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Alekssandra [29.7K]
3 years ago
8

Intensive properties can be: Select the correct answer below:

Physics
1 answer:
jarptica [38.1K]3 years ago
4 0

Answer: The answer is 3 either physical or chemical properties

Explanation:

because all properties of matter are either physical or chemical

Hope this helps :)

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Imagine a 10kg block moving with a velocity of 20m/s to the left.
lara31 [8.8K]
Kinetic energy , KE= [1/2]m*v^2

m = 10 kg
v=20m/s

KE = [1/][(10kg)(20m/s)^2 = [1/2](10kg)(400m^2/s^2) = 2000 joule

Answer: 2000 joule
3 0
4 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
-2
photoshop1234 [79]

Answer:-2

Explanation:

7 0
2 years ago
In Speed Study Number 1, we looked at two cars traveling the same distance at different speeds on city streets. Car "A" traveled
Ganezh [65]

Answer:

230.4 s

Explanation:

The speed of car A is

v_A = 35 mi/h

and the distance travelled is

d = 10 mi

so the time taken for car A is

t_A = \frac{d}{v_A}=\frac{10 mi}{35 mi/h}=0.286 h

The speed of car B is

v_B = 45 mi/h

and the distance travelled is

d = 10 mi

so the time taken for car B is

t_B = \frac{d}{v_B}=\frac{10 mi}{45 mi/h}=0.222 h

So the difference in time is

\Delta t = t_A - t_B = 0.286 h -0.222 h=0.064 h

Which corresponds to

\Delta t = 0.064 h \cdot 3600 s/h = 230.4 s

so car B arrived 230.4 s before car A.

5 0
3 years ago
5. How will you separate water from petrol?
ohaa [14]

Answer:

by a seprating funnel or let it stand it will settle on its own

Explanation:

5 0
3 years ago
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