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Serggg [28]
2 years ago
11

A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1

869 N to the right. What is the acceleration of the crate?
a=(6.0 or 6.00 or 8.10 or 8.1)=m/s^2
Physics
1 answer:
Daniel [21]2 years ago
6 0

The acceleration of the steel crate, given the data from the question is 6.0 m/s²

<h3>How to determine the net force</h3>
  • Force to the left (Fբ) = 480 N
  • Force to the right (Fᵣ) = 1869 N
  • Net force (Fₙ) = ?

Fₙ = Fᵣ - Fբ

Fₙ = 1869 - 480

Fₙ = 1389 N

<h3>How to determine the acceleration</h3>
  • Mass (m) = 230 Kg
  • Net force (Fₙ) = 1389 N
  • Acceleration (a) = ?

Fₙ = ma

Divide both sides by m

a = Fₙ / m

a = 1389 / 230

a = 6.0 m/s²

Learn more about acceleration:

brainly.com/question/491732

#SPJ1

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http://www.rmiia.org/auto/teens/Teen_Driving_Statistics.asp

(I just corrected the question. Sorry if it is still incorrect.)
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Period of brightness variation and luminosity.

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Which of the following is an example of how humans can increase biodiversity?
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The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr
Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by  the following equation:

dsin\theta_{n}=n\lambda   (1)

Where:

d is the width of the slit

\lambda  is the wavelength of the light

n is an integer different from zero.

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda   (2)

Now we have to find the value of \theta_{1}:

sin\theta_{1}=\frac{\lambda}{d}  

\theta_{1}=arcsin(\frac{\lambda}{d})   (3)

We know:

\lambda=104nm=104(10)^{-9}m

In addition we are told the diffraction grating has 5000 slits per mm, this means:

d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}

Substituting the known values in (3):

\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})

\theta_{1}=arcsin(0.52)

<u>Finally:</u>

\theta_{1}=31.33\º >>>This is the first-order diffraction angle

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