All categories

2 years ago

How does kinetic energy affect the stopping distance of a small vehicle compared to a large vehicle?

1 answer:

8 0

<span>The kinetic energy is the work done by the object due to its motion. It is represented by the formula of the half the velocity squared multiply by the mass of the object. In this problem, you have two vehicles, the other one is large and the other one is small. Let us assume that they travel with the same velocity. Note that the kinetic energy is proportional to the mass of the object. So when you increase the mass of the other, it also increases the kinetic energy of that object. The same holds true for the two vehicles. The larger the vehicle, its kinetic energy is also large and therefore its stopping distance will be longer than that of the smaller vehicle.</span>

You might be interested in

Calculate the amount of work done if you use a 100N force to push a 50kg box 5m across the kitchen floor.

irinina [24]

So if the formula for work is force times displacement times cosine(theta), you'd plug in the numbers

100x5 (since there's no angle in the problem, cosine(theta) isn't used

100x5 = 500

So the answer would be B.

Hope that helps!

6 0

2 years ago

Read 2 more answers

After all that you have learned in this unit, construct a pamphlet (brochure) in Microsoft Publisher helping new freshman to enc

aleksandr82 [10.1K]

You're not going to like this answer, but it's the only one possible:. It wasn't I who learned anything in this unit. If it was either of us, it was YOU. I can't even tell from reading the question what the topic of the unit was. Was it pamphlets ? Microsoft Publisher ? Freshmen ? Getting Through High School ? This is a lot like asking me to write something "in your own words".

5 0

2 years ago

A 5 meter long ladder leans against a wall. The bottom of the ladder slides away from the wall at the constant rate of 1 3 m/s.

Oksana_A [137]

Answer:9.75 m/s

Explanation:

Given

Length of ladder $(L)=5 m$

Foot the ladder is moving away with speed of $\frac{\mathrm{d} x}{\mathrm{d} t}=13 m/s$

From diagram

$x^2+y^2=L^2$ ------1

at $x=3$

$y^2=25-9=16$

$y=4 m$

Now differentiating equation 1 w.r.t time

$2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0$

$x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}$

$3\times 13=-4\times \frac{\mathrm{d} y}{\mathrm{d} t}$

$\frac{\mathrm{d} y}{\mathrm{d} t}=-\frac{3\times 13}{4}=-9.75 m/s$

negative indicates distance is decreasing with time

5 0

2 years ago

If you ran 15 km/h for 20 min, how much distance would you cover?

nikitadnepr [17]

Since we have 15 kilometers per hour, and we're looking for 20 minutes, let's set up proportions.
20/60 minutes = x/15
20/60 = 1/3, so let's leave that simplified.
1/3 = x/15
Look at the denominators, 3 to 15 is a factor of 5, so multiply the numerator by 5.
1 • 5 = 5, so you will cover 5 kilometers in 20 minutes.

I hope this helps!

4 0

2 years ago

A box sits on a table. A short arrow labeled F subscript N points up. A short arrow labeled F subscript g points down. A long ar

Aloiza [94]

Answer:

unbalanced and right

Explanation:

7 0

2 years ago