The duckbill platypus and spiny anteater are the only two known

Which property help to explain differences in the specific heat capabilities of two substances

OLga [1]

Answer:

Forces between molecules

Explanation:

The tensions between molecules are the characteristic that explains variances in the specific heat capacity of two substances.

This means that a substance's specific heat capacity will increase or be higher the closer its atoms are bound together. As a result, it differs for the different states of matter, such as solid, liquid, and gas.

3 0

1 year ago

A 9.6 cm diameter circular loop of wire is in a 1.10 T magnetic field perpendicular to the plane of the loop. The loop is remove

Natali5045456 [20]

Answer:

Thus induced emf is 0.0531 V

Solution:

As per the question:

Diameter of the loop, $d = 9.6\ cm = 0.096\ m$

Thus the radius of the loop, R = 0.048 m

Time in which the loop is removed, t = 0.15 s

Magnetic field, B = 1.10 T

Now,

The average induced emf, e is given by Lenz Law:

$e = - \frac{\Delta \phi_{B}}{\Delta t}$

where

$\phi_{B}$ = magnetic flux = $A\Delta B$

where

A = cross sectional area

Also, we know that:

$e = - \frac{A\Delta B}{\Delta t}$

$e = - \frac{\pi r^{2}\times (0 - 1.10)}{0.15}$

$e = - \frac{\pi \times 0.048^{2}\times (0 - 1.10)}{0.15}$

e = 0.0531 V

The sketch is shown in the figure, where I indicates the direction of the induced current.

3 0

3 years ago

Read 2 more answers

Hi If the coefficient of kinetic friction between the 5.0 kg mass and the table is 0.305, what is the tension in the string?

statuscvo [17]

Answer:

18 N

Explanation:

Draw a free body diagram for each block.

There are four forces acting on block I:

Weight force Mg pulling down

Normal force N pushing up

Tension force T pulling right

Friction force Nμ

There are two forces acting on block II:

Weight force mg pulling down

Tension force T pulling up

Sum of forces on block I in the +y direction:

∑F = ma

N − Mg = 0

N = Mg

Sum of forces on block I in the +x direction:

∑F = ma

T − Nμ = Ma

T − Mgμ = Ma

Sum of forces on block II in the -y direction:

∑F = ma

mg − T = ma

Solve for a in the first equation, then substitute into the second.

a = (T − Mgμ) / M

mg − T = m (T − Mgμ) / M

mMg − MT = mT − mMgμ

mMg + mMgμ = mT + MT

mMg (1 + μ) = (m + M) T

T = mMg (1 + μ) / (m + M)

T = (2) (5) (9.8) (1 + 0.305) / (2 + 5)

T = 18.27

Rounding to two significant figures, the tension is 18 N.

4 0

3 years ago

What can be the maximum value of the original kinetic energy of disk AA so as not to exceed the maximum allowed value of the the

timurjin [86]

The question is incomplete. The complete question is :

In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?

Solution :

Let M.I. of disk A = $I_0$