According to recent research, ice skaters are able to glide smoothly across the ice because ____

Consider a mixture of air and gasoline vapor in a cylinder with a piston and an original volume of 42.5 cm³. The combustion of t

Butoxors [25]

Answer:

volume of the gas expands to 8.74 Litres

Explanation:

The amount of work done by the gas is given by;

W = P•ΔV

Where;

W = work done by gas

P = Pressure

ΔV = change in volume i.e V_f - V_i

Where V_f is final volume and V_i is initial volume

Thus;

W = P(V_f - V_i)

We are given;

W = 927J

V_i = 42.5 cm³ = 42.5 x 10^(-6) m³

P = 1.05 atm = 106391.25 N/m²

Thus;

925 = 106391.25(V_f - (42.5 x 10^(-6))

925/106391.25 = V_f - (42.5 x 10^(-6))

0.00869432402 = V_f - (42.5 x 10^(-6))

V_f = 0.00869432402 + (42.5 x 10^(-6)))

V_f = 0.00874 m³

Final volume = 0.00874 m³

Converting to Litres, gives;

V_f = 0.00874 x 1000 = 8.74 Litres

8 0

3 years ago

A magnifying glass is an example of which type of microscope?

gogolik [260]

A magnifying glass is a simple microscope

5 0

3 years ago

Read 2 more answers

Find h , the maximum height attained by the projectile. express the maximum height in terms of v 0 , Î¸ , and g .

lutik1710 [3]

solution:

the utmost height would be comprehensive while y'(t)=0. (on the suitable of the trajectory, the y speed is 0.) $y(t)=(88sin20)t-16t^2 y'(t)=88sin20-32t So y'(t)=0 while t=(88sin20)/32$

3 0

4 years ago

A system of two objects has ΔKtot = 6 J and ΔUint = -5 J. Part A How much work is done by interaction forces? Express your answe

Elina [12.6K]

A) +5 J

B) +1 J

Explanation:

A)

The internal forces (interaction forces) acting on a system do not change the mechanical energy (sum of potential and kinetic energy) of the system.

However, these forces are responsible for converting the energy from one form into another; the work done by these forces is equal to the amount of energy converted from one form into the other.

In this problem, we have:

$\Delta U=-5 J$ is the loss in potential energy of the system

$\Delta K=+6 J$ is the gain in kinetic energy of the system

By looking at these numbers, this means that the internal forces have converted 5 J of energy from potential energy into kinetic energy (while the additional +1 J missing is due to external forces, as explained in part B).

Therefore, the work done by internal forces is

W = +5 J

B)

First of all, we calculate the change in mechanical energy of the system.

The mechanical energy of a system is the sum of its kinetic energy (K) and its potential energy (U):

$E=K+U$

So, the change in mechanical energy is equal to the sum of the changes of kinetic energy and the changes of potential energy:

$\Delta E= \Delta K + \Delta U$

In this problem:

$\Delta K=+6 J$

$\Delta U=-5 J$

So, the change in mechanical energy is:

$\Delta E=+6+(-5)=+1 J$

According to the work-energy theorem, the work done by external forces on a system is equal to the change in mechanical energy of the system: therefore in this case, the work done by external forces is

$W=\Delta E=+1 J$

5 0

4 years ago

A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodie

vivado [14]

Answer:

The length of the rod for the condition on the question to be met is $L = 1.5077 \ m$

Explanation:

The Diagram for this question is gotten from the first uploaded image

From the question we are told that

The mass of the rod is $M = 3.41 \ kg$

The mass of each small bodies is $m = 0.249 \ kg$

The moment of inertia of the three-body system with respect to the described axis is $I = 0.929 \ kg \cdot m^2$

The length of the rod is L

Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

$I = I_r + 2 I_m$

Where $I_r$ is the moment of inertia of the rod about the describe axis which is mathematically represented as

$I_r = \frac{ML^2 }{12}$

And $I_m$ the moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented as