If the PLATE SEPARATION of an isolated charged parallel-plate capacitor is doubled: A. the electric field is doubled
B. the potential difference is halved
C. the charge on each plate is halved
D. the surface charge density on each plate is doubled
E. none of the above

Answer:

a) 1/4 what it is now

Explanation:

As we know that force of gravitation between two planets at some distance "r" from each other is given as

[tex]F_g = \frac{Gm_1m_2}{r^2}[/tex]

now since we know that if the distance between Earth and Sun is changed

So the force of gravity will be given as

[tex]\frac{F_g'}{F_g} = \frac{r_1^2}{r_2^2}[/tex]

now we know that the distance between sun and earth is changed to twice the initial distance between them

so we have

[tex]r_2 = 2r_1[/tex]

so new gravitational force between sun and earth is given as

[tex]F_g' = \frac{r_1^2}{(2r_1)^2}F_g[/tex]

[tex]F_g' = \frac{1}{4}F_g[/tex]

Answer:

[tex]intensity = 12.98 W/m^2[/tex]

[tex]Energy = 3.93 J[/tex]

Explanation:

As we know that the magnitude of electric field intensity is given as

[tex]E = 98.9 V/m[/tex]

now we know that intensity of the wave is given as the product of energy density and speed of the wave

[tex]intensity = \frac{1}{2}\epsilon_0 E^2 c[/tex]

[tex]intensity = \frac{1}{2}(8.85 \times 10^{-12})(98.9)^2(3\times 10^8)[/tex]

[tex]intensity = 12.98 W/m^2[/tex]

so intensity is the energy flow per unit area per unit of time

so the energy that flows through the area of 0.0259 m^2 in 11.7 s is given as

[tex]Energy = Area \times time \times intensity[/tex]

[tex]Energy = 0.0259(11.7)(12.98)[/tex]

[tex]Energy = 3.93 J[/tex]

Answer:

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

so here we have

[tex]f = \frac{4.04}{60} =0.067 rev/s[/tex]

here we know that angular frequency is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(0.067)[/tex]

[tex]\omega = 0.42 rad/s[/tex]

now this is the angular speed of the disc and this speed will remain same for all points lying on the disc

Angular speed do not depends on the distance from the center but it will be same for all positions of the disc

Answer:

A primitive solid is a 'building block' that you can use to work with in 3D. Rather than extruding or revolving an object, AutoCAD has some basic 3D shape commands at your disposal.

Explanation:

Answer:

0.686 Ohm

Explanation:

Heat energy H = 50 cal = 50 x 4.2 J = 210 J, time t = 1 second, V = 12 V

Let R be the resistance.

Heat energy = V^2 x t / R

210 = 12 x 12 x 1 / R

210 = 144 / R

R = 144 / 210 = 0.686 Ohm

The resistance of the heater wire is approximately 0.688 ohms.

To calculate the resistance of the heater wire, we need to use the information that 50 cal of heat is generated per second when a potential difference of 12 V is applied across the leads. The first step is to convert the heat from calories to joules, which is the standard unit of energy in Physics. Since 1 calorie is equivalent to 4.184 joules, we can calculate the power (P) in joules per second (or watts) by multiplying the heat generated per second by this conversion factor:

P = 50 cal/s ×4.184 J/cal = 209.2 J/s = 209.2 W

Now, we can use the formula for electric power P = V^2/R, where P is the power, V is the potential difference, and R is the resistance. By rearranging the formula to solve for R, we get R = V^2/P.

Plugging our values into this equation gives us:

R = (12 V)^2 / (209.2 W) = 144 V^2 / 209.2 W = approx 0.688 ohms.

Answer:

The time taken for ATP to diffuse across an average cell is 0.66 seconds

Explanation:

It is given that,

Diffusion constant of ATP is, [tex]D=3\times 10^{-10}\ m^2s^{-1}[/tex]

Distance to be diffused across is, [tex]x=20\ \mu m=20\times 10^{-6}\ m[/tex]

We need to find the time taken for ATP to diffuse across an average cell. It is given by :

[tex]t=\dfrac{x^2}{2D}[/tex]

[tex]t=\dfrac{(20\times 10^{-6}\ m)^2}{2\times 3\times 10^{-10}\ m^2s^{-1}}[/tex]

t = 0.66 seconds

So, the time taken for ATP to diffuse across an average cell is 0.66 seconds. Hence, this is the required solution.

Answer:

The acceleration and time are 1.95 m/s and 12.5 s.

Explanation:

Given that,

Speed = 80 ft/s =24.384 m/s

Distance = 500 ft =152.4 m

We need to calculate the acceleration

Using third equation of motion

[tex]v^2-u^2= 2as[/tex]

[tex]a = \dfrac{v^2-u^2}{2s}[/tex]

Where, u = initial velocity

v = final velocity

a = acceleration

s = distance

Put the value in the equation

[tex]a=\dfrac{(24.384)^2-0}{2\times152.4}[/tex]

[tex]a=1.95\ m/s^2[/tex]

We need to calculate the time

Using first equation of motion

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

[tex]t =\dfrac{v-u}{a}[/tex]

[tex]t=\dfrac{24.384-0}{1.95}[/tex]

[tex]t =12.5\ s[/tex]

Hence, The acceleration and time are 1.95 m/s and 12.5 s.

Answer:

  Collision will occur.

  Speed of red train when they collide = 0 m/s.

  Speed of green train when they collide = 10 m/s.

Explanation:

Speed of red train = 72 km/h = 20 m/s

Speed of green train = 144 km/h = 40 m/s.

Deceleration of trains = 1 m/s²

For red train:-

    Equation of motion v = u + at

              u = 20 m/s

              v = 0 m/s

              a = -1 m/s²

    Substituting

             0 = 20 - 1 x t

             t = 20 s.

    Equation of motion s = ut + 0.5at²

              u = 20 m/s

              t = 20 s

              a = -1 m/s²    

    Substituting

             s = 20 x 20 - 0.5 x 1 x 20² = 200 m

   So red train travel 200 m before coming to stop.

For green train:-

    Equation of motion v = u + at

              u = 40 m/s

              v = 0 m/s

              a = -1 m/s²

    Substituting

             0 = 40 - 1 x t

             t = 40 s.

    Equation of motion s = ut + 0.5at²

              u = 40 m/s

              t = 40 s

              a = -1 m/s²    

    Substituting

             s = 40 x 40 - 0.5 x 1 x 40² = 800 m

   So green train travel 800 m before coming to stop.

 Total distance traveled = 800 + 200 = 1000 m>950 m.

  So both trains collide.

  Distance traveled by green train when red train stops(t=20s)

     Equation of motion s = ut + 0.5at²

              u = 40 m/s

              t = 20 s

              a = -1 m/s²    

    Substituting

             s = 40 x 20 - 0.5 x 1 x 20² = 600 m

    Total distance after 20 s = 600 + 200 = 800 m< 950m . So they collide after red train stops.

  Speed of red train when they collide = 0 m/s.

  Distance traveled by green train when they collide = 950 - 200 = 750 m

  Equation of motion v² = u² + 2as

              u = 40 m/s

              s= 750 m

              a = -1 m/s²    

    Substituting  

              v² = 40² - 2 x 1 x 750 = 100

               v = 10 m/s

  Speed of green train when they collide = 10 m/s.

Final answer:

The red train traveling at 72 km/h and green train at 144 km/h will collide because their combined stopping distances exceed their initial separation. The speeds at impact are not provided, but the collision is inevitable due to their insufficient stopping distance.

Explanation:

To determine if a collision occurs between the red train traveling at 72 km/h and the green train traveling at 144 km/h, we need to convert their speeds into meters per second and calculate the stopping distance for both trains based on their deceleration.

The red train is traveling at 72 km/h, which is equivalent to 20 m/s (since 72 km/h / 3.6 = 20 m/s). The green train is traveling at 144 km/h, which is equivalent to 40 m/s (since 144 km/h / 3.6 = 40 m/s).

To calculate the stopping distance, use the equation d = v2 / (2a), where d is the stopping distance, v is the initial velocity, and a is the deceleration. So, for the red train, the stopping distance is (20 m/s)2 / (2 × 1.0 m/s2) = 200 m. For the green train, the stopping distance is (40 m/s)2 / (2 × 1.0 m/s2) = 800 m.

Adding both stopping distances, we get a total of 200 m + 800 m = 1000 m. Since the trains are only 950 m apart, their combined stopping distance exceeds the separation, meaning they will collide.

Just before the collision, both trains have been decelerating for the same amount of time. Given that the total stopping time can be found from v = at where v is the final velocity and t is time, we find that their time to stop (if unobstructed) would be t = v / a. Since the red train decelerates from 20 m/s, its stopping time is 20 m/s / 1 m/s2 = 20 s. For the green train, 40 m/s / 1 m/s2 = 40 s. Since they have not yet reached 20 s before collision, we know they will still be moving upon impact.

The collision occurs before either train can come to a complete stop, and thus we would use the physics of constant deceleration to determine their speeds at the moment of impact, but as the full calculation is not provided in this answer, it would require additional work to determine exact speeds.

Answer: The freezing point of solution is -3.34°C

Explanation:

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of calcium nitrate follows:

[tex]Ca(NO_3)_2(aq.)\rightarrow Ca^{2+}(aq.)+2NO_3^-(aq.)[/tex]

The total number of ions present in the solution are 3.

[tex]Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}[/tex]

Where,

[tex]m_{solute}[/tex] = Given mass of solute [tex](Ca(NO_3)_2)[/tex] = 11.3 g

[tex]M_{solute}[/tex] = Molar mass of solute [tex](Ca(NO_3)_2)[/tex] = 164  g/mol

[tex]W_{solvent}[/tex] = Mass of solvent (water) = 115 g

Putting values in above equation, we get:

[tex]\text{Molality of }Ca(NO_3)_2=\frac{11.3\times 1000}{164\times 115}\\\\\text{Molality of }Ca(NO_3)_2=0.599m[/tex]

[tex]\Delta T=iK_fm[/tex]

where,

i = Vant hoff factor = 3

[tex]K_f[/tex] = molal freezing point depression constant = 1.86°C/m.g

m = molality of solution = 0.599 m

Putting values in above equation, we get:

[tex]\Delta T=3\times 1.86^oC/m.g\times 0.599m\\\\\Delta T=3.34^oC[/tex]

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

[tex]\Delta T=\text{freezing point of water}-\text{freezing point of solution}[/tex]

[tex]\Delta T[/tex] = 3.34 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

[tex]3.34^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-3.34^oC[/tex]

Hence, the freezing point of solution is -3.34°C

Final answer:

The freezing point of a solution made by dissolving 11.3 g of Ca(NO3)2 in 115 g of water is -3.34 °C. To find this, we calculate molality, account for the dissociation of ions and use the freezing point depression constant for water.

Explanation:

To calculate the freezing point depression of a solution of Ca(NO3)2 in water, we first determine the molality of the solution. With the provided mass of Ca(NO3)2 (11.3 g) and its formula weight (164 g/mol), we find there are 0.0689 moles of Ca(NO3)2. Since we only have 115 g of water, to convert to kilograms, we have 0.115 kg. The molality (m) is then 0.0689 moles / 0.115 kg = 0.599 m. Since Ca(NO3)2 dissociates into three ions (Ca2+, 2NO3-), the van't Hoff factor (i) is 3.

The depression of the freezing point is determined using the formula ΔTf = i * Kf * m, where Kf is the molal freezing point depression constant for water (1.86 °C/m). So the depression is ΔTf = 3 * 1.86 °C/m * 0.599 m = 3.34 °C.

The freezing point of the solution is then 0 °C - 3.34 °C = -3.34 °C, which is the answer.

Answer:

a)frequency of the oscillation = 1/(2*pi)*square root (k/x) =1/(2*pi)*square root (450/(15*10^-2))=8.72 cycle/second

b)spring potential energy = 0.5*k*(x)^2

=0.5*450*(15*10^-2)^2 =5.0625 joule

maximum kinetic energy =spring potential energy

c)

maximum kinetic energy=5.0625

kinetic energy=0.5*m*v^2

v=square toot ((5.0625/(0.5*2.5)) =2 m/s

Answer:

a)  Initial angular speed = 30 rad/s

b) Final angular speed = 70 rad/s        

Explanation:

a) We have equation of motion s = ut + 0.5at²

    Here s = 400 radians

              t = 8 s

              a = 5 rad/s²

    Substituting

             400 = u x 8 + 0.5 x 5 x 8²

              u = 30 rad/s

   Initial angular speed = 30 rad/s

b) We have equation of motion v = u + at

     Here u = 30 rad/s

               t = 8 s

              a = 5 rad/s²  

    Substituting

             v = 30 + 5 x 8 = 70 rad/s    

   Final angular speed = 70 rad/s        

Answer:

22.89 N

Explanation:

F = 37 N

Let the acceleration in the system is a and f be the force between the tewo blocks.

Apply the Newton's second law

By the free body diagrams

F - f = 7.4 x a .... (1)

f = 12 x a ...... (2)

Adding both of them

37 = 19.4 x a

a = 1.9 m/s^2

Put in equation (2)

f = 12 x 1.9 = 22.89 N

The work done on the particle as it moves from x = b to x = a is k(1/a - 1/b).

To calculate the work done on a particle by the attractive force, we need to find the integral of the force function over the distance the particle moves. In this case, the force function is given by F(x) = k/x^2, where k is the constant. The work done when the particle moves from x = b to x = a is given by:

Work = ∫(k/x^2) dx from x = b to x = a

To evaluate this integral, we need to use the power rule of integration. The result will be:

Work = k(1/a - 1/b)

Therefore, the work done on the particle as it moves from x = b to x = a is k(1/a - 1/b).

Answer:

[tex]T_2[/tex] = [tex]90.667K[/tex]

Explanation:

Given:

For the first solenoid

Number of turns, n₁ = 1200 turns/m

Current, I₁ = 2.5 A

Paramagnetic material temperature, T₁ = 320 K

Now for the second solenoid

Number of turns, n₂ = 1000 turns/m

Current, I₂ = 0.85 A

Paramagnetic material temperature = T₂

The magnetic flux (B) is given as

[tex]B=\frac{c\mu_onI}{T}[/tex]

where,

c = curie's constant

μ₀ = arithmetic constant

also it is given that the magnetization in both the cases are same

therefore the magnetic flux will also be equal

thus,

[tex]\frac{c\mu_on_1I_1}{T_1}[/tex] = [tex]\frac{c\mu_on_2I_2}{T_2}[/tex]

or

[tex]\frac{n_1I_1}{T_1}[/tex] = [tex]\frac{n_2I_2}{T_2}[/tex]

or

[tex]\frac{1200\times 2.5}{320}[/tex] = [tex]\frac{1000\times 0.85}{T_2}[/tex]

or

[tex]9.375[/tex] = [tex]\frac{850}{T_2}[/tex]

or

[tex]T_2[/tex] = [tex]\frac{850}{9.375}[/tex]

or

[tex]T_2[/tex] = [tex]90.667K[/tex]

To calculate the temperature difference between the inner and outer surfaces of the glass bulb in a 100 W bulb that generates 95W of heat, we use a formula from the principles of heat conduction where we input parameters including the heat generated by the bulb, the thermal conductivity of the glass, the surface area of the glass and the thickness of the glass.

In order to calculate the temperature difference between the inner and outer surfaces of the glass bulb, we need to use the formula for heat conduction, which is given by the formula Q = (k*A*ΔT)/d, where Q is the heat generated by the bulb, k is the thermal conductivity of the glass, A is the surface area of the glass, ΔT is the temperature difference, and d is the thickness of the glass. In this case, we know that the light bulb generates 95W of heat, the radius of the bulb is 3.0 cm, the thickness of the glass is 0.50 mm. Assuming the thermal conductivity of the glass to be 0.8 W/m.K, we can substitute these values into the formula to calculate the temperature difference ΔT = Qd / (k*A). Note that here, the surface area of the glass, A = 4πr².

As you can see, the calculation requires a clear grasp of the concepts of heat conduction, thermal conductivity, and physical constants of materials. Understanding how these factors interact is key to solving problems about heat transfer in Physics.

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Answer:

1.53 x 10^-5 m^2

Explanation:

use the Stefan's law

Energy per unit time = σ x A x T^4

σ = 5.67 x 10 -8 W/m^2 K^4

96 = 5.67 x 10^-8 x A x (3242)^4

A = 1.53 x 10^-5 m^2

The required tension to produce the transverse waves with a speed of 50.0 m/s on a 0.06 kg, 5.00 m long string is approximately 30.0 N. If the string has a tension of 8.00 N, the wave speed will be roughly 26.0 m/s.

This question is about the physics of waves on strings and involves the concepts of wave speed, tension, and linear mass density. Let's handle this in two parts.

(a) To determine the required tension to produce the transverse waves with a speed of 50.0 m/s, we first need to calculate the string's linear mass density, which is the mass of the string divided by its length. So, the linear mass density (μ) would be 0.06 kg / 5.00 m = 0.012 kg/m. Now there is a formula that determines the wave speed (v) on a string: v = sqrt(FT/μ) where FT is the tension in the string. Rearranging to solve for FT, we get FT = μv^2. Substituting the values we have,

(b) If the tension is 8.00 N, we can use the same formula to calculate the wave speed. This time, rearranging for v, we get v = sqrt(FT/μ). Substituting the values we have, v = sqrt((8.00 N)/(0.012 kg/m)) which gives us approximately 26.0 m/s. Therefore, the wave speed with a tension of 8.00 N is roughly 26.0 m/s.

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The linear mass density of the string is calculated first, which is 0.012 kg/m. Using this in the wave speed equation, the required tension to produce a wave speed of 50.0 m/s is 30.0 N. If the tension is 8.00 N, then the resulting wave speed would be around 25.82 m/s.

The student is attempting to produce transverse waves on a string. To determine the required tension to achieve a wave speed of 50.0 m/s, we should first calculate the linear mass density of the string, using the formula μ = m/L, where m is the mass of the string and L is its length. So, for a string 5.0 m long and a mass of 0.060 kg, μ = 0.060 kg / 5.00 m = 0.012 kg/m.

For (a), the formula for wave speed is v = √(FT/μ), where FT is the tension in the string and μ is the linear mass density. We need to rearrange it to calculate the required tension: FT = μ * v^2 = 0.012 kg/m * (50.0 m/s)^2 = 30.0 N.

For (b), using the same formula and given new tension value, the wave speed is v = √(FT/μ) = √(8.00 N / 0.012 kg/m) = approx. 25.82 m/s.

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Answer:

The formula is Work = Force x distance moved

so that would be 0.600 x 5 = 3 Joules which is 0.003 kilojoules

The work done in by the attendant on a can of soup is 3 joules or 0.00072 Kilocalories

Work in physics is defined as the dot product of force and the displacement produced by it.

Given is an attendant who pushed 0.600 m horizontally with a force of 5.00 N.

From the definition of work done, we can write -

W = F.d

W = Fd cosФ

Then angle between force and displacement is 0°. Therefore, cosФ is equal to 1.

W = Fd

W = 5 x 0.6

W = 3 joules = 3/4184 = 0.00072 Kilocalories

Therefore, the work done in by the attendant on a can of soup is 3 joules or 0.00072 Kilocalories

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Answer:

Force constant, K = 101.49 N/m

Explanation:

It is given that,

Mass, m = 30.6 kg

Time period of oscillation, T = 3.45 s

We need to find the force constant of the spring. The time period of the spring is given by :

[tex]T=2\pi\sqrt{\dfrac{m}{K}}[/tex]

[tex]K=\dfrac{4\pi^2m}{T^2}[/tex]

[tex]K=\dfrac{4\pi^2\times 30.6\ kg}{(3.45\ s)^2}[/tex]

K = 101.49 N/m

So, the force constant of the spring is 101.49 N/m. Hence, this is the required solution.

Answer:

a)

85.05 N/m

b)

179.81 rad/s

Explanation:

a)

k = spring constant of the spring

m = mass of the block = 0.473 kg

x = stretch caused in the spring = 0.109 m

h = height dropped by the block = 0.109 m

Using conservation of energy

Spring potential energy gained by the spring = Potential energy lost by the block

(0.5) k x² = mgh

(0.5) k x² = mgx

(0.5) k x = mg

(0.5) k (0.109) = (0.473) (9.8)

k = 85.05 N/m

b)

angular frequency is given as

[tex]w = \sqrt{\frac{k}{m}}[/tex]

[tex]w = \sqrt{\frac{85.05}{0.473}}[/tex]

[tex]w [/tex] = 179.81 rad/s

The spring constant of the spring is 42.54 N/m, and the angular frequency of the block's vibrations is 4.88 rad/s.

To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is proportional to its displacement. In this case, the weight of the block is equal to the force provided by the spring at the equilibrium position.

Using the equation F = kx, where F is the force, k is the spring constant, and x is the displacement, we can solve for k. Since the block momentarily comes to rest after dropping 0.109 m, we can set the force provided by the spring equal to the weight of the block and solve for k.

Given:

Mass of the block (m) = 0.473 kg

Displacement of the block (x) = 0.109 m

Using the equation F = kx, we can rewrite it as k = F/x. The weight of the block is equal to its mass multiplied by the acceleration due to gravity (9.8 m/s^2), so the force provided by the spring is 0.473 kg * 9.8 m/s^2 = 4.6354 N. Substituting these values into the equation, we find the spring constant (k) to be:

k = 4.6354 N / 0.109 m = 42.54 N/m

To find the angular frequency of the block's vibrations, we can use the equation:

ω = sqrt(k/m)

Substituting the values of k and the mass of the block (m) = 0.473 kg into the equation, we can calculate the angular frequency (ω):

ω = sqrt(42.54 N/m / 0.473 kg) = 4.88 rad/s

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Answer:

0.091 sec

Explanation:

f = frequency of oscillation of the mass attached to the spring = 11 Hz

T = Time period of oscillation of the mass attached to the spring = ?

Time period and frequency of oscillation of the mass attached to the end of spring are related as

[tex]T = \frac{1}{f}[/tex]

Inserting the values

[tex]T = \frac{1}{11}[/tex]

T = 0.091 sec

Answer:

option (A)

Explanation:

Angle between the plane of coil and the magnetic field = 60 degree

Angle between the normal of coil and the magnetic field = 90 - 60 = 30 degree

N = 7, B = 0.4 T, T = 0.06 s , w = 2 pi / T, A = 900 cm^2 = 0.09 m^2

peak emf, e0 = N x B x A x (2 x 3.14 / T) x Sin 30

e0 = 7 x 0.4 x 0.09 x 6.28 x 0.5  / 0.06 = 13.18 Volt = 13 volt

Answer:

NO

Explanation:

The answer is NO because when the particle reaches the center of square there is a net force acting on the particle dew to various other charges and this net force gives acceleration to the particle. Moreover, For particle or object to be in equilibrium the net force acting on it should be zero and hence no acceleration. Although velocity can be zero or non zero at equilibrium state.

Answer:

The context is missing here, but ill try to explain a general case.

Something is in equilibrium if it is in a valley of the potential energy, this is because things in life try to be in the minimal energy state possible. Think for example in a thing that is away from the ground, the object will try to reach the ground, in this way minimizing the potential energy.

Now, if once the particle reaches the center of the square it remains at rest, it means that the total forces acting on the particle are zero and this is why the particle stays at rest, this would mean that the particle is in equilibrium, and if someone moves it a little bit of the center, some of the forces will increase and others will decrease, and then the equilibrium will be broken and the particle will move again.

In another case, if the particle is momentarily at rest (just for a few seconds) it may be because the forces acting on it are affecting the particle in such way that is moving is fully stopped in one direction, and the new forces are accelerating the particle in the opposite direction (in the same way that if you throw something upside when it reaches the maximum height it has for a brief moment a velocity equal to zero)  

Answer:

Heat energy needed = 3036.17 kJ

Explanation:

We have

     heat of fusion of water = 334 J/g

     heat of vaporization of water = 2257 J/g

     specific heat of ice = 2.09 J/g·°C

     specific heat of water = 4.18 J/g·°C

     specific heat of steam = 2.09 J/g·°C

Here wee need to convert 1 kg ice from -13°C to vapor at 100°C

First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.

Mass of water = 1000 g

Heat energy required to change ice temperature from -13°C to 0°C

          H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ

Heat energy required to change ice from 0°C to water at 0°C

          H₂ = mL = 1000 x 334 = 334 kJ

Heat energy required to change water temperature from 0°C to 100°C  

          H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ    

Heat energy required to change water from 100°C to steam at 100°C  

          H₄ = mL = 1000 x 2257 = 2257 kJ    

Total heat energy required

          H = H₁ +  H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ

Heat energy needed = 3036.17 kJ

Answer:

The acceleration and charge are 17.122 m/s² and [tex]8.6\times10^{-5}\ C[/tex]

Explanation:

Given that,

Mass of object = 0.8 g

Electric field = 159 N/C

Distance = 72 m

Time = 2.9 s

We know that,

The electric force is

[tex]F = Eq[/tex]....(I)

The newton's second law

[tex]F=ma[/tex]

Put the value of F in the equation (I)

[tex]ma=Eq[/tex]...(II)

We calculate the acceleration

Using equation of motion

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

[tex]a =\dfrac{2s}{t^2}[/tex]

[tex]a=\dfrac{2\times72}{(2.9)^2}[/tex]

[tex]a=17.122\ m/s^2[/tex]

From equation (II)

[tex]q=\dfrac{ma}{E}[/tex]

[tex]q=\dfrac{0.8\times10^{-3}\times17.122}{159}[/tex]

[tex]q=0.000086148427673\ C[/tex]

[tex]q=8.6\times10^{-5}\ C[/tex]

Hence, The acceleration and charge are 17.122 m/s² and [tex]8.6\times10^{-5}\ C[/tex]

Answer:

147,099,713.4 km

152,096,286.6 km

Explanation:

a = 149598000 km

e = 0.0167

The formula to find the perihelion

Rp = a ( 1 - e) = 149598000 ( 1 - 0.0167) = 147,099,713.4 km

The formula for aphelion

Ra = a ( 1 + e) = 149598000 ( 1 + 0.0167) = 152,096,286.6 km

To find the minimum and maximum distances from Earth to the Sun (perihelion and aphelion), we calculate them using Earth's semi-major axis of 149,598,000 kilometers and the eccentricity of 0.0167. The perihelion is 147,099,014 kilometers, and the aphelion is 152,096,986 kilometers.

The student's question revolves around finding the minimum (perihelion) and maximum (aphelion) distances from the Earth to the Sun, given the length of half of the major axis — also known as the semi-major axis — and the eccentricity of Earth's orbit. The semi-major axis (a) is 149,598,000 kilometers and the eccentricity (e) is 0.0167. The distance from the center of the ellipse, where Earth's orbit is, to the focus (c) is equal to the product of the semi-major axis and the eccentricity (c = ae).

The perihelion distance is the semi-major axis minus the distance c, resulting from the Earth being at the closest point to the Sun in its orbit. Conversely, the aphelion distance is the semi-major axis plus the distance c, when Earth is farthest from the Sun. Therefore, the perihelion (rp) can be calculated as rp = a - c, and the aphelion (ra) as ra = a + c.

Using the formula c = ae, we find that c is approximately 2,498,986 kilometers (149,598,000 km * 0.0167). Thus:

Answer:

45.3°C

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = mC (T − T₀)

Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:

305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)

T = 45.3°C

The final temperature of the metal, after adding 305 J of heat to 94.0 g of metal initially at 20.0 °C, is 45.34 °C. The calculation uses the specific heat capacity formula and involves solving for the change in temperature.

To find the final temperature, we can use the formula:

where:

First, solve for ΔT:

Next, find the final temperature:

Therefore, the final temperature of the metal is 45.34 °C.

Answer:

Gravitational potential energy, PE = 2.8 J

Explanation:

It is given that,

Mass of the bluebird, m = 34 g = 0.034 kg

It flies from the ground to the top of an 8.5-m tree, h = 8.5 m

We need to find the change in the bluebird's gravitational potential energy as it flies to the top of the tree. It can be calculated as :

[tex]PE=m\times g\times h[/tex]

[tex]PE=0.034\ kg\times 9.8\ m/s^2\times 8.5\ m[/tex]

PE = 2.83 J

or

PE = 2.8 J

So, the gravitational potential energy as it flies to the top of the tree is 2.8 J. Hence, this is the required solution.

Answer:

reeeeeeeeeeeeeeeeeeeeeeee

Answer:

(i) W = 8.918 N

(ii) [tex]V = 9.1 \times 10^{-4} m^3[/tex]

(iii) d = 9.1 cm

Explanation:

Part a)

As we know that weight of cube is given as

[tex]W = mg[/tex]

[tex]W = \rho V g[/tex]

here we know that

[tex]\rho = 0.91 g/cm^3[/tex]

[tex]Volume = L^3[/tex]

[tex]Volume = 10^3 = 1000 cm^3[/tex]

now the mass of the ice cube is given as

[tex]m = 0.91 \times 1000 = 910 g[/tex]

now weight is given as

[tex]W = 0.910 \times 9.8 = 8.918 N[/tex]

Part b)

Weight of the liquid displaced must be equal to weight of the ice cube

Because as we know that force of buoyancy = weight of the of the liquid displaced

[tex]W_{displaced} = 8.918 N[/tex]

So here volume displaced is given as

[tex]\rho_{water}Vg = 8.918[/tex]

[tex]1000(V)9.8 = 8.918[/tex]

[tex]V = 9.1 \times 10^{-4} m^3[/tex]

Part c)

Let the cube is submerged by distance "d" inside water

So here displaced water weight is given as

[tex]W = \rho_{water} (L^2 d) g[/tex]

[tex]8.918 = 1000(0.10^2 \times d) 9.8[/tex]

[tex]d = 0.091 m[/tex]

so it is submerged by d = 9.1 cm inside water

The ice cube weighs approximately 8.9 N. It would displace a volume of 910 cm³ of water, which means that 91% of the ice cube would be under the water surface.

The first part of the question asks what is the weight of the ice cube. To find the weight, we need to first calculate the volume of the cube, which is 10 cm × 10 cm × 10 cm = 1000 cm³.

Given the density of ice is 0.91 g/cm³, we can multiply this by the volume to find the mass of the cube: 0.91 g/cm³ × 1000 cm³ = 910 g. The weight is then calculated by multiplying the mass by gravity, roughly 9.8 m/s², which gives us 8.9 N.

For the second part, about what volume of liquid water is displaced, the amount of water displaced by the ice cube equals the volume of the ice cube that's beneath the water surface.

Because the cube's density is less than that of water (0.91 g/cm³ vs 1.00 g/cm³), it will displace an amount of water with equal mass. Given that 1 cm³ of water has a mass of 1 g, the ice cube will displace 910 cm³ of water.

In the last part, we need to find how much of the cube is under the water surface. This is the ratio of the mass of the water displaced to the total mass of the ice cube, which is 910 g / 1000 g = 0.91 or 91% of the cube.

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