:

$\delta = \frac{\pi dy}{\lambda D}$ (2)

where

$\delta = total phase angle$

$\theta = angle of deviation$

Using eqn (2):

$\delta = \frac{\pi \times 0.42\times 10^{-3}\times 4.1\times 10^{-3} }{546.1\times 10^{-9}\times 1.3} = 7.6202 radians$

Now, using eqn (1):

$\frac{I}{I_{max}} = \frac{sin^{2}(7.6202) }{(\frac{7.6202}{2})^{2}} = 0.0146$

4 0

4 years ago

A car slows down at -5.00 m/s² until it comes to a stop after travelling 15.0 m. What was the initial speed of the car?

lapo4ka [179]

<h2>Answer: 12.24m/s</h2>

According to <u>kinematics</u> this situation is described as a uniformly accelerated rectilinear motion. This means the acceleration while the car is in motion is constant.

Now, among the equations related to this type of motion we have the following that relates the velocity with the acceleration and the distance traveled:

$V_{f}^{2}-V_{o}^{2}=2.a.d$ (1)

Where:

$V_{f}$ is the Final Velocity of the car. We are told "the car comes to a stop after travelling", this means it is 0.

$V_{o}$ is the Initial Velocity, the value we want to find

$a$ is the constant acceleration of the car (the negative sign means the car is decelerating)

$d$ is the distance traveled by the car

Now, let's substitute the known values in equation (1) and find $V_{o}$ :

$0-V_{o}^{2}=2(-5m/{s}^{2})(15m)$ (2)

$-V_{o}^{2}=-150{m}^{2}/{s}^{2}$ (3)

Multiplying by -1 on both sides of the equation:

$V_{o}^{2}=150{m}^{2}/{s}^{2}$ (4)

$V_{o}=\sqrt{150{m}^{2}/{s}^{2}}$ (5)

Finally:

$V_{o}=12.24m/s$ >>>This is the Initial velocity of the car

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3 years ago

The manipulated variable is plotted on the horizontal axis true or false

AlladinOne [14]

True, the manipulated variable is plotted on the horizontal axis

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How dose energy I've through an ecosystem

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3 years ago