water contracts on freezing is it incorrect or conrrect

Answer:

4 s

Explanation:

We are asked to calculate time taken (t).

By using the first equation of motion,

[tex]\longrightarrow[/tex] v = u + at

[tex]\longrightarrow[/tex] 160 = 100 + 15t

[tex]\longrightarrow[/tex] 160 - 100 = 15t

[tex]\longrightarrow[/tex] 60 = 15t

[tex]\longrightarrow[/tex] 60 ÷ 15 = t

[tex]\longrightarrow[/tex] 4 s = t

Answer:

The drop time ball 1 is less than the drop time of ball 2. A further explanation is provided below.

Explanation:

The net force acting on the ball will be:

⇒ [tex]F_{net}=mg-F_r[/tex]

Here,

F = Force

m = mass

g = acceleration

Now,

According to the Newton's 2nd law of motion, we get

⇒ [tex]F_{net} = ma[/tex]

To find the value of "a", we have to substitute "[tex]F_{net}=ma[/tex]" in the above equation,

⇒ [tex]ma=mg-F_r[/tex]

⇒    [tex]a=g-\frac{F_r}{m}[/tex]

We can see that, the acceleration is greater for the greater mass of less for the lesser mass. Thus the above is the appropriate solution.

Answer:

Both the ball takes equal time to reach to the ground.  

Explanation:

Two balls of same diameter

Let the height is h.

Mass of ball 1 is more than the mass of ball 2.

The second equation of motion is

[tex]h = u t +0.5 gt^2[/tex]

Here, the buoyant force due to air is same. So, the time of fall is independent of the mass.

So, both the ball takes equal time to reach to the ground.  

Answer:

a) Initial Temperature = 609.4 K and Final Temperature = 325.7 K

b) the change in internal energy is -18279.78 J

c) heat lost by the gas is zero or 0

d) the work done on the gas is -18279.78 J

Explanation:

Given the data in the question;

P[tex]_i[/tex] = 1 atm = 101325 pascal

P[tex]_f[/tex] = ?

V[tex]_i[/tex] = 0.1550 m³

V[tex]_f[/tex] = 0.742 m³

we know that for an adiabatic process  γ = 1.4

P[tex]_i[/tex]V[tex]_i^Y[/tex] = P[tex]_f[/tex]V[tex]_f^Y[/tex]

P[tex]_f[/tex] = P[tex]_i[/tex][tex]([/tex] V[tex]_i[/tex] / V[tex]_f[/tex] [tex])^Y[/tex]

we substitute

P[tex]_f[/tex] = 1 × [tex]([/tex] 0.1550  / 0.742  [tex])^{1.4[/tex]

= [tex]([/tex] 0.2088948787 [tex])^{1.4[/tex]

= 0.11166 atm

a) the initial and final temperatures

Initial temperature

T[tex]_i[/tex] = P[tex]_i[/tex]V[tex]_i[/tex] / nR

given that n = 3.10 mol

= ( 101325 × 0.1550 ) / ( 3.10 × 8.314 )

= 15705.375 / 25.7734

T[tex]_i[/tex]  = 609.4 K

Final temperature

T[tex]_f[/tex] = P[tex]_f[/tex]V[tex]_f[/tex] / nR

= ( 0.11166 × 101325 × 0.742 ) / ( 3.10 × 8.314 )

= 8394.95 / 25.7734

= 325.7 K

Therefore, Initial Temperature = 609.4 K and Final Temperature = 325.7 K

b) the change in internal energy

ΔE[tex]_{int[/tex] = nC[tex]_v[/tex]ΔT

here, C[tex]_v[/tex] = ( 5/2 )R

ΔE[tex]_{int[/tex] = 3.10 × ( 5/2 )8.314 × ( 325.7 - 609.4 )

= -18279.78 J

Therefore, the change in internal energy is -18279.78 J

c) the heat lost by the gas

Since its an adiabatic process,

Q = 0

Therefore, heat lost by the gas is zero or 0

d)  the work done on the gas

W = ΔE[tex]_{int[/tex] - Q

= -18279.78 J - 0

W = -18279.78 J

Therefore, the work done on the gas is -18279.78 J

a) The Initial Temperature and Final Temperature of gas are 601.68 K and 321.61 K respectively.

b) The change in internal energy is -18279.78 J.

c) The heat lost by the gas is zero.

d) The work done on the gas is -18279.78 J.

Given data:

The moles of sample is, n = 3.10 mol.

The initial volume of sample is, [tex]V_{1}=0.1550 \;\rm m^{3}[/tex].

The final volume of sample is, [tex]V_{2}=0.742 \;\rm m^{3}[/tex].

The initial pressure of the sample is, [tex]P_{1}=1.00 \;\rm atm[/tex].

(a)

We know that the relation between the pressure and volume for an adiabatic process is as follows,

[tex]P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}[/tex]

Here, [tex]\gamma[/tex]  is a adiabatic index. And for air, its value is 1.41.

Solving as,

[tex]P_{2}=P_{1} \times\dfrac{V_{1}^{\gamma}}{V_{2}^{\gamma}}\\\\\\P_{2}=1.00 \times\dfrac{0.1550^{1.41}}{0.742^{1.41}}\\\\\\P_{2} = 0.11166 \;\rm atm[/tex]

Now, calculate the final temperature using the ideal gas equation as,

[tex]P_{2}V_{2}=nRT_{2}\\\\T_{2}= \dfrac{P_{2} \times V_{2}}{nR}\\\\T_{2}= \dfrac{0.11166 \times 10^{5}\times 0.742}{3.10 \times 8.31}\\\\T_{2}=321.61 \;\rm K[/tex]

Similarly, calculate the initial temperature as,

[tex]P_{1}V_{1}=nRT_{1}\\\\T_{1}= \dfrac{P_{1} \times V_{1}}{nR}\\\\T_{1}= \dfrac{1 \times 10^{5}\times 0.1550}{3.10 \times 8.31}\\\\T_{1}=601.68 \;\rm K[/tex]

Thus, we can conclude that the initial and final temperature of the gas is 601.68 K and 321.61 K respectively.

(b)

The change in internal energy is given as,

ΔE = nCΔT

here, C = ( 5/2 )R

ΔE = 3.10 × ( 5/2 )8.314 × ( 325.7 - 609.4 )

      = -18279.78 J

Therefore, the change in internal energy is -18279.78 J.

c)

The heat lost by the gas . Since its an adiabatic process, so there will be no heat interaction.  

Q = 0

Therefore, heat lost by the gas is zero or 0

d)  

The work done on the gas

W = ΔE - Q

W = -18279.78 J - 0

W = -18279.78 J

Therefore, the work done on the gas is -18279.78 J.

Learn more about the adiabatic process here:

https://brainly.com/question/14930930

Answer:

self-indulgence of coil is the property by virtue of wich is tends to maintain magnatic flux link with it and opposed any any change in the flux inducing current in it

Answer and Explanation:

a. An oxygen-filled balloon is not able to float in the air, because the oxygen inside the balloon is of the same density, that is, the same "weight" as the oxygen outside the balloon and present in the atmosphere. The balloon can only float if the gas inside it is less dense than atmospheric oxygen. Helium gas is less dense than atmospheric gas, so if a balloon is filled with helium gas, that balloon will be able to float because of the difference in density.

b. The ship is able to float in the water because its steel construction is hollow and full of air. This makes the average density of this ship less than the density of water, which makes the ship lighter than water and for this reason, this ship is able to float. In addition, the ship is partially immersed, allowing the weight of the ship on the water to counteract the buoyant force that the water promotes on the ship. Weight and buoyant are two opposing forces that keep the ship afloat.

Answer:

461.73 K

Explanation:

Given that,

The mass of a bullet, m = 5.7 g

Speed of the bullet, v = 490 m/s

Half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree.

Using the conservation of energy,

[tex]\dfrac{1}{2}(\dfrac{1}{2}mv^2)=mc\Delta T\\\\\Delta T=\dfrac{v^2}{4c}[/tex]

Where

x is the specific heat of lead, c = 130 J/kg K

So,

[tex]\Delta T=\dfrac{(490)^2}{4\times 130}\\\\=461.73\ K[/tex]

So, the increase in temperature of the bullet is 461.73 K.

Answer:

h = 2755102 m = 2755.102 km

Explanation:

According to the given condition:

Potential Energy = Energy Consumed by Bulb

[tex]mgh = Pt\\\\h = \frac{Pt}{mg}[/tex]

where,

h = height = ?

P = Power of bulb = 75 W

t = time = (2 h)(3600 s/1 h) = 7200 s

m = mass of bulb = 20 g = 0.02 kg

g = acceleration due to gravity = 9.8 m/s²

Therefore,

[tex]h = \frac{(75\ W)(7200\ s)}{(0.02\ kg)(9.8\ m/s^2)}[/tex]

h = 2755102 m = 2755.102 km

Answer:

temperature

Explanation:

temperature change forms different elements and different element sustain different colour

Answer:

When an object is dropped, the "principal" force that acts on that object is the gravitational force.

Thus, in the absence of air resistance and such, the acceleration of the object will be equal to the gravitational acceleration:

g = 9.8m/s^2

So, when we drop objects in the moon (where there is no air) the acceleration of every object will be exactly the same. (so there is no dependence in the mass or shape of the object)

Thus, if we drop a coin and a feather in the moon, both objects will fall with the same acceleration, and then both objects will hit the ground at the same time.

But if we are in Earth, we can not ignore the air resistance (a force that acts in the opposite direction than the movement of the object)

And this force depends on the shape and mass of the object (for example, something with a really larger surface and really thin, like a sheet of paper will be more affected by this force than a small rock)

Then here, when the air resistance applies, we should expect that the heavier and smaller object (the coin) to be less affected by this force, then the resistance that the coin experiences is smaller, then the coin falls "faster" than the feather.

Explanation:

15.556 metres per second

Answer:

t = 2.7 s

Explanation:

This is a kinematics problem.

How the elevator reduces its speed to zero by a distance equal to the length of the cylinder

        y = 2π r = 2 π d / 2

        y = π d

        y = π 1.1

        y = 3,456 m

now we can look for the acceleration of the system

        v² = v₀² - 2 a y

        0 = v₀² - 2 a y

        a = v₀² / 2y

        a = 2.6² / 2 3.456

        a = 0.978 m / s²

now let's calculate the time

        v = v₀ - a t

        0 = v₀ - at

         t = v₀ / a

         t = 2.6 /0.978

         t = 2.658 s

ask for the result with two significant figures

         t = 2.7 s

Answer: 2.4 km/hr

Explanation:

Distance = 600m

Time= 15 minutes = 15 x 60 second/minute = 900 seconds

Speed =  [tex]\frac{Distance}{Time}[/tex]  =  [tex]\frac{600}{900}[/tex]  = [tex]\frac{2}{3}[/tex] m/sec

⇒ [tex]\frac{2}{3}[/tex] x [tex]\frac{18}{5}[/tex] = 2.4 km/hr (1 m/sec = 3.6 km/hr)

Answer:

[tex]P=17658Pa[/tex]

Explanation:

From the question we are told that:

Dimension

 [tex]L*B=27.0*8.9[/tex]

Depth [tex]d=1.8m[/tex]

Generally the equation for Volume of water is mathematically given by

 [tex]V=L*B*D[/tex]

 [tex]V=27.0*8.9*1.8[/tex]

 [tex]V=432.54m^3[/tex]

Therefore

Force at the bottom of the Pool

 [tex]F=\rho Vg[/tex]

Where

 [tex]\rho \ density\ of \ water(1000kg/m^3)[/tex]

 [tex]F=1000*432.54m^3*9.81[/tex]

 [tex]F=4.2*10^{6}N[/tex]

Generally the equation for Pressure at the bottom is mathematically given by

 [tex]P=\frac{Forece }{Area}[/tex]

 [tex]P=\frac{4.2*10^{6}N}{27.0*8.9}[/tex]

 [tex]P=17658Pa[/tex]

Answer:

n × t_1/2

Exmplanation:

The age of meteorite is calculated by multiplying it's quantity n with the half life . This means that the formula is for age of this meteorite is;

Age of meteorite= n × t_1/2

where;

n = quantity of the meteorite

t_/2 = half life of the meteorite

Thus:

The correct formula is; n × t_1/2

Answer:

4 kg of force

Explanation:

Force = (mass x distance to fulcrum) / length of fulcrum to end

Subsitute values

F = (10 x 20)/50

F =4

Answer: The correct statement is:

--> The voltage across each resistor is the same.

Explanation:

RESISTORS are defined as the components of an electric circuit which are capable of creating resistance to the file of electric current in the circuit. They work by converting electrical energy into heat, which is dissipated into the air. These resistors can be divided into two according to their arrangements in the electric cell. It include:

--> Resistors in parallel and

--> Resistors in series

RESISTORS are said to be in parallel when two or more resistance or conductors are connected to common terminals so that the potential difference ( voltage) across each conductor IS THE SAME but with different current flow through each of them. Also, Individual resistances diminish to equal a smaller total resistance rather than add to make the total.

Answer:

if anyone is reading my coment your oarents will became very fine soon

Answer:

(a) 8.57 x 10^8 Hz

(b) 23.3 cm

Explanation:

Wavelength = 35 cm = 0.35 m

speed =3 x10^8 m/s

Let the frequency is f.

(a) The relation is

speed  = frequency x wavelength

3 x 10^8 = 0.35 x f

f = 8.57 x 10^8 Hz

(b) refractive index of glass  is 1.5

The relation for the refractive index and the wavelength is

wavelength in glass= wavelength in air/ refractive index.

Wavelength in glass= 35/1.5 = 23.3 cm

Answer:

a)  [tex]v_1=-0.833m/s[/tex]

b)  [tex]V_2=12.5m/s[/tex]

Explanation:

From the question we are told that:

Hockey puck Mass [tex]m_1=0.495kg[/tex]

Hockey puck Speed [tex]u_1=4.50m/s[/tex]

Puck Mass [tex]m_2=0.720kg[/tex]

Assuming

Initial speed of Puck [tex]u_2=0[/tex]

Generally the equation for Speed of First Puck is mathematically given by

 [tex]v_1=(\frac{m_1-m_2}{m_1+m_2})*v_1+(\frac{2m_2}{m_1+m_2})u_2[/tex]

 [tex]v_1=(\frac{0.495-0.720}{0.495+0.720})*4.50+0[/tex]

 [tex]v_1=-0.833m/s[/tex]

Generally the equation for Speed of Second Puck is mathematically given by

 [tex]V_2=(\frac{2m_1}{m_1+m_2})u_2-(\frac{m_1-m_2}{m_1+m_2})v[/tex]

 [tex]V_2=(\frac{2*0.495}{0.495*0.720})*4.50-0[/tex]

 [tex]V_2=12.5m/s[/tex]

Answer:

 B = 1.1413 10⁻² T

Explanation:

We use energy concepts to calculate the proton velocity

starting point. When entering the electric field

        Em₀ = U = q V

final point. Right out of the electric field

        em_f = K = ½ m v²

energy is conserved

       Em₀ = Em_f

       q V = ½ m v²

       v = [tex]\sqrt{2qV/m}[/tex]

we calculate

       v = [tex]\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }[/tex]

       v = [tex]\sqrt{632.3353 \ 10^8}[/tex]

       v = 25.15 10⁴ m / s

now enters the region with magnetic field, so it is subjected to a magnetic force

        F = m a

the force is

       F = q v x B

as the velocity is perpendicular to the magnetic field

       F = q v B

acceleration is centripetal

       a = v² / r

we substitute

       qvB =1/2  m v² / r

       B =  v[tex]\frac{m v}{2 q r}[/tex]

we calculate

       B = [tex]\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}[/tex]

       B = 1.1413 10⁻² T

Answer:

The height is the same

Explanation:

Because they were at the same height but they fell at different velocities

Answer:

E_total = - 50 N / A

Explanation:

The electric field is a vector magnitude whereby

          E_total = Eₐ + E_b

where the bold letters indicate vectors, in this case the charges of the two objects A and B are the same and they are on the same line

         E_total = - E_a - E_b

the electric field for a point charge is

        E_a = [tex]k \ \frac{q_a}{r_a^2 }[/tex]

        qₐ= Eₐ rₐ² / k

indicates that Eₐ = 40.0 N / C

        qₐ = 40.0 0.250²/9 10⁹

        qₐ = 2.777 10⁻¹⁰ C

indicates that the charge of the two points is the same

        qₐ = q_b

      E_total = - k qₐ / rₐ² - k qₐ / (2 rₐ)²

      E_total = [tex]-k \ \frac{q_a}{r_a^2} \ ( 1 + \frac{1}{4} )[/tex]

we calculate

       E_total = - 40.0 (5/4)

       E_total = - 50 N / A

Answer:

7.60× 10^6 V/m

Explanation:

electric field strength can be determined as ratio of potential drop and distance, I.e

E=V/d

Where E= electric field

V= potential drop= 74.0 mV= 0.07 V

d= distance= 9.20 nm = 9.2×10^-9 m

Substitute the values

E= 0.07/ 9.2×10^-9

= 7.60× 10^6 V/m

Answer:

have an increased resistance

Answer:

[tex]c.) \: systematic \: error \\ \\ = > it \: is \: the \: error \: caused \: \\ \\ due \: to \: the \: procedure \\ \\ \: and \: used \: apparatuses \\ \\ \huge\mathfrak\red{Hope \: it \: helps}[/tex]

Answer:

No, it will not affect the results.

Explanation:

For elastic collisions in an isolated system, when a collision occurs, it means that the systems objects total momentum will be conserved under the condition that there will be no net external forces that act upon the objects.

What that means is that if the pucks start spinning after the collision, we are not told that there was any net external force acting on the puck and thus momentum will be conserved because momentum before collision will be equal to the momentum after the collision.

Answer: 66.67 m, 44.44 s

Explanation:

Given

Velocity of flow is [tex]u=1.5\ m/s[/tex]

Mary can row with speed [tex]v=3.6\ m/s[/tex]

Width of the river [tex]y=160\ m[/tex]

Flow will drift the Mary towards east, while Mary boat will cause it to travel in North direction

time taken to cross river

[tex]\Rightarrow t=\dfrac{160}{3.6}\\\\\Rightarrow t=\dfrac{400}{9}\ s[/tex]

Flow will drift Mary by

[tex]\Rightarrow x=ut\\\\\Rightarrow x=1.5\times \dfrac{400}{9}\\\\\Rightarrow x=66.67\ m[/tex]

Velocity w.r.t shore is

[tex]\Rightarrow v_{net}=\sqrt{3.6^2+1.5^2}\\\Rightarrow v_{net}=\sqrt{15.21}\\\Rightarrow v_{net}=3.9\ m/s[/tex]