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Triss [41]
3 years ago
7

It is easier to climb up a slanted slope than a vertical slope

Physics
2 answers:
deff fn [24]3 years ago
4 0

The correct answer is it is easier to climb a slanted slope rather then a vertical slope. Vertical is straight up versus a nice slant.

V125BC [204]3 years ago
3 0

IT IS EASIER TO CLIMB A SLANTED SLOPE

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Explain why a moving object cannot come to a stop instantaneously (in zero seconds). Hint: Think about the acceleration that wou
gizmo_the_mogwai [7]
To stop instantly, you would need infinite deceleration. This in turn, requires infinite force, as demonstrable with this equation:F=ma<span>So when you hit a wall, you do not instantly stop (e.g. the trunk of the car will still move because the car is getting crushed). In a case of a change in momentum, </span><span><span>m<span>v⃗ </span></span><span>m<span>v→</span></span></span>, we can use the following equation to calculate force:F=p/h<span>However, because the force is nowhere close to infinity, time will never tend to zero either, which means that you cannot come to an instantaneous stop.</span>
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You are driving at 40 miles per hour and you speed up to 60 miles per hour over the course of 20 miles. How long were
notka56 [123]

Answer:

120 hours

Explanation:

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3 years ago
What is voltage, and what is its relationship to amperage and power?
ziro4ka [17]

Explanation:

Amperage is the unit of electric current. It describes the strength of the electric current in a circuit.

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    Voltage as V

    Power as P

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learn more:

Voltage brainly.com/question/6949231

#learnwithBrainly

 

8 0
3 years ago
A Mercedes-Benz 300SL (m = 1700 kg) is parked on a road that rises 15 degrees above the horizontal. What are the magnitudes of (
kogti [31]

Answer: See below

Explanation:

<u>Given:</u>

Mass of the Mercedes-Benz (m) = 1700 kg

Inclination of the road (θ) = 15.0

<em>The free body diagram is shown in figure attached below</em>

<em />

a) The normal force is equal to the cos component of the weight of the car.

\begin{aligned}&f=m g \cos \theta \\&f=1700 \times 9.81 \times \cos 15 \\&f=16108.74 \mathrm{~N}\end{aligned}

b) The static force will be equal to the weight's sin component.

\begin{aligned}&f=m g \sin \theta \\&f=1700 \times 9.81 \times \sin 15 \\&f=4316.32 \mathrm{~N}\end{aligned}

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