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san4es73 [151]
2 years ago
12

What is the average acceleration of a car that goes from rest to 60 km/h in 8 seconds?  Be sure to show all work to support your

answer.
Physics
2 answers:
Ghella [55]2 years ago
7 0
60 divided by 8 = 7.5 mps
wel2 years ago
5 0
The average acceleration can be found by dividing the final speed by the time taken to reach said point so in this case you divide 60 by 8 resulting in 7.5 which will be your answer
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Streams pick up a greater load of sand and gravel as they leave a mountain range and flow across a lowland.
inessss [21]
B

because it will pick it up while coming down not just flowing on low land<span />
6 0
3 years ago
Chord progressions that move to resting points that release tension are called
Sav [38]

Cadences.

These cadences are the resulting tensions that chords release from their resting points. This movement is classified from a unstable chord progression to a stable one.  Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
6 0
2 years ago
Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi
Bad White [126]
A) Agreed. 
<span>b) Value agreed but units should be W (watts). </span>

<span>c) Here's one method... </span>

<span>15 miles = 24140 m </span>

<span>1 gallon of gasoline contains 1.4×10⁸ J. </span>

<span>So moving a distance of 24140m requires gasoline containing 1.4×10⁸ J </span>

<span>Therefore moving a distance of 1m requires gasoline containing 1.4×10⁸/24140 = 5800 J </span>

<span>Overcoming rolling resitance for 1m requires (useful) work = force x distance = 1000x1 = 1000J </span>

<span>So 5800J (in the gasoline) provides 1000J (overcoming rolling resistance) of useful work for each metre moved. </span>

<span>Efficiency = useful work/total energy supplied </span>
<span>= 1000/5800 </span>
<span>= 0.17 (=17%) </span>
8 0
3 years ago
Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the ce
jasenka [17]

Answer:

\lambda= 506.25 nm

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

y=\frac{m \lambda D}{a}

Where:

y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1

Solving for λ:

\lambda=\frac{y*a}{mD}

Replacing the data provided by the problem:

\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm

7 0
3 years ago
A jet airliner moving initially at 889 mph
Eduardwww [97]

Answer:

1500 mph

Explanation:

Take east to be +x and north to be +y.

The x component of the velocity is:

vₓ = 889 cos 0° + 830 cos 59°

vₓ = 1316.5 mph

The y component of the velocity is:

vᵧ = 889 sin 0° + 830 sin 59°

vᵧ = 711.4 mph

The speed is found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1316.5 mph)² + (711.4 mph)²

v = 1496 mph

Rounded to two significant figures, the jet's speed relative to the ground is 1500 mph.

8 0
3 years ago
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