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Serggg [28]
3 years ago
5

there are two slides at the park between which you are deciding. Both start at the height of 6 meter. One is short and Steve whi

le the other is long and shallow. If both sides are frictionless, which one should you choose if you want to be moving as fast as possible at the bottom? A. short and steep. B. long and shallow. C. it doesn't matter.
Physics
1 answer:
kumpel [21]3 years ago
6 0

It doesn't matter.  If the slides are truly frictionless, then
your kinetic energy at the bottom will be equal to the
potential energy you had at the top, no matter what kind
of route you took getting down.
___________________________

The only way I can think of that it would make a difference
would be if the shallow slide were REALLY REALLY long,
and you didn't have anything to eat all the way down. 
Then you might lose some weight while you're on the slide,
and your mass might be less at the bottom than it was at the
top.  Then, in order to have the same kinetic energy at the
bottom, you'd need to be going a little bit faster.

But if it takes less than, say, two or three days, to go down the
long, shallow slide, then this effect would probably be too small
to make any difference.

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The two basic units of weight in the metric system is the___?
Alika [10]

Answer:

The two basic units of weight in the metric system is the?

The gram and kilogram is the Correct Answer...

3 0
3 years ago
The purpose of striking the ball in a volleyball game is to: 1. Place the ball in motion 2. Change the direction of the ball's m
zhenek [66]

Answer:

4. All of the above

Explanation:

The purpose of striking the ball in a volleyball game:

From the serve you could state that you need to place the ball in motion.

When returning a shot of, you normally want to change the direction of the ball's motion.

During a dropshot, you purposely want to slow down the ball's motion.

The correct answer must be all of the above.

8 0
3 years ago
The potential difference between the plates of a capacitor is 145 V. Midway between the plates, a proton and an electron are rel
aniked [119]

Answer:

= 2.52 x 10^ 6 m/s        

Explanation:

The force that acts on charged particles between capacitor plates =

F = (q) (Δv)  ÷ d

Here,  d = distance between the two plates

          q = charge of the charged particle

         Δv = voltage

Normally, the force that makes both proton and electron released from rest, giving the charge acceleration is F=m X a. where m= mass and a = acceleration

Poting this equation with the first one, we have:

m X a =  (q) (Δv)  ÷ d

So, the acceleration of a proton when moving towards a negatively charged plate is

a = (q) (Δv)  ÷ (d) (m) {proton}

Likewise, the acceleration of an electron when moving towards a positively charged plate is

a = (q) (Δv)  ÷ (d) (m) {electron}

Dividing the proton acceleration formula by the electron acceleration formula we have:

a (proton) / a (electron) = m (proton) / m(electron)

inserting equation of motion to get distance, s

s = ut + 1/2 at^2

recall that electron travel distance, d/2

d/2 = 1/2 at^2

making t the subject of the formula

we have, t =√(d ÷ a(electron))

The distance of proton:

d/2 =  ut + 1/2 at^2 [proton}

put d/2 =  ut + 1/2 at^2 [proton} into t =√(d ÷ a(electron))

Initial speed, ui = √(d ÷ a(electron)) = (d/2) - (1/2) x (d) (a(proton) + a(electron))

since acceleration wasn't given in the question, lets use mass(elect

ron)  ÷ mass(proton) rather than use (a(proton) + a(electron))

Therefore, intial speed= 1/2√((e X Δv) ÷ m(electron)) (1- m(electron)/ m(proton))

   Note, e = 1.60 x 10^-19

           m(electron) = 9.11 X 10^-31

            m(proton) = 1.67  X 10^-27

Input these values into the formula above, initial speed, UI =  

           = 2.52 x 10^ 6 m/s          

7 0
3 years ago
A solid sphere of weight 42.0 N rolls up an incline at an angle of 36.0°. At the bottom of the incline the center of mass of the
Alecsey [184]

Answer:

Part a)

KE = 77.95 J

Part b)

L = 3.16 m

Part c)

distance L is independent of the mass of the sphere

Explanation:

Part a)

As we know that rotational kinetic energy of the sphere is given as

KE = \frac{1}{2}I\omega_2 + \frac{1}{2}mv^2

so we will have

KE = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 + \frac{1}{2}mv^2

so we will have

KE = \frac{1}{5} mv^2 + \frac{1}{2}mv^2

KE = \frac{7}{10} mv^2

KE = \frac{7}{10}(\frac{42}{9.81})(5.10^2)

KE = 77.95 J

Part b)

By mechanical energy conservation law we know that

Work done against gravity = initial kinetic energy of the sphere

So we will have

mgLsin\theta = KE

\frac{42}{9.81}(9.81)L sin36 = 77.95

L = 3.16 m

Part c)

by equation of energy conservation we know that

\frac{7}{10}mv^2 = mgL sin\theta

so here we can see that distance L is independent of the mass of the sphere

7 0
3 years ago
The speed of light is greater in a vacuum than in air or water.<br><br> True or false
Dima020 [189]
True !! Hope I helped you out a bit!
7 0
3 years ago
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