Which is an example of speed?

Physics

2 answers:

agasfer [191]3 years ago

6 0

A dog runs an average of 8 mph
or
cindy biked westward at 30 km/h

geniusboy [140]3 years ago

4 0

<span>Cindy biked westward at 30 km/h.
Velocity. (We know both the speed AND the direction.)

A rock is to the left of a flowerpot.
Relative location.

A dog runs an average of 8 mph.
Speed (distance divided by time)

Leo lives 2 miles north of Andre.
Relative location.</span>

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A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 200 m long,

Svetlanka [38]

Answer:

d) 289.31 m

Explanation:

Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m

Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .

Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d

To equate

357.18 m -144.54 m = .735 m d

d = 289.31 m .

4 0

3 years ago

a 2.35 water bucket is swung in a full cirlce of radius 0.824 m just fast enough so that the water doesn't fall out the top mean

tiny-mole [99]

Answer:

2.84 m/s

Explanation:

At the top position of the circular trajectory, the normal reaction is zero:

N = 0

So it means that the only force that is providing the centripetal force is the gravitational force (the weight of the bucket). Therefore we have:

$mg = m \frac{v^2}{r}$

where

m is the mass of the water bucket

g = 9.8 m/s^2 is the acceleration of gravity

v is the speed of the bucket

r = 0.824 m is the radius of the circle

Solving for v,

$v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(0.824 m)}=2.84 m/s$

4 0

3 years ago

Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini

amid [387]

Answer:

Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

$E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}$

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

$E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}$

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$