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Flauer [41]
2 years ago
14

A hiker travels 5.0 km north and then 2.5 km west What is the resultant displacement vector?​

Physics
1 answer:
tigry1 [53]2 years ago
5 0

Answer: The resultant displacement vector is 5.6km

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ANSWER QUICK 30 POINTS
Lelu [443]
What do you need help on
4 0
3 years ago
what is the magnitude of g at a height above earth surface where free fall acceleration equal 6.5m/s2
ZanzabumX [31]
The magnitude of gravity is expressed in terms of its acceleration. So the magnitude of ' g ' at that altitude is exactly 6.5 m/s^2.
8 0
3 years ago
A father is trying to teach his child to ice skate. As the child stands still, the father pushes him forward with an acceleratio
rjkz [21]

Hi there!

We can use Newton's Second Law:

\Sigma F = ma

∑F = net force (N)

m = mass (kg)

a = acceleration (m/s²)

We are given the mass and acceleration, so:

∑F = 20 · 2 = <u>40 N</u>

4 0
2 years ago
HURRY Which change is an example of transforming potential energy to kinetic energy
Jet001 [13]

Answer:

C.  changing nuclear energy to radiant energy

Explanation:

Nuclear energy takes atoms in their potential state, split them (fission) or fuse them (fusion)  creating chain reactions of radiant energy.  Most nuclear electrical power plants use fission, radiant energy heats water making steam to spin turbines.

Or think of the atom bomb.  Definitely potential energy until the fuse starts detonation and chain reactions.  The radiant kinetic energy and shock waves were horrendous.

3 0
3 years ago
Read 2 more answers
#1 Not sure where to start. This is for AP Physics!
yaroslaw [1]

First,

\rho=\dfrac mV

where \rho is density, m is mass, and V is volume. We can compute the volume of the roll:

2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V

\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness x. Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

V=Ax

where A is the given area, so

4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x

\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}

If we're taking significant digits into account, the volume we found would have been V=470\,\mathrm m^3, in turn making the thickness x=250\,\mathrm{mm}.

8 0
3 years ago
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