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Sophie [7]
3 years ago
11

If you drive 30 miles in 40 minutes, and then 15 miles in 20 minutes, do you have a constant speed?

Physics
1 answer:
kompoz [17]3 years ago
7 0

Answer:

A) yes

Explanation:

First section of trip: 30 miles in 40 minutes

Second section of trip: 15 miles in 20 minutes

The formula for speed is distance over time k=\frac{d}{t}

Calculate the speeds for each section of the trip.

First:

k = d/t

k = 30miles/40minutes  <= reduce fraction by 10 (30÷10 and 40÷10)

k = 3 miles / 4 minutes

Second:

k = d/t

k = 15miles/20minutes  <= reduce fraction by 5 (15÷5 and 20÷5)

k = 3 miles / 4 minutes

Therefore there is a constant speed because both sections of the trip are driving at "3 miles / 4 minutes".

3 miles / 4 minutes can be also formatted as:

0.75 miles per minute.

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An Air Force plane lands with a velocity of 125 m/s and accelerates at a maximum rate of -6.5 m/s^2.
Alisiya [41]

Answer:

a) 19.2 s

b) No

Explanation:

Given:

v₀ = 125 m/s

a = -6.5 m/s²

v = 0 m/s

a) Find: t

v = at + v₀

(0 m/s) = (-6.5 m/s²) t + (125 m/s)

t ≈ 19.2 s

b) Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (125 m/s)² + 2 (-6.5 m/s²) Δx

Δx ≈ 1200 m

An aircraft carrier that's 850 meters long won't be long enough.

8 0
3 years ago
A car changes velocity at a constant acceleration of 2.5m/s to reach 43.7m/s in 2.7 s how fast was the car moving when it began
Montano1993 [528]

The formula we can use in this case is:

v = v0 + a t

where v is final velocity, v0 is initial velocity, a is acceleration and t is time

So finding for v0:

v0 = v – a t

v0 = 43.7 – (2.5) 2.7

v0 =  36.95 m/s

8 0
3 years ago
The radius of a sphere is increasing at a rate of 4 mm/s. how fast is the volume increasing when the diameter is 40 mm?
marin [14]

Using <span>r </span> to represent the radius and <span>t </span> for time, you can write the first rate as:

<span><span><span><span>dr</span><span>dt</span></span>=4<span>mms</span></span> </span>

or

<span><span>r=r<span>(t)</span>=4t</span> </span>

The formula for a solid sphere's volume is:

<span><span>V=V<span>(r)</span>=<span>43</span>π<span>r3</span></span> </span>

When you take the derivative of both sides with respect to time...

<span><span><span><span>dV</span><span>dt</span></span>=<span>43</span>π<span>(3<span>r2</span>)</span><span>(<span><span>dr</span><span>dt</span></span>)</span></span> </span>

...remember the Chain Rule for implicit differentiation. The general format for this is:

<span><span><span><span><span>dV<span>(r)</span></span><span>dt</span></span>=<span><span>dV<span>(r)</span></span><span>dr<span>(t)</span></span></span>⋅<span><span>dr<span>(t)</span></span><span>dt</span></span></span> </span>with <span><span>V=V<span>(r)</span></span> </span> and <span><span>r=r<span>(t)</span></span> </span>.</span>

So, when you take the derivative of the volume, it is with respect to its variable <span>r </span> <span><span>(<span><span>dV<span>(r)</span></span><span>dr<span>(t)</span></span></span>)</span> </span>, but we want to do it with respect to <span>t </span> <span><span>(<span><span>dV<span>(r)</span></span><span>dt</span></span>)</span> </span>. Since <span><span>r=r<span>(t)</span></span> </span> and <span><span>r<span>(t)</span></span> </span> is implicitly a function of <span>t </span>, to make the equality work, you have to multiply by the derivative of the function <span><span>r<span>(t)</span></span> </span> with respect to <span>t </span> <span><span>(<span><span>dr<span>(t)</span></span><span>dt</span></span>)</span> </span>as well. That way, you're taking a derivative along a chain of functions, so to speak (<span><span>V→r→t</span> </span>).

Now what you can do is simply plug in what <span>r </span> is (note you were given diameter) and what <span><span><span>dr</span><span>dt</span></span> </span> is, because <span><span><span>dV</span><span>dt</span></span> </span> describes the rate of change of the volume over time, of a sphere.

<span><span><span><span><span>dV</span><span>dt</span></span>=<span>43</span>π<span>(3<span><span>(20mm)</span>2</span>)</span><span>(4<span>mms</span>)</span></span> </span><span><span>=6400π<span><span>mm3</span>s</span></span> </span></span>

Since time just increases, and the radius increases as a function of time, and the volume increases as a function of a constant times the radius cubed, the volume increases faster than the radius increases, so we can't just say the two rates are the same.

7 0
2 years ago
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