a bicyclist travels with an averahe velocity of 15km/h north, for 20 minutes .what is his displacement ?

What is the amount of heat, in Sl units, necessary to melt 1 lb of ice?

likoan [24]

Answer:

Heat required to melt 1 lb of ice is 151.469 KJ

Explanation:

We have given mass of ice = 1 lb

We know that 1 lb = 0.4535 kg

Latent heat of fusion for ice =334 KJ/kg

Amount if heat for fusion of ice is given by

$Q=mL$ , here m is mass of ice and L is latent heat of fusion

So heat $Q=mL=0.4535\times 334=151.469kj$

So heat required to melt 1 lb of ice is equal to 151.469 KJ

3 0

3 years ago

A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If

vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
The traslational part can be written as follows:

$K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)$

The rotational part can be expressed as follows:

$K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)$

where I = moment of Inertia regarding the axis of rotation.
ω = angular speed of the rotating object.
If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

$v = \omega * R (3)$

For a solid cylinder, I = M*R²/2 (4)
Replacing (3) and (4) in (2), we get:

$K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)$

Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

$K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)$

Repeating the same steps for the spherical shell:

$I_{sph} = \frac{2}{3} * M* R^{2} (7)$

$K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)$

$K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)$

Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
Rearranging, and taking square roots on both sides, we get:

$\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)$

This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.

3 0

3 years ago

This 80 kg car is moving at 20m/sec at the top where the hills radius is 100m. What is the centrifugal force?

earnstyle [38]

100 seconds is the right thing

3 0

3 years ago

Please help, I do not understand

Anettt [7]

I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.

We're given two angular speeds, and we need to solve for a time.

Outer (slower) planet:
Angular speed = ω rad/sec
Time per unit angle = (1/ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .

Inner (faster) planet:
Angular speed = 2ω rad/sec
Time per unit angle = (1/2ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.

So far so good. We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed. Perfect !

At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:

They're in line, SOMEwhere on the circles, when

   (a fraction of one orbit) = (the same fraction of the other orbit)
AND
   the total elapsed time is a common multiple of their periods.

Wait ! Ignore all of that. I'm doing a good job of confusing myself, and
probably you too. It may be simpler than that. (I hope so.) Throw away
those last few paragraphs.

The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed. We're just looking for the Least
Common Multiple of the two periods.

      K (2π/ω seconds) = (K+1) (π/ω seconds)

   2Kπ/ω   =    Kπ/ω + π/ω

Subtract Kπ/ω :    Kπ/ω = π/ω

Multiply by ω/π :    K = 1

(Now I have a feeling that I have just finished re-inventing the wheel.)

And there we have it:

   In the time it takes the slower planet to revolve once,
   the faster planet revolves twice, and catches up with it.

   It will be 2π/ω seconds before the planets line up again.

   When they do, they are again in the same position as shown
   in the drawing.

To describe it another way . . .

   When Kanye has completed its first revolution ...

   Bieber has made it halfway around.

   Bieber is crawling the rest of the way to the starting point while ...

   Kanye is doing another complete revolution.

   Kanye laps Bieber just as they both reach the starting point ...

   Bieber for the first time, Kanye for the second time.

You're welcome. The generous bounty of 5 points is very gracious,
and is appreciated. The warm cloudy water and green breadcrust
are also delicious.

5 0

3 years ago

What is the source of all mechanical waves?

frutty [35]

Answer:

The correct answer is A. Vibration.

Explanation:

Mechanical waves is formed by the oscillation of matter and therefore transfer energy from one medium to the other. Unlike electromagnetic waves, mechanical waves need some medium to propagate. It requires an initial energy input and thus carries this energy when it propagates. There are three types of mechanical waves namely transverse waves, longitudinal waves and surface waves. Examples of such waves are sound waves, water waves and seismic waves.

6 0

3 years ago

a bicyclist travels with an averahe velocity of 15km/h north, for 20 minutes .what is his displacement ?​

a bicyclist travels with an averahe velocity of 15km/h north, for 20 minutes .what is his displacement ?