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3 years ago

Please answer correctlyWill give the brainliest!

Physics

1 answer:

Oksi-84 [34.3K]3 years ago

3 0

In the first case:

when we heat any gas, the Kinetic Energy of the molecules increases, making it collide more frequently with the surface, increasing the pressure

more collisions with the surface means more force applied on it, which would push the piston harder than before, moving it outwards.

In the second case:

since the molecules inside the beaker have no way to escape, they would keep compressing the more you push the beaker downwards.

since there is the same number of molecules and lesser volume to cover, the molecules will start colliding with the surfaces more frequently, which would resist the downward force.

another way to think about it is to imagine yourself where the trapped air is. you would be happy when the room is spacious but if the wall starts moving towards you, you would resist the change by your body because you need space to exist. making it harder for the wall to move.

pushing the beaker downwards will keep getting harder and harder the more you push until you reach a point where the molecules will be completely compact. applying even more force forces the molecules to enter water, removing the air that was resisting it all and making you able to get the beaker in water.

Third case:

just like in the first case, the heated air will apply force on the surface, including the cork. which would pop off when enough force is applied.

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A flat sheet of paper of area 0.365 m2 is oriented so that the normal to the sheet is at an angle of 60 ∘ to a uniform electric

andriy [413]

Answer:

A.) 3.65 N*m²/C B) No C) 0º D) 90º

Explanation:

A) The electric flux, when the electric field is uniform across a gausssian surface, can be calculated as the dot product of the electric field vector, and the vector representing the area of the surface (normal to the surface and directed outward it by convention), as follows:

Flux = E*A*cos φ

where E = 20 N/C, A = 0.365 m², φ = 60º.

Replacing by the values, we can get the value of the electric flux, as follows:

Flux = 20 N/C* 0.365 m²*0.5 = 3.65 N*m²/C

B) While the area remains constant, and doesn't change orientation, the value of the flux will be the same, regardless the shape of the sheet.

C) When the normal to the sheet and the electric field are parallel each other, the surface will intercept the maximum number of field lines, i.e. the flux will be directly E*A*cos 0º = E*A (maximum value possible).

D) When the electric field is tangent to the surface, this means that no field lines will be intercepted by the sheet, so the flux is zero.

In this case, φ = 90º, cos φ = 0

⇒ E*A*cos 90º = E*A*0 = 0

5 0

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

An unknown material, m1 = 0.49 kg, at a temperature of T1 = 92 degrees C is added to a Dewer (an insulated container) which cont

erastova [34]

Answer:

$c_u=1540.5J/kg^{\circ}K$

Explanation:

We know that heat relates to mass, specific heat and variation of temperature experimented because of this heat through the equation $Q=mc\Delta T=mc(T_f-T_i)$ . The heat released by the unknown material is absorbed by water, so we have $Q_u=-Q_w$ , and we can write:

$m_uc_u(T_{uf}-T_{ui})=-m_wc_w(T_{wf}-T_{wi})$

Since thermal equilibrium is reached we know that $T_{cf}=T_{wf}=T_f=31^{\circ}C=304^{\circ}K$ , where we have added $273^{\circ}$ to convert the temperature from Celsius to Kelvin, as we must do. Since we want the specific heat of the unknown material, we do: