All categories

4 years ago

Which of the following statements accurately describes the properties of gases?

Physics

2 answers:

alexira [117]4 years ago

6 0

THE CORRECT ANSWER IS B

Ray Of Light [21]4 years ago

4 0

<span>A change in the pressure of a gas results in a more significant change in volume than it would in a liquid. is the statement that accurately describes the property of gas. Gas only depends on how you store it. the bigger the space the wider gas can expand, the smaller the space, the more compress the gas can become.</span>

You might be interested in

What are the laws of physics?

Ray Of Light [21]

Classic mechanics and atomic physics

3 0

4 years ago

The specific heat of Aluminum is 0.9 J/g K. How much heat is lost as a 100 gram sample of Aluminum is dropped into a beaker of w

otez555 [7]

Energy=mass*SHC*temp change
=100*0.9*75
=6750J

6 0

4 years ago

Mercury, Venus, and Earth are the three planets closest to the Sun. The force of gravity on the surface of each planet is differ

LUCKY_DIMON [66]

Earths surface gravity: 9.7

Venus surface gravity: 8.8

Mercury's surface gravity: 3.7

Your answer would be D: Earth, Venus, Mercury

Hope this helps! (:

4 0

3 years ago

An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis

Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

$W=\Delta K$ (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

$W=\Delta K + E_{lost}$

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane ( $\Delta K$ ) and part is lost because of the air resistance ( $E_{lost}$ ).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

$F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N$

Now we can calculate the acceleration of the plane, by using Newton's second law:

$a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2$

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

$v^2- u^2 = 2aS$

where

$v=?$ is the final velocity

$u=66.0 m/s$ is the initial velocity

$a=1.69 m/s^2$ is the acceleration

$S=5.00 \cdot 10^2 m$ is the distance travelled

Solving for v, we find

$v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s$

8 0

3 years ago

A spring (k=15.19kN/m)is is compresses 25cm and held in place on a 36.87° incline. A block (M=10kg) is placed on the spring. Whe

Savatey [412]

Answer:

The maximum vertical displacement is 2.07 meters.

Explanation:

We can solve this problem using energy. Since there is a frictional force acting on the block, we need to consider the work done by this force. So, the initial potential energy stored in the spring is transferred to the block and it starts to move upwards. Let's name the point at which the block leaves the ramp "1" and the highest point of its trajectory in the air "2". Then, we can say that:

$E_0=E_1\\\\U_e_0=K_1+U_g_1+W_f_1$

Where $U_e_0$ is the elastic potential energy stored in the spring, $K_1$ is the kinetic energy of the block at point 1, $U_g_1$ is the gravitational potential energy of the block at point 1, and $W_f_1$ is the work done by friction at point 1.

Now, rearranging the equation we obtain:

$\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgh_1+\mu Ns_1$

Where $k$ is the spring constant, $x$ is the compression of the spring, $m$ is the mass of the block, $v_1$ is the speed at point 1, $g$ is the acceleration due to gravity, $h_1$ is the vertical height of the block at point 1, $\mu$ is the coefficient of kinetic friction, $N$ is the magnitude of the normal force and $s_1$ is the displacement of the block along the ramp to point 1.

Since the force is in an inclined plane, the normal force is equal to:

$N=mg\cos\theta$

Where $\theta$ is the angle of the ramp.

We can find the height $h_1$ using trigonometry:

$h_1=s_1\sin\theta$

Then, our equation becomes:

$\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgs_1\sin\theta+\mu mgs_1\cos\theta\\\\\implies v_1=\sqrt{\frac{2(\frac{1}{2}kx^{2}-mgs_1\sin\theta-\mu mgs_1\cos\theta)}{m}}=\sqrt{\frac{kx^{2}}{m}-2gs_1(\sin\theta+\mu \cos\theta)}$

Plugging in the known values, we get:

$v_1=\sqrt{\frac{(15190N/m)(0.25m)^{2}}{10kg}-2(9.8m/s^{2})(1.12m)(\sin36.87\°+(0.300) \cos36.87\°)}\\\\v_1=8.75m/s$

Now, we can obtain the height from point 1 to point 2 using the kinematics equations. We care about the vertical axis, so first we calculate the vertical component of the velocity at point 1:

$v_1_y=v_1\sin\theta=(8.75m/s)\sin36.87\°=5.25m/s$

Now, we have:

$y=\frac{v_1_y^{2}}{2g}\\\\y=\frac{(5.25m/s)^{2}}{2(9.8m/s^{2})}\\\\y=1.40m$

Finally, the maximum vertical displacement $h_2$ is equal to the height $h_1$ plus the vertical displacement $y$ :

$h_2=h_1+y=s_1\sin\theta +y\\\\h_2=(1.12m)\sin36.87\°+1.40m\\\\h_2=2.07m$

It means that the maximum vertical displacement of the block after it becomes airborne is 2.07 meters.

7 0

3 years ago