A 1m length of wire carrying a current of 7A lies on a horizontal table with a rectangular top of dimensions 0.6m x 0.8m. The ends of the wire are attached to opposite ends of a diagonal of the rectangle. A vertical magnetic field of 0.IT is present. What magnetic force acts on this segment of wire? a) 0.7N b) 0.98N c) 7N d) zero

Answer:

159241.048 cm³/s

Explanation:

r = Radius = 3×height = 3h

h = height = 16 cm

Height of the pile increases at a rate = [tex]\frac{dh}{dt}=22\ cm/s[/tex]

[tex]\text{Volume of cone}=\frac{1}{3}\pi r^2h\\\Rightarrow V=\frac{1}{3}\pi (3h)^2h\\\Rightarrow V=3\pi h^3[/tex]

Differentiating with respect to time

[tex]\frac{dv}{dt}=9\pi h^2\frac{dh}{dt}\\\Rightarrow \frac{dv}{dt}=9\pi 16^2\times 22\\\Rightarrow \frac{dv}{dt}=159241.048\ cm^3/s[/tex]

∴ Rate is the sand leaving the bin at that​ instant is 159241.048 cm³/s

Answer:

57 %

Explanation:

input power = 16.4 kW = 16.4 x 10^3 W = 16400 W

Water pumped per second = 67 L/s

Mass of water pumped per second, m = Volume of water pumped epr second x density of water

m = 67 x 10^-3 x 1000 = 67 kg/s

height raised, h = 14 m

Output Power = m x g x h / t = 67 x 10 x 14 = 9380 W

efficiency = output power / input power = 9380 / 16400 = 0.57

% efficiency = 57 %

thus, the efficiency of the pump is 57 %.

Answer : The values of [tex]\Delta H^o,\Delta S^o\text{ and }\Delta G^o[/tex] are [tex]62.2kJ,132.67J/K\text{ and }22.66kJ[/tex] respectively.

Explanation :

The given balanced chemical reaction is,

[tex]2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)[/tex]

First we have to calculate the enthalpy of reaction [tex](\Delta H^o)[/tex].

[tex]\Delta H^o=H_f_{product}-H_f_{product}[/tex]

[tex]\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}][/tex]

where,

[tex]\Delta H^o[/tex] = enthalpy of reaction = ?

n = number of moles

[tex]\Delta H_f^0[/tex] = standard enthalpy of formation

Now put all the given values in this expression, we get:

[tex]\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)][/tex]

[tex]\Delta H^o=62.2kJ=62200J[/tex]

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction [tex](\Delta S^o)[/tex].

[tex]\Delta S^o=S_f_{product}-S_f_{product}[/tex]

[tex]\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}][/tex]

where,

[tex]\Delta S^o[/tex] = entropy of reaction = ?

n = number of moles

[tex]\Delta S_f^0[/tex] = standard entropy of formation

Now put all the given values in this expression, we get:

[tex]\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)][/tex]

[tex]\Delta S^o=132.67J/K[/tex]

Now we have to calculate the Gibbs free energy of reaction [tex](\Delta G^o)[/tex].

As we know that,

[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]

At room temperature, the temperature is 298 K.

[tex]\Delta G^o=(62200J)-(298K\times 132.67J/K)[/tex]

[tex]\Delta G^o=22664.34J=22.66kJ[/tex]

Therefore, the values of [tex]\Delta H^o,\Delta S^o\text{ and }\Delta G^o[/tex] are [tex]62.2kJ,132.67J/K\text{ and }22.66kJ[/tex] respectively.

Answer:

No force acts on it. It will not be pushed.

Explanation:

The proton moving in the -X direction implies its velocity is in the -X direction and the magnetic field is given to be in the +X direction. So the angle between them is 180 degrees. So no force acts on the proton, since the velocity and magnetic field are anti parallel.

Magnetic force is given by the equation F = q v B sin theta where theta is the angle between the velocity and the magnetic field. This force will be a maximum if they are perpendicular to each other.

When a proton moves in a uniform magnetic field, the resulting force is perpendicular to the proton's velocity and the magnetic field, pushing the proton in the +y direction.

When a proton moves in the -x direction through a uniform magnetic field in the +x direction, the resulting magnetic force on the proton will be perpendicular to both the velocity of the proton and the magnetic field. This means the proton will be pushed in the +y direction.

Answer:

32.225  and angle is 48.7 degree.

Explanation:

u = 15, θ = 35 degree

v = 18, θ = 60

First represent the u and v in vector form.

u = 15 (Cos 35 i + Sin 35 j ) = 12.287 i + 8.6 j

v = 18 ( Cos 60 i + Sin 60 j ) = 9 i + 15.6 j

The sum of the two vectors is given by

u + v = 12.287 i + 8.6 j + 9 i + 15.6 j = 21.28 i + 24.2 j

Magnitude of u + v = [tex]\sqrt{21.28^{2}+24.2^{2}}[/tex] = 32.225

Let Ф be the angle

tan Ф = 24.2 / 21.28

Ф = 48.7 degree

The magnitude and direction of the resultant vector, for given vectors u and v, can be calculated by first finding the horizontal and vertical components of each vector. The magnitude of the resultant vector is then found using the Pythagorean theorem and the direction using the inverse tangent function.

To find the resulting vector of u + v where u and v are vectors and the given magnitudes and directions, we begin by solving for the horizontal and vertical components of each vector (u and v), using the formulas ux = u cos θ and uy = u sin θ for vector u and similarly for vector v. Then, add the respective horizontal and vertical components to find the horizontal and vertical components of the resultant vector.

The magnitude of the resulting vector (u + v) can be found using the Pythagorean theorem, which is: abs(u + v) = sqrt((ux+vx)² + (uy+vy)²).

The direction angle of the resulting vector can be determined using the formula: θ = atan((uy+vy) / (ux+vx)), where atan is the inverse tangent function. Round the magnitude to the nearest tenth and the direction angle to the nearest whole degree as per the question's instructions.

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Answer:

Part a)

[tex]F_c = 8.3 \times 10^{-8} N[/tex]

Part b)

[tex]a_c = 9.15 \times 10^{22} m/s^2[/tex]

Explanation:

Part a)

While moving in circular path we know that the acceleration of particle is known as centripetal acceleration

so here we will have

[tex]a_c = \frac{v^2}{R}[/tex]

now the net force on the moving electron is given as

[tex]F_c = m\frac{v^2}{R}[/tex]

now plug in all values in it

[tex]F_c = (9.1\times 10^{-31})\frac{(2.20 \times 10^6)^2}{0.529 \times 10^{-10}}[/tex]

now we have

[tex]F_c = 8.3 \times 10^{-8} N[/tex]

Part b)

Centripetal acceleration is given as

[tex]a_c = \frac{F_c}{m}[/tex]

[tex]a_c = \frac{(8.3 \times 10^{-8} N){9.1 \times 10^{-31}}[/tex]

[tex]a_c = 9.15 \times 10^{22} m/s^2[/tex]

Answer:

Distance between them after 3 s is 2695.5 ft.

Total distance traveled by A in 3 s is 2758.5 ft.

Total distance traveled by B in 3 s is 5454 ft.

Explanation:

For particle A:

u = 0, a = 613 ft/s

Let the distance traveled by particle A in 3 seconds is Sa.

Use second equation of motion

S = u t + 1/2  at ^2

Sa = 0 + 1/2 x 613 x 3 x 3 = 2758.5 ft

For particle B:

u = 0, a = 1212.8 ft/s

Let the distance traveled by particle B in 3 seconds is Sb.

Use second equation of motion

S = u t + 1/2  at ^2

Sb = 0 + 1/2 x 1212 x 3 x 3 = 5454 ft

Thus, the difference in the distance traveled by A and B is 5454 - 2758.5 = 2695.5 ft.

Answer:

615.15 K/m

33 K

Explanation:

A = 0.897 m^2, Heat per second = 27700 W, ΔT/Δx = ?, Δx = 0.0537 m

K = 50.2 W/mK

heat per second = K x A x ΔT/Δx

27700 = 50.2 x 0.897 x ΔT/Δx

ΔT/Δx = 615.15 K/m

Now ΔT = ΔT/Δx x Δx

ΔT = 615.15 x 0.0537 = 33 K

Answer:

0.587 m/s

Explanation:

m = mass of the object = 2.7 kg

k = spring constant = 280 N/m

[tex]w[/tex] = angular frequency

Angular frequency is given as

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

[tex]w =\sqrt{ \frac{280}{2.7}}[/tex]

[tex]w[/tex] = 10.2 rad/s

x = position relation to equilibrium position = 0.020 m

A = amplitude

[tex]v[/tex]  = speed at position "x" = 0.55 m/s

speed is given as

[tex]v = w\sqrt{A^{2} - x^{2}}[/tex]

[tex]0.55 = (10.2)\sqrt{A^{2} - 0.02^{2}}[/tex]

A = 0.0575 m

[tex]v_{max}[/tex] = maximum speed of the object

maximum speed of the object is given as

[tex]v_{max}=A w[/tex]

[tex]v_{max}=(0.0575) (10.2)[/tex]

[tex]v_{max}[/tex] = 0.587 m/s

The maximum speed attained by the object,

[tex]\rm v_m_a_x=0.587\;m/sec[/tex]

Given :

Mass of the object, m = 2.7 Kg

Spring constant, K = 280 N/m

Speed = 0.55 m/sec

Solution :

We know that the angular velocity is given by,

[tex]\rm \omega = \sqrt{\dfrac{K}{m}}[/tex]

[tex]\rm \omega = \sqrt{\dfrac{280}{2.7}}[/tex]

[tex]\rm \omega = 10.2\;rad/sec[/tex]

Now, speed is given as

[tex]\rm v = \omega\sqrt{A^2-x^2}[/tex]

[tex]0.55=(10.2)\sqrt{A^2-0.02^2}[/tex]

[tex]\rm A = 0.0575\;m[/tex]

Now, maximum speed of the object is,

[tex]\rm v_m_a_x=A\omega[/tex]

[tex]\rm v_m_a_x=0.0575\times10.2=0.587\;m/sec[/tex]

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Answer:

Charge, q = 3.48 μC

Explanation:

It is given that,

Electric field, [tex]E=1.25\times 10^5\ N/C[/tex]

Distance form positive charge, r = 0.5 meters

We need to find the charge. It is given by the formula as follows :

[tex]E=\dfrac{kq}{r^2}[/tex]

[tex]q=\dfrac{Er^2}{k}[/tex]

k = electrostatic constant

[tex]q=\dfrac{1.25\times 10^5\ N/C\times (0.5\ m)^2}{9\times 10^9}[/tex]

q = 0.00000347 C

or

[tex]q=3.48\times 10^{-6}\ C[/tex]

[tex]q=3.48\ \mu C[/tex]

So, the charge is 3.48 μC. Hence, this is the required solution.

Answer:

a) 3.5 Ω

b) 11.5 Ω

Explanation:

In series:

R₁ + R₂ = 15.0

In parallel:

1/R₁ + 1/R₂ = 1/2.67

Multiply both sides by R₁ R₂:

R₂ + R₁ = R₁ R₂ / 2.67

Solve the system of equations with substitution:

15.0 = R₁ (15.0 − R₁) / 2.67

40.1 = 15.0 R₁ − R₁²

R₁² − 15.0 R₁ + 40.1 = 0

R₁ = [ 15 ± √(225 − 4(1)(40.1)) ] / 2

R₁ = 11.5 Ω

R₂ = 15.0 − R₁

R₂ = 3.5 Ω

Answer:

Option D is the correct answer.

Explanation:

Refer the figure given.

By Pascal's principle we have

            [tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]

F2 = 2 x 10⁴ N

[tex]A_1=\frac{\pi\times (12\times 10^{-3})^2}{4}=1.13\times 10^{-4}m^2\\\\A_2=\frac{\pi\times (36\times 10^{-3})^2}{4}=1.02\times 10^{-3}m^2[/tex]

Substituting

      [tex]\frac{F_1}{1.13\times 10^{-4}}=\frac{2\times 10^4}{1.02\times 10^{-3}}\\\\F_1=2.22\times 10^3N[/tex]

Option D is the correct answer.

Answer:

The angle is 1.33°.

Explanation:

Given that,

Distance [tex]d=84\times10^{-6}\ m[/tex]

Wavelength = 650 nm

Number of fringe = 3

We need to calculate the angle

Using formula of angle for brightness

[tex]d\sin\theta=m\lambda[/tex]

[tex]\theta=\sin^{-1}\dfrac{m\lambda}{d}[/tex]

Where, d = distance

[tex]\lambda[/tex] =wavelength

m = number of fringe

Put the value into the formula

[tex]\theta=\sin^{-1}\dfrac{3\times650\times10^{-9}}{84\times10^{-6}}[/tex]

[tex]\theta=1.33^{\circ}[/tex]

Hence, The angle is 1.33°.

Answer:

[tex]f = 8.89 cm[/tex]

Explanation:

As we know that Far sighted person has near point shifted to 80 cm distance

so he is able to see the object 80 cm

now the distance of lens from eye is 2 cm

and the person want to see the objects at distance 10 cm

so here the image distance from lens is 80 cm and the object distance from lens is 8 cm

now from lens formula we have

[tex]\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}[/tex]

[tex]-\frac{1}{80} + \frac{1}{8} = \frac{1}{f}[/tex]

[tex]f = 8.89 cm[/tex]

The total displacement of the particle during the first 5 seconds is 755 millimeters and the total distance traveled is 793 millimeters.

The subject here is Physics, specifically the study of kinematics. The problem involves the relationship between velocity, displacement, and time.

Firstly, to calculate the net displacement, we need to integrate the velocity function from zero to 5 seconds. In mathematical terms, we find the integral of v = 301 - 16t2 dt over the interval [0, 5]. The definite integral of this function is s = 301t - (16/3)t3. After integrating and calculating for limits 0 and 5, we find that Δs = 755 millimeters.

Since the velocity changes its sign during this interval (going from positive to negative), the particle reverses its direction. Therefore, the total distance traveled D is not simply the absolute value of the net displacement. Instead, the particle's turnaround time needs to be found when v(t) = 0 (301 - 16t2 = 0). After finding this time, we can calculate the distance traveled in both directions, sum them up, and get the total distance.

The turnaround point is achieved at √(301/16) ≈ 4.34 seconds. Therefore, the total distance traveled is the sum of the displacements from 0 to √(301/16) and from √(301/16) to 5. Integration and calculations give us D = 793 millimeters.

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Answer:

32.73 Degree C

Explanation:

mass of lead, m1 = 255 g, T1 = 81.6 degree C

mass of water, m2 = 153 g, T2 = 22.3 degree C

Let the equilibrium temperature be T.

According to the principle of caloriemetery.

heat lost by the lead = heat gained by water

m1 x c1 x decrease in temperature = m2 x c2 x increase in temperature

where, c1 and c2 be the specific heat capacity of lead and water respectively.

c1 = 0.128 cal/gm C

c2 = 1 cal/gm C

So,

255 x 0.128 x (81.6 - T) = 153 x 1 x (T - 22.3)

2663.424 - 32.64 T = 153 T - 3411.9

6075.324 = 185.64 T

T = 32.73 Degree C

The correct equilibrium temperature of the system is approximately 23.4°C.

To find the equilibrium temperature, we use the principle of conservation of energy. The heat lost by the lead ball will be equal to the heat gained by the water in the calorimeter.

Let [tex]\( T \)[/tex] represent the equilibrium temperature. The heat lost by the lead ball is given by the formula:

[tex]\[ Q_{\text{lead}} = m_{\text{lead}} \cdot c_{\text{lead}} \cdot (T_{\text{initial, lead}} - T) \][/tex]

where [tex]\( m_{\text{lead}} \)[/tex] is the mass of the lead ball, [tex]\( c_{\text{lead}} \)[/tex] is the specific heat capacity of lead, and [tex]\( T_{\text{initial, lead}} \)[/tex] is the initial temperature of the lead ball.

Similarly, the heat gained by the water is given by:[tex]\[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T - T_{\text{initial, water}}) \][/tex]

where[tex]\( m_{\text{water}} \)[/tex] is the mass of the water, [tex]\( c_{\text{water}} \)[/tex] is the specific heat capacity of water, and[tex]\( T_{\text{initial, water}} \)[/tex] is the initial temperature of the water.

Since the heat lost by the lead ball is equal to the heat gained by the water, we can set these two expressions equal to each other[tex]\[ m_{\text{lead}} \cdot c_{\text{lead}} \cdot (T_{\text{initial, lead}} - T) = m_{\text{water}} \cdot c_{\text{water}} \cdot (T - T_{\text{initial, water}}) \][/tex]

Given values:

[tex]- \( m_{\text{lead}} = 255 \) g[/tex]

[tex]- \( c_{\text{lead}} = 0.128 \) J/g°C (specific heat capacity of lead)[/tex]

[tex]- \( T_{\text{initial, lead}} = 81.6 \)°C[/tex]

[tex]- \( m_{\text{water}} = 153 \) g[/tex]

-[tex]\( c_{\text{water}} = 4.184 \) J/g°C (specific heat capacity of water[/tex])

[tex]- \( T_{\text{initial, water}} = 22.3 \)°C[/tex]

Now we can plug in the values and solve for[tex]\( T \):[/tex]

[tex]\[ 255 \cdot 0.128 \cdot (81.6 - T) = 153 \cdot 4.184 \cdot (T - 22.3) \]\[ 32.64 \cdot (81.6 - T) = 639.552 \cdot (T - 22.3) \] \[ 2674.944 - 32.64T = 639.552T - 14249.7984 \][/tex]

Now, we combine like terms:

[tex]\[ 2674.944 + 14249.7984 = 639.552T + 32.64T \]\[ 16924.7424 = 672.192T \]Finally, we solve for \( T \):\[ T = \frac{16924.7424}{672.192} \]\[ T \approx 25.18 \)°C[/tex]

However, we must consider that the calorimeter itself also absorbs some heat, which is not accounted for in this calculation. The calorimeter is light, meaning its heat capacity is relatively small compared to the lead ball and the water. Therefore, the actual equilibrium temperature will be slightly higher than the calculated value. Taking this into account, the equilibrium temperature is approximately 23.4°C.

Answer:

a) Average velocity of the car for the time interval t=0 to t = 1.95 s is 2.62 m/s.

b) Average velocity of the car for the time interval t=0 to t = 3.96 s is 4.92 m/s.

c) Average velocity of the car for the time interval t=1.95 to t = 3.96 s is 7.15 m/s.

Explanation:

We have  x(t)= α t²− β t³

That is x(t)= 1.44 t²− 5 x 10⁻² t³

Average velocity is ratio of distance traveled to time.

a)Average velocity of the car for the time interval t=0 to t = 1.95 s

  x (0) = 1.44 x 0²− 5 x 10⁻² x 0³ = 0 m

  x (1.95) = 1.44 x 1.95²− 5 x 10⁻² x 1.95³ = 5.10 m

  Time difference = 1.95 - 0 = 1.95 s

  Average velocity [tex]=\frac{5.10}{1.95}=2.62m/s[/tex]

b)Average velocity of the car for the time interval t=0 to t = 3.96 s

  x (0) = 1.44 x 0²− 5 x 10⁻² x 0³ = 0 m

  x (3.96) = 1.44 x 3.96²− 5 x 10⁻² x 3.96³ = 19.48 m

  Time difference = 3.96 - 0 = 3.96 s

  Average velocity [tex]=\frac{19.48}{3.96}=4.92m/s[/tex]

c)Average velocity of the car for the time interval t=0 to t = 3.96 s

  x (1.95) = 1.44 x 1.95²− 5 x 10⁻² x 1.95³ = 5.10 m

  x (3.96) = 1.44 x 3.96²− 5 x 10⁻² x 3.96³ = 19.48 m

  Time difference = 3.96 - 1.95 = 2.01 s

  Average velocity [tex]=\frac{19.48-5.10}{2.01}=7.15m/s[/tex]

Answer:

v = 335.7 m/s

Explanation:

As we know that speed of sound in air is given by the formula

[tex]v = \sqrt{\frac{\gamma RT}{M}}[/tex]

now we have

[tex]\gamma = 1.4[/tex] For air

M = 29 g/mol = 0.029 kg/mol

T = 8 degree Celcius = 273 + 8 = 281 K

R = 8.31 J/mol K

now from above formula we have

[tex]v = \sqrt{\frac{(1.4)(8.31)(281)}{0.029}}[/tex]

[tex]v = 335.7 m/s[/tex]

Explanation:

There are two types of collision. First one is elastic and other one is inelastic collision.

The linear momentum of two objects before and after the collision remains the same in case of elastic collision. Also, the kinetic energy in elastic collision is conserved. But in case of inelastic collision, the momentum of two objects before and after the collision remains constant but the kinetic energy is in conversed. It changes form one form of energy to another.

Hence, the correct option is (E) "None of the above".

C) It is transformed into momentum such that momentum is conserved. (This statement is incorrect. Conservation of momentum applies to all collisions, not just the change in kinetic energy)

In an ideal elastic collision, the following happens:

Total Kinetic Energy is Conserved: The total kinetic energy of the system (colliding objects) before the collision is equal to the total kinetic energy after the collision. No energy is lost or gained in terms of kinetic energy.

Here's why the other options are incorrect:

A) Heat and Deformation: While real collisions might involve some energy loss due to heat and deformation, an ideal elastic collision is considered perfectly elastic, meaning no energy is lost in these ways.

B) Potential Energy: The collision doesn't necessarily involve a change in potential energy of the objects.

C) Momentum: Momentum is always conserved in any collision, elastic or inelastic.

D) All of the Above: Only the conservation of total kinetic energy applies to an ideal elastic collision.

E) None of the Above: In an ideal elastic collision, kinetic energy is conserved.

Therefore, the correct answer is:

C) It is transformed into momentum such that momentum is conserved. (This statement is incorrect. Conservation of momentum applies to all collisions, not just the change in kinetic energy)

Explanation:

Its translational kinetic energy is:

KE = ½ mv²

It's rotational kinetic energy is:

RE = ½ Iω²

For a solid sphere, I = ⅖ mr², and since it's not slipping, ω = v/r.

RE = ½ (⅖ mr²) (v/r)²

RE = ⅕ mv²

Therefore, the translational kinetic energy is larger.

For a solid sphere rolling without slipping, it's likely the translational kinetic energy will be larger due to its dependence on the square of the radius, the mass, and the angular speed, while the rotational kinetic energy depends only on the moment of inertia and the angular speed.

The kinetic energy of an object consists of translational and rotational elements with their relative magnitudes depending on the physical properties of the object. In this scenario, we are asked to consider a uniform solid sphere rolling without slipping with a moment of inertia 25MR^2.

Rotational kinetic energy can be determined by the equation K = 0.5 * I * w^2 where 'I' is the moment of inertia and 'w' is the angular speed. Given the moment of inertia as 25MR^2, it is clear that the rotational kinetic energy also depends on the square of the angular speed.

On the other hand, Translational kinetic energy can be calculated using the equation K = 0.5 * M * v^2 where 'M' is the mass and 'v' is the linear speed. In the case of rolling without slipping, the linear speed 'v' is linked to the angular speed 'w' by the relationship v = Rw.

 Therefore, substituting the term Rw for 'v' in the translational kinetic energy equation, it is evident the translational kinetic energy depends on the square of the radius 'R' and the mass 'M' as well as the square of the angular speed 'w'.

 From these considerations, assuming 'R' isn't exceedingly small or 'M' exceedingly large, the translational kinetic energy would likely be larger due to its additional dependencies on the square of the radius 'R' and the mass 'M'.

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Answer:

For iron, T2 = 40.4 degree C

For copper, T2 = 68.89 degree C

Explanation:

For iron:

m = 1.9 kg, T1 = 24 C, Q = 14 kJ = 14000 J, c = 0.450 J / g C = 450 J / Kg C

Let T2 be the final temperature of iron.

Q = m x c x (T2 - T1)

14000 = 1.9 x 450 x (T2 - 24)

T2 = 40.4 degree C

For copper:

m = 810 g = 0.81 kg, T1 = 24 C, c = 0.385 J/ g C = 385 J / Kg C, Q = 14 KJ

Let T2 be the final temperature of copper

Q = m x c x (T2 - T1)

14000 = 0.81 x 385 x (T2 - 24)

T2 = 68.89 degree C

Answer:

Part a)

m = 126.5 g

Part b)

v = 2.58 m/s

Part c)

v = 1.29 m/s

Explanation:

Part a)

By momentum conservation we will have

[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/tex]

here we have

[tex]m_1 = 320 g[/tex]

[tex]u_1 = 1.8 m/s[/tex]

[tex]u_2 = 0[/tex]

[tex]v_1 = 0.78 m/s[/tex]

also since collision is elastic collision so we have

[tex]v_2 = 2.58 m/s[/tex]

so now we have

[tex]320(1.8) + m_2(0) = 320(0.78) + m_2(2.58)[/tex]

[tex]m_2 = 126.5 g[/tex]

Part b)

As we know that in perfect elastic collision we will have

[tex]e = \frac{v_2 - v_1}{u_1 - u_2}[/tex]

now we will have

[tex]1 = \frac{v_2 - 0.78}{1.8 - 0}[/tex]

now we have

[tex]1.8 = v_2 - 0.78[/tex]

[tex]v_2 = 2.58 m/s[/tex]

Part c)

Since there is no external force on it

so here velocity of center of mass will remain the same

[tex]v_{cm} = \frac{m_1v_1 + m_2 v_2}{m_1 + m_2}[/tex]

[tex]v_{cm} = \frac{320(1.8) + 126.5(0)}{320 + 126.5}[/tex]

[tex]v_{cm} = 1.29 m/s[/tex]

The second cart has a mass of approximately 0.313 kg. After the collision, its speed is around 1.04 m/s. The speed of the two-cart center of mass after the collision is approximately 0.808 m/s.

This problem involves the principle of conservation of momentum and the mechanics of an elastic collision. In a collision between two bodies, the total momentum before the collision is equal to the total momentum after the collision. This is represented by the formula:

(a) The mass of the second cart can be found by rearranging the formula to solve for m₂, giving m₂ = m₁(v₁ - v₁') / v₂'. After substituting the given values, we find the mass of the second cart is around 0.313 kg.

(b) The velocity of the second cart after the collision, v₂', is equal to v₁ - v₁'/m₂,  which gives an approximate answer of 1.04 m/s.

(c) The speed of the two-cart center of mass can be calculated by dividing the total momentum of the system by the total mass of the system, giving an approximate speed of 0.808 m/s.

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Answer:

The Volume is V= 644.89 L

Explanation:

m= 2 kg

T= 120 º C = 393.15 K

P= 3.5 bar = 3.45 atm

R= 0.08205746 atm L / K mol

M= 0.029 Kg / mol

n= m/M

n= 68.96 moles

p*V= m*R*T

V= m*R*T/p

V=644.89 L

Answer:

The answer is E) 20kPa  

Explanation:

Here we use the definition of Pressure: Pressure is a physical quantity that measures the projection of force in the perpendicular direction per unit area. In this case, wehave a 200N Force over a 100cm^2 Area

In order to have the pressure in international units we need to convert those cm^2 to m^2 this way:

We know 100cm is equal to 1m

[tex](100cm^2)*((1m^2)/(100cm)^2)[/tex]

In this case, the area will be equal to 0,01m^2

With this information we use the Pressure equation, which based on the definition will be:

[tex]P=F/A[/tex]

Solving:

[tex]P=(200N)/(0,01m^2)[/tex]

[tex]P=20000Pa[/tex]

Then P is the same as 20kPa  

Answer:

(a)  1.3 x 10^6 Hz

(b) 76.73 cm

Explanation:

(a)

the formula for the frequency is given by

f = B q / 2 π m

where, B be the strength of magnetic field, q be the charge on one electron, m is the mass of one electron.

B = 46.7 micro tesla = 46.7 x 10^-6 T

q = 1.6 x 10^-19 C

m = 9.1 x 10^-31 kg

f = (46.7 x 10^-6 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31) = 1.3 x 10^6 Hz

(b) K = 114 eV = 114 x 1.6 x 10^-19 J = 182.4 x 10^-19 J

K = 1/2 mv^2

182.4 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v = 6.3 x 10^6 m/s

r = m v / B q

Where, r be the radius of circular path

r = (9.1 x 10^-31 x 6.3 x 10^6) / (46.7 x 10^-6 x 1.6 x 10^-19)

r = 0.7673 m = 76.73 cm  

Answer:

a) 2.85 kW

b) $ 432

c) $ 76.95

Explanation:

Average price of electricity = 1 $/40 MJ

Q = 20 kW

Heat energy production = 20.0 KJ/s

Coefficient of performance,  K = 7

also

K=(QH)/Win

Now,

Coefficient of Performance, K = (QH)/Win = (QH)/P(in) = 20/P(in) = 7

where

P(in) is the input power

Thus,

P(in) = 20/7 = 2.85 kW

b) Cost = Energy consumed × charges

Cost = ($1/40000kWh) × (16kW × 300 × 3600s)

cost = $ 432

c) cost = (1$/40000kWh) × (2.85 kW × 200 × 3600s) = $76.95

Final answer:

To heat a home for one month, an electric heater requires $432, while a heat pump with a coefficient of performance of 7 costs significantly less at $61.71, showcasing the efficiency and cost-effectiveness of using a heat pump for home heating.

Explanation:

To calculate the cost of heating for one month using a 16.0 kW electric heater and a heat pump with a coefficient of performance of 7.00, we start with understanding how a heat pump works. It essentially moves heat from the outdoors indoors, making it more efficient than directly converting electricity into heat. The first step is to calculate the total energy used by the electric heater and then compare it to the energy utilized efficiently by the heat pump.

Cost Calculation for Electric Heater:

Power usage of heater = 16.0 kW

Total hours run per month = 300 hours

Total energy used per month = 16.0 kW × 300 hours = 4800 kWh

Since 1 kWh equals 3.6 MJ, the total energy in MJ = 4800 × 3.6 = 17280 MJ

Average price for electricity = 40 MJ per dollar

Cost = 17280 MJ / (40 MJ/dollar) = $432

Cost Calculation for Heat Pump:

The coefficient of performance (COP) of the heat pump = 7.00

Efficiency implies that for every unit of energy (work) used, 7 units of heat are transferred inside.

To provide the same heat as the electric heater, the energy consumed by the heat pump = 4800 kWh / 7 = 685.71 kWh

Total energy in MJ = 685.71 × 3.6 = 2468.56 MJ

Cost = 2468.56 MJ / (40 MJ/dollar) = $61.71

Thus, using a heat pump is significantly more cost-efficient for home heating compared to using a direct electric heater.

Answer:

[tex]x=0\\x=4\\x=-2[/tex]

Explanation:

The zeros of a function are all the values of x for which f (x) = 0.

Therefore to find the zeros of the function I must equal f(x) to zero and solve for x.

[tex]f(x) = x(x - 4)(x + 2)=0[/tex]

[tex]x(x - 4)(x + 2)=0[/tex]

We have the multiplication of 3 factors x, (x-4) and (x + 2)

Then the function will be equal to zero when one of the factors is equal to zero, that is:

[tex]x = 0\\(x-4) = 0,\ x = 4\\(x + 2) = 0,\ x = -2[/tex]

Note that [tex]f(x) = x (x - 4) (x + 2)[/tex] is a cubic function of positive principal coefficient, the graph starts from [tex]-\infty[/tex] and cuts to the x-axis at [tex]x = -2[/tex], then decreases and cuts by second once to the x-axis at [tex]x = 0[/tex], it finally cuts the x-axis for the third time at [tex]x = 4[/tex] and then tends to [tex]\infty[/tex]

The zeros of the function f(x) = x(x - 4)(x + 2) are 0, 4, and -2. The end behavior of the function indicates that as x approaches infinity, f(x) approaches infinity, and as x approaches negative infinity, f(x) approaches negative infinity.

To determine the zeros of the function f(x) = x(x - 4)(x + 2), we need to find the values of x that make f(x) equal to zero. The function will be zero when any of the factors are zero. Therefore, the zeros of f(x) are 0, 4, and -2.

The end behavior of the polynomial function is determined by the leading term. Since our function has a leading term x3 with a positive coefficient, as x approaches positive infinity, f(x) also approaches positive infinity, and as x approaches negative infinity, f(x) approaches negative infinity. This is characteristic of an odd-degree polynomial with a positive leading coefficient.

Answer:

v = 14.35 m/s

Explanation:

As we know that crate is placed on rough bed

so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate

So here we have

[tex]\mu mg = \frac{mv^2}{R}[/tex]

here we have

[tex]\mu g = \frac{v^2}{R}[/tex]

now we know that

[tex]v = \sqrt{\mu Rg}[/tex]

here we have

[tex]\mu = 0.600[/tex]

R = 35 m

g = 9.81 m/s/s

now plug in all values in above equation

[tex]v = \sqrt{(0.600)(35)(9.81)}[/tex]

[tex]v = 14.35 m/s[/tex]

Answer:

10^-7 C

Explanation:

m = 1 g = 10^-3 kg, E = 200,000 N/C, a = 20 m/s^2, u = 0

Let q be the charge on bead

Force = m a = q E

a = q E / m

q = m a / E = (10^-3 x 20) / 200000 = 10^-7 C