The motion of a particle is defined by the relation x = t3 – 12t2 +36t +32, where x and t are expressed in feet and seconds, respectively. Determine the time, position, and acceleration of the particle when v = 0 ft/s.

Answer:

36//

Explanation:

(½)^t

(½)^72

72/2=36//

Answer:

The boat's acceleration are:  a= 0.3 m/s² in west direction.

Explanation:

Fm= 4100 N

Fwi= 800 N

Fwa= 1200 N

Fm - Fwi - Fwa= m *a

clearing a:

a= 0.3 m/s²

Answer:

[tex]B_{net} = \frac{2\sqrt2 \mu_0 i}{\pi d}[/tex]

Explanation:

Magnetic field due to straight current carrying wire is given by the formula

[tex]B = \frac{\mu_0 i}{4\pi r}(sin\theta_1 + sin\theta_2)[/tex]

now we will have for one side of the square at its center position given as

[tex]B = \frac{\mu_0 i}{4\pi (\frac{d}{2})}(sin45 + sin45)[/tex]

[tex]B = \frac{2\sqrt2 \mu_0 i}{4 \pi d}[/tex]

now for the we have for complete square loop it will become 4 times of the one side

[tex]B_{net} = 4 B[/tex]

[tex]B_{net} = 4 \frac{2\sqrt2 \mu_0 i}{4 \pi d}[/tex]

[tex]B_{net} = \frac{2\sqrt2 \mu_0 i}{\pi d}[/tex]

Answer:

2 kg

Explanation:

Assuming the rod's mass is uniformly distributed, the center of mass is at half the length.

Sum of the moments at the balance point:

-(Mg)(L/3) + (mg)(L/2 − L/3) = 0

(Mg)(L/3) = (mg)(L/2 − L/3)

(Mg)(L/3) = (mg)(L/6)

2M = m

M = 1 kg, so m = 2 kg.

The mass of the rod is 2 kg.

By equating the torques due to the ball and the rod, we determine the mass of the rod is 6 kg.

To determine the mass of the rod, we need to use the concept of torque balance. Here, we have a 1 kg ball hanging at the end of a 1 m long rod.

The system balances at a point one-third of the distance from the end holding the mass. This means the pivot point is located 1/3 meter from the end with the ball.

The torque due to the ball:

The torque due to the rod:

Since the system is in balance, the torques must be equal:

Torque_ball = Torque_rod

(1 kg) * (9.8 m/s²) * (1 m) = (mass of the rod) * (9.8 m/s²) * (1/6 m)

Solving for the mass of the rod:

(1 kg) * (9.8 m/s²) * (1 m) = (mass of the rod) * (9.8 m/s²) * (1/6)

mass of the rod = 6 kg

Thus, the mass of the rod is 6 kg.

Answer:

98.1 J

Explanation:

GPE = mass × g × height

= 2 × 9.81 × 5

= 98.1 J

Answer:

2.5 x 10^-3 T

Explanation:

n = 100 turns / cm = 10000 turns / m, R = 60 ohm, V = 12 V

i = V / R = 12 / 60 = 0.2 A

Magnetic field due to solenoid is

B = μ0 x n x i

B = 4 x 3.14 x 10^-7 x 10000 x 0.2

B = 2.5 x 10^-3 T

Explanation:

It is given that,

Charge, [tex]Q_1=5.7\ nC=5.7\times 10^{-9}\ C[/tex]

Charge, [tex]Q_2=-2.3\ nC=-2.3\times 10^{-9}\ C[/tex]

Distance between charges, d = 45 cm = 0.45 m

We need to find the electric potential at a point midway between the charges. It is given by :

[tex]V=\dfrac{kQ_1}{r_1}+\dfrac{kQ_2}{r_2}[/tex]

[tex]V=k(\dfrac{Q_1}{r_1}+\dfrac{Q_2}{r_2})[/tex]

Midway between the charges means, r = 22.5 cm = 0.225 m

[tex]V=9\times 10^9\times (\dfrac{5.7\times 10^{-9}\ C}{0.225\ m}+\dfrac{-2.3\times 10^{-9}\ C}{0.225\ m})[/tex]

[tex]V=136\ volts[/tex]

So, the electric potential at a point midway between the charges is 136 volts. Hence, this is the required solution.

Final answer:

The electric field strength at the specified point is calculated by finding the vector sum of the electric fields due to each charge, using Coulomb's law and the superposition principle.

Explanation:

The electric field strength at a point due to a single point charge can be calculated using Coulomb's law. The electric field E is the force F per unit charge q, and it acts in the direction away from a positive charge or towards a negative charge. For two point charges, Q₁ and Q₂, each will create an electric field at the point of interest, and the net electric field is the vector sum of these two fields.

To find the electric field at a point that is 20 cm from Q₁ (a positive 3 nC charge) and 50 cm from Q₂ (a negative 4 nC charge), we calculate the electric field from each charge individually and then combine them. The formula for the electric field due to a point charge is E = k * |Q| / r², where k = 8.99 x 10⁹ Nm²/C² is the Coulomb's constant, Q is the charge, and r is the distance from the charge to the point of interest.

Since the charges are opposite in sign and different in magnitude, and the point is closer to Q₁, the electric fields due to each charge will not cancel out, and the net electric field will have a direction based on the vector addition of the individual fields.

Final answer:

A system can sustain a nonzero velocity with zero net external force in accordance with Newton's first law of motion, such as a hockey puck moving at a constant velocity on an ice surface without friction.

Explanation:

A system can definitely have a nonzero velocity even when the net external force acting upon it is zero. This situation aligns with Newton's first law of motion, also known as the law of inertia, which states that an object will maintain its state of rest or uniform motion unless acted upon by a net external force.

An example of this scenario can be seen when an object is moving at a constant velocity on a frictionless surface, where no external forces are acting to accelerate or decelerate it. In real-world terms, consider a hockey puck gliding across a smooth ice surface with no additional forces applied to it; it will continue to move at the same speed and in the same direction.

Answer:

The net gravitational force exerted by these objects is [tex]5.712\times10^{-8}\ N[/tex]

Explanation:

Given that,

Mass of another object m = 33.0 kg

Mass M₁ = 160 kg

Mass M₂ = 460 kg

Distance = 3.40 m

We need to calculate the net gravitational force

Using formula,

[tex]F=\dfrac{GM_{2}m}{r^2}-\dfrac{GM_{1}m}{r^2}[/tex]

[tex]F=Gm(\dfrac{M_{2}-M_{1}}{r^2})[/tex]

Where, G = gravitational constant

m = mass of another object

[tex]M_{1}[/tex]= mass of first object

[tex]M_{2}[/tex]= mass of second object

r = distance

Put the value into the formula

[tex]F=6.67\times10^{-11}\times33.0\times\dfrac{460-160}{(3.40)^2}[/tex]

[tex]F=5.712\times10^{-8}\ N[/tex]

Hence, The net gravitational force exerted by these objects is [tex]5.712\times10^{-8}\ N[/tex]

To find the magnitude and direction of the electric field, let us find the horizontal and vertical components of the field separately, then we will use those values to calculate the total magnitude and direction.

The tension in the thread is 6.57×10⁻²N and the thread is aligned horizontally, so the tension force is directed entirely horizontally. The sphere is in static equilibrium, therefore the horizontal component of the electrostatic force acting on the sphere, Fx, must act in the opposite direction of the tension and have a magnitude of 6.57×10⁻²N. We know this equation relating a charge, an electric field, and the force that the field exerts on the charge:

F = Eq

F is the electric force, E is the electric field, and q is the charge

Let us adjust the equation for only the horizontal components of the above quantities:

Fx = (Ex)(q)

Fx is the horizontal component of the electric force and Ex is the horizontal component of the electric field.

Given values:

F = 6.57×10⁻²N

q = 6.80×10³C

Plug in these values and solve for Ex:

6.57x10⁻² = Ex(6.80×10³)

Ex = 9.66×10⁻⁶N/C

Since the sphere is in static equilibrium, the vertical component of the electrostatic force acting on the sphere, Fy, must have the same magnitude and act in the opposite direction of the sphere's weight. If we assume the weight to act downwards, then Fy must act upward.

We know the weight of the sphere is given by:

W = mg

W is the weight, m is the mass, and g is the acceleration of objects due to earth's gravity field near its surface.

We also know this equation:

F = Eq

Let us adjust for the vertical components:

Fy = (Ey)(q)

Set Fy equal to W and we get:

(Ey)(q) = mg

Given values:

q = 6.80×10³C

m = 0.018kg

g = 9.81m/s²

Plug in the values and solve for Ey:

(Ey)(6.80×10³) = 0.018(9.81)

Ey = 2.60×10⁻⁵N/C

Let's now use the Pythagorean theorem to find the total magnitude of the electric field:

E = [tex]\sqrt{Ex^{2}+Ey^{2}}[/tex]

E = 2.77×10⁻⁵N/C

The direction of the electric field is given by:

θ = tan⁻¹(Ey/Ex)

θ = 20.4° off the horizontal

Answer:

Rate of change of magnetic flux

Explanation:

The induced current is equal to the ratio of induced emf to the resistance of the conductor.

According to the Faraday's law of electromagnetic induction, the induced emf is proportional to the rate of change of magnetic flux.

Answer:

1387908 lbm/h

Explanation:

Air flowing into jet engine = 70 lbm/s

ρ = Exhaust gas density = 0.1 lbm/ft³

r = Radius of exit with a circular cross section = 1 ft

v = Exhaust gas velocity = 1450 ft/s

Exhaust gas mass (flow rate)= Air flowing into jet engine + Fuel

Q = (70+x) lbm/s

Area of exit with a circular cross section = π×r² = π×1²= π m²

Now from energy balance

Q = ρ×A×v

⇒70+x = 0.1×π×1450

⇒70+x = 455.53

⇒ x = 455.53-70

⇒ x = 385.53 lbm/s

∴ Mass of fuel which is supplied to the engine each minute is 1387908 lbm/h

To find the mass of fuel supplied to the engine per minute, calculate the exhaust gas volume flow rate, then the mass flow rate using gas density, and subtract the air mass flow rate from it. Convert the resulting fuel mass flow rate from seconds to minutes by multiplying by 60.

The student's question is about calculating the mass flow rate of fuel supplied to the engine of a jet, using the conservation of mass and given data about the exhaust gases. The density of the exhaust gas, the velocity of the exhaust, and the cross-sectional area of the outlet are provided.

First, calculate the volume flow rate of exhaust gases by multiplying the area of the outlet by the velocity of the exhaust. Since the radius is given, the area (A) can be calculated with the formula A = πr^2, where r is the radius. Then, the volume flow rate (V) is found with the formula V = A × exhaust velocity.

Next, multiply the volume flow rate by the exhaust gas density to find the mass flow rate of the exhaust gas. Finally, the difference between the mass flow rate of the incoming air and the mass flow rate of the exhaust gas will give the mass flow rate of the fuel injected into the engine. Since the question asks for the mass of fuel per minute, this rate should be converted from seconds to minutes by multiplying by 60.

Answer:

Heat capacity this bomb calorimeter =  [tex] 41.74 KJ/^oC [/tex]

Explanation:

Given data we have in the question:

The heat of combustion for the benzene, [tex]Q_{C_6H_6} = - 41.74 KJ/g[/tex]  (Here the negative sign depicts the release of heat and only magnitude is considered fro the heat capacity)

The mass of benzene, [tex]M_{C_6H_6} = 4.50g[/tex]

The change in temperature due to the combustion, [tex]\Delta T = 4.50^oC[/tex]

Now, the formula for the heat capacity is given as:

Heat capacity (C) = [tex]\frac{Q_{C_6H_6}\times M_{C_6H_6}}{\Delta T }[/tex]

or

C =  [tex]\frac{ 41.74 KJ/g\times 4.50g}{4.50^oC }[/tex]

or

C = [tex] 41.74 KJ/^oC [/tex]

The heat capacity of the bomb calorimeter, using the given values and the formula: (heat = heat capacity * change in temperature), is 41.74 kJ/°C.

In bomb calorimetry, the heat capacity of the bomb calorimeter, or the calorimeter constant (C), can be calculated by using the formula: q = C * change in temperature. The heat (q) released by burning the benzene is the product of the mass burned and the heat of combustion, which is -41.74 kJ/g * 4.5 g, giving us -187.83 kJ. Since a negative q indicates that heat is released, we're actually dealing with 187.83 kJ of heat. The heat capacity (C) of the calorimeter can now be calculated by rearranging the formula to C = q / delta T = 187.83 kJ / 4.5°C = 41.74 kJ/°C. This means the heat capacity of the bomb calorimeter is 41.74 kJ/°C.

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Answer:

14.64% of the volume of the glass, more water can be added

Explanation:

given:

Initial temperature, = 100°C = 373K

Final temperature = 20°C = 293K

Coefficient of volume expansion for glass, =  2.7 x 10⁻⁵ K⁻¹

Coefficient of volume expansion for Water, =  2.1 x 10⁻⁴ K⁻¹

The apparent Coefficient of volume expansion for Water, γ = 2.1 x 10⁻⁴ K⁻¹ - 2.7 x 10⁻⁵ K⁻¹ = 1.83 × 10⁻⁴ K⁻¹

Change in temperature, ΔΘ = Final Temperature - Initial Temperature = 293K - 373K = -80K

Now, the Coefficient of volume expansion is given as:

[tex]\Delta V = -{\gamma}{V \Delta \theta}\\[/tex]

where,

V = initial Volume

ΔV = change in the volume

Thus,

[tex]\Delta V = -{1.83\times 10^{-4} K^{-1} }\times {V \times -80K}\\[/tex]

or

[tex]\Delta V = 146.4\times 10^{-4}\times V[/tex]

or

[tex]\frac{\Delta V}{V} = 146.4\times 10^{-4}[/tex]

Multiplying both sides by 100

we get

[tex]\frac{\Delta V}{V}\times 100 = 146.4\times 10^{-4}\times 100[/tex]

or

change in volume with respect to the initial volume = 1.464%

thus, 1.464 % of the original volume water can be added.

Upon cooling from 100 °C to 20 °C, the thermal contraction of the water in the glass will be greater than that of the glass itself due to higher coefficient of volume expansion for water, thereby creating some extra space in the glass. This extra space represents the amount of additional water that can be added.

In order to find out how much more water can be added to the glass as it cools from 100 °C to 20 °C, we first need to know the volume of water and glass that will change with this temperature drop. The formula we use to calculate this is ΔV = βV₀ΔT, where β is the coefficient of volume expansion, V₀ is the initial volume, and ΔT is the change in temperature.

First, we calculate this for the glass and then for the water, using their respective coefficients of volume expansion. The difference in the volume changes for water and glass will give us the amount of water that can be added to the glass as it cools. Since the coefficient of volume expansion for water is greater than glass, its volume will contract more due to the decrease in temperature, leaving some extra space in the glass.

Note that we make an assumption here that the glass and water were initially at thermal equilibrium at 100 °C and hence have the same initial volume. This problem illustrates the principle of thermal expansion which is a property of matter to change its volume with a change in temperature, and is dependent on the material's coefficient of volume expansion.

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The final speed of the sled is calculated by first calculating the frictional force using the coefficient of kinetic friction and weight of the sled. This force is then used to calculate the deceleration due to friction. Finally, an equation of motion is employed to calculate the sled's final speed after it has traveled the given distance.

To answer this question, we first need to calculate the force of friction, which can be found using the formula F_friction = µN, where µ is the coefficient of kinetic friction and N is the normal force. In this case, as the sled is moving on a horizontal surface, the normal force is equal to the weight of the sled, N = mg. So the frictional force F_friction = µmg.

Next, due to Newton's second law (F = ma), the deceleration (negative acceleration) is calculated by a = F_friction/m. With this acceleration, we can now use one of the equations of motion, V_f = V_i + 2a*d, to calculate the final velocity, where V_f is the final velocity, V_i is the initial velocity, a is acceleration and d is the distance.

This method allows us to calculate the final speed of the sled after traveling a certain distance considering kinetic friction.

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The final speed of the sled is calculated by finding the work done by friction and applying the work-energy theorem to find the change in kinetic energy. The kinetic energy is equated to its initial value minus the work done by friction, and solved for the final speed.

To tackle this question, we should first calculate the total work done by the friction force (kinetic friction) which is known to act in the opposite direction of the sled's movement. The work done by friction is given by the equation W = Fd in which 'F' stands for friction force and 'd' for distance. The friction force can be calculated by the formula F = μN, where 'μ' is the coefficient of friction and 'N' is the normal force. Since the sled moves across a horizontal surface, N is equal to the weight of the sled (mg, where 'm' is the mass and 'g' is the acceleration due to gravity).

Then, using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy, we can solve for the final speed. The kinetic energy is given by KE = 0.5mv², where 'm' is the mass and 'v' is the speed. So, the initial and the final kinetic energies of the sled are KE_initial = 0.5 * 2.12kg * (5.49 m/s)² and KE_final = KE_initial - Work_done_by_friction. We can solve the latter for the final speed of the sled, v_final = sqrt((2*KE_final)/m).

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Answer:

9.07 x 10^-3 N/C

Explanation:

The mobility of electrons is defined as the ratio of drift velocity to the applied electric field.

Vd = 4.08 x 10^-5 m/s

Mobility = 4.5 x 10^-3 m/s

Let the electric field is E.

Mobility = Vd / E

E = Vd / mobility

E = 4.08 x 10^-5 / (4.5 x 10^-3)

E = 9.07 x 10^-3 N/C

The magnitude of the electric field inside the copper wire is approximately 9.067 — 10-3 N/C.

To calculate the magnitude of the electric field inside the copper wire, we can use the relationship between drift speed (v_d), mobility (μ), and electric field (E):

[tex]\[ v_d = \mu \cdot E \][/tex]

Given the drift speed (v_d) of 4.08 — 10^-5 m/s and the mobility (μ) of 4.5 — 10^-3 (m/s)/(N/C), we can rearrange the equation to solve for the electric field (E):

[tex]\[ E = \frac{v_d}{\mu} \][/tex]

Substituting the given values:

[tex]\[ E = \frac{4.08 \times 10^{-5} \text{ m/s}}{4.5 \times 10^{-3} \text{ (m/s)/(N/C)}} \][/tex]

[tex]\[ E = \frac{4.08}{4.5} \times 10^{-5 + 3} \text{ N/C} \][/tex]

[tex]\[ E = 0.9067 \times 10^{-2} \text{ N/C} \][/tex]

[tex]\[ E \approx 9.067 \times 10^{-3} \text{ N/C} \][/tex]

Therefore, the magnitude of the electric field inside the copper wire is approximately 9.067 — 10^-3 N/C.

Answer: 0.258

Explanation:

The resistance [tex]R[/tex] of a wire is calculated by the following formula:

[tex]R=\rho\frac{l}{s}[/tex]    (1)

Where:

[tex]\rho[/tex] is the resistivity of the material the wire is made of. For aluminium is [tex]\rho_{Al}=2.65(10)^{-8}m\Omega[/tex]  and for copper is [tex]\rho_{Cu}=1.68(10)^{-8}m\Omega[/tex]

[tex]l[/tex] is the length of the wire, which in the case of aluminium is [tex]l_{Al}=12m[/tex], and in the case of copper is [tex]l_{Cu}=30m[/tex]

[tex]s[/tex] is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

[tex]s=\pi{(\frac{d}{2})}^{2}[/tex]  (2) Where [tex]d[/tex]  is the diameter of the circumference.

For aluminium wire the diameter is  [tex]d_{Al}=2.5mm=0.0025m[/tex]  and for copper is [tex]d_{Cu}=1.6mm=0.0016m[/tex]

So, in this problem we have two transversal areas:

For aluminium:

[tex]s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}[/tex]

[tex]s_{Al}=0.000004908m^{2}[/tex]   (3)

For copper:

[tex]s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}[/tex]

[tex]s_{Cu}=0.00000201m^{2}[/tex]    (4)

Now we have to calculate the resistance for each wire:

Aluminium wire:

[tex]R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}[/tex]     (5)

[tex]R_{Al}=0.0647\Omega[/tex]     (6)  Resistance of aluminium wire

Copper wire:

[tex]R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}[/tex]     (6)

[tex]R_{Cu}=0.250\Omega[/tex]     (7)  Resistance of copper wire

At this point we are able to calculate the  ratio of the resistance of both wires:

[tex]Ratio=\frac{R_{Al}}{R_{Cu}}[/tex]   (8)

[tex]\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}[/tex]   (9)

Finally:

[tex]\frac{R_{Al}}{R_{Cu}}=0.258[/tex]  This is the ratio

Answer:

83.33 C

Explanation:

T1 = 111 C, m1 = 2m

T2 = 28 C, m2 = m

c  = 0.387 J/gK

Let the final temperature inside the calorimeter of T.

Use the principle of calorimetery

heat lost by hot body = heat gained by cold body

m1 x c x (T1 - T) = m2 x c x (T - T2)

2m x c X (111 - T) = m x c x (T - 28)

2 (111 - T) = (T - 28)

222 - 2T = T - 28

3T = 250

T = 83.33 C

Thus, the final temperature inside calorimeter is 83.33 C.

Answer:

The horizontal distance traveled when the stone returns to its original height = 220.81 m

Explanation:

Considering vertical motion of catapult:-

At maximum height,

Initial velocity, u =  50 sin30 = 25 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v = u + at

Substituting

    v = u + at

    0 = 25  - 9.81 x t

    t = 2.55 s

Time of flight = 2 x Time to reach maximum height = 2 x 2.55 = 3.1 s

Considering horizontal motion of catapult:-

Initial velocity, u =  50 cos30 = 43.30 m/s

Acceleration , a = 0 m/s²

Time, t = 5.10 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

   s = 43.30 x 5.10 + 0.5 x 0 x 5.10²

  s = 220.81 m

The horizontal distance traveled when the stone returns to its original height = 220.81 m

Answer:

0.368 m^3

Explanation:

3.28 feet = 1 m

Cube on both the sides

35.29 feet^3 = 1 m^3

So, 13 cubic feet = 13 / 35.29 m^3

= 0.368 m^3

Explanation:

This is because the drag force suffered by the aircraft is proportional to the speed at which it travels. The thrust of the engines prints a speed to the plane and this speed prints a drag force, always reaching an equilibrium point of these two forces where the speed of the plane is constant and the acceleration is equal to zero.

Therefore, by reducing the thrust, the drag force is greater and the plane begins to decrease its speed, until it reaches the point where the new drag force is matched with the new thrust force, giving it a new final speed , without acceleration.

An airplane flying at a constant speed maintains a balance between engine thrust and drag force. Reducing thrust lessens this force, leading to a new equilibrium at a lower speed due to the drag force being proportional to the square of the velocity.

When an airplane moves through air, its engines provide thrust to counteract drag, which is a force opposing the motion. The drag force increases with the speed of the airplane, following a relationship where the magnitude of the drag force is proportional to the square of its speed. According to this principle, if an airplane reduces its thrust, the drag force eventually balances the reduced thrust at a new, lower equilibrium speed, which allows the plane to continue flying at a new constant speed, but slower.

This balance between thrust and drag is a direct application of Newton's second law of motion, where the net force applied to an object is equal to its mass times its acceleration (F=ma). In the steady flight of an airplane, when thrust equals drag and lift equals weight, the acceleration is zero, resulting in constant velocity flight. Reducing the thrust results in an initial slowing down of the airplane until the drag force decreases to match the new lower thrust, and the plane achieves a new constant (lower) velocity.

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Answer:

[tex]l=222.803mm[/tex]

Explanation:

Given:

Elastic modulus, E = 102 GPa

Diameter, d  = 3.8mm = 0.0038 m

Applied tensile load = 2440N

Maximum allowable elongation, = 0.47mm = 0.00047

Now,

The cross-sectional area of the specimen,[tex]A_o=\frac{\pi d^2}{4}[/tex]

substituting the values in the above equation we get

[tex]A_o=\frac{\pi 0.0038^2}{4}[/tex]

or

[tex]A_o=1.134\times 10^{-5}[/tex]

now

the stress (σ) is given as:

[tex]\sigma=\frac{Force}{Area}[/tex]

and[tex]E=\frac{\sigma}{\epsilon}[/tex]

where,

[tex]\epsilon =\ Strain[/tex]

also,

[tex]\epsilon=\frac{\Delta l}{l}[/tex]

where,

[tex]l=initial \ length[/tex]

thus,

[tex]E=\frac{\frac{F}{A_o}}{\frac{\Delta l}{l}}[/tex]

or on rearranging we get,

[tex]l=\frac{E\times \Delta l\times A}{F}[/tex]

substituting the values in the above equation we get

[tex]l=\frac{102\times 10^9\times 0.00047\times 1.134\times 10^{-5}}{2440}[/tex]

or

[tex]l=0.222803m[/tex]

or

[tex]l=222.803mm[/tex]

Explanation:

A charge alters the space around it. This alteration of space is called the electric field. It is also defined as the electric force acting on a charged particle per unit test charge. It is given by :

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

Where

F is the electric force, [tex]F=\dfrac{kq_1q_2}{r^2}[/tex]

The direction of electric field is in the direction of electric force. For a positive charge, the direction of electric field lines are outwards and for a negative charge, the direction of field lines are inwards.

Hence, the correct option is (c) "electric field".

Using  the formula for power:

P = V^2 / R  

130 W = (9.00 V)^2 / R  

Solve for r:

R = 81/130

R = 0.623 ohms

Now solve for the length of wire:

R = rho L / A  

A must be in m^2 - 2.32 mm^2 * 1 m^2/10^6 mm^2 = 2.132x10^-6 m^2  

Now you have:

0.623 = (6.47x10^-8) L / (2.32x10^-6)  

L = 22.34 m (Round answer as needed.)

Answer:

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Answer:

       stress   = 366515913.6 Pa

Explanation:

given data:

density of alloy = 8.5 g/cm^3 = 8500 kg/m^3

length turbine blade = 10 cm =  0.1 m

cross sectional area = 15 cm^2 = 15*10^-4 m^2

disc radius = 70 cm = 0.7 m

angular velocity = 7500 rpm = 7500/60 rotation per sec

we know that

stress = force/ area

force = m*a

where a_{c} is centripetal acceleration =

[tex]a_{c} =r*\omega ^{2}= r*(2*\pi*\omega)^{2}[/tex]

         =[tex]0.70*(2*\pi*\frac{7500}{60})^{2}[/tex]

         = 431795.19 m/s^2

mass = [tex]\rho* V[/tex]

Volume = area* length = 15*10^{-5}  m^3

[tex]mass = m = \rho*V = 8500*15*10^{-5} kg[/tex]

force = m*a_{c}

         [tex]=8500*15*10^{-5}*0.70*(2*\pi*\frac{7500}{60})^{2}[/tex]

force = 549773.87 N

stress = force/ area

          = [tex]\frac{549773.87}{15*10^{-5}}[/tex]

       stress   = 366515913.6 Pa

Final answer:

The question requires the calculation of stress on jet turbine blades using physics principles involving centrifugal force and material stress.

Explanation:

The question involves calculating the stress on the turbine blades of a jet engine, which requires a knowledge of physics concepts, particularly mechanics and dynamics. The given data include the turbine's rotational velocity (7,500 rpm), radius of the turbine disc (70 cm), the cross-sectional area of the blades (15 cm2), the length of the blades (10 cm), and the alloy density (8.5 g/cm3). To solve this, one would need to calculate the centrifugal force acting on the blades due to rotation and then divide that force by the cross-sectional area to find the stress. However, the calculation involves steps and concepts not provided in the information above, so the direct calculation cannot be completed without additional physics formulae and explanation.

Answer:

(a)

P₂ = 7.13 atm

(b)

T₂ = 157.14 K

Explanation:

(a)

V₁ = initial volume = 3.7 L = 3.7 x 10⁻³ m³

V₂ = final volume = 0.85 L = 0.85 x 10⁻³ m³

P₁ = Initial Pressure of the gas = 0.91 atm = 0.91 x 101325 = 92205.75 Pa

P₂ = Final Pressure of the gas = ?

Using the equation

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

[tex](92205.75) (3.7\times 10^{-3})^{1.4 } = P_{2} (0.85\times 10^{-3})^{1.4 }[/tex]

[tex]P_{2}[/tex] = 722860 Pa

[tex]P_{2}[/tex] = 7.13 atm

(b)

T₁ = initial temperature =283 K

T₂ = Final temperature = ?

using the equation

[tex]P{_{1}}^{1-\gamma } T{_{1}}^{\gamma } = P{_{2}}^{1-\gamma } T{_{2}}^{\gamma }[/tex]

[tex](92205.75)^{1-1.4 } (283)^{1.4 } = (722860)^{1-1.4 } T{_{2}}^{1.4 }[/tex]

T₂ = 157.14 K

Answer:

[tex]N = 1.62 \times 10^{15}[/tex]

Explanation:

Energy of each photon is given as

[tex]E = \frac{hc}{\lambda}[/tex]

here we will have

[tex]\lambda = 632 nm[/tex]

now we will have

[tex]E = \frac{(6.626\times 10^{-34})(3 \times 10^8)}{632 \times 10^{-9}}[/tex]

[tex]E = 3.14 \times 10^{-19} J[/tex]

now let say there is N number of photons per second

so power due to photons is

[tex]P = N(3.14 \times 10^{-19})[/tex]

now intensity is given as power received per unit area

so we have

[tex]I = \frac{P}{A}[/tex]

[tex]825 = \frac{N(3.14 \times 10^{-19})}{\pi (1.40 \times 10^{-3})^2}[/tex]

[tex]825 = N(5.11 \times 10^{-14})[/tex]

[tex]N = 1.62 \times 10^{15}[/tex]

The capacitance of a capacitor when the charge increases by 20 µC and the voltage increases from 84 V to 121 V is approximately 0.541 µF.

The question asks to determine the capacitance of a capacitor. To find the capacitance when the charge (ΔQ) increases by 20 µC as the voltage increases from 84 V to 121 V, we can use the capacitor charge formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. The change in voltage (ΔV) can be calculated as 121 V - 84 V = 37 V. The capacitance (C) can then be found using the change in charge and the change in voltage:

C = ΔQ / ΔV

Plugging in the values, we get:

C = 20 µC / 37 V ≈ 0.541 µF

Therefore, the capacitance of the capacitor is approximately 0.541 µF.

Answer:

(a) 17.37 rad/s^2

(b) 12479

Explanation:

t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0

w = v / r = 99 / 0.06 = 1650 rad/s

(a) Use first equation of motion for rotational motion

w = w0 + α t

1650 = 0 + α x 95

α = 17.37 rad/s^2

(b) Let θ be the angular displacement

Use third equation of motion for rotational motion

w^2 = w0^2 + 2 α θ

1650^2 = 0 + 2 x 17.37 x θ

θ = 78367.87 rad

number of revolutions, n = θ / 2 π

n = 78367.87 / ( 2 x 3.14)

n = 12478.9 ≈ 12479

The mass of a sphere of gold with a radius of 12.5 cm is 1.5789752 x 10^5 g in scientific notation. A cube of platinum with an edge length of 0.049 mm has a mass of 2.517269 x 10^-3 g. The mass of 67.1 mL of ethanol with a density of 0.798 g/mL is 53.5458 g.

To calculate the mass of a sphere of gold with a radius of 12.5 cm using the density formula (density = mass/volume), first we need to find the volume of the sphere using the formula V = (4/3)πr^3. Then we can multiply the volume by the density of gold, which is 19.3 g/cm^3.

(a) Volume of gold sphere = (4/3)π(12.5 cm)^3 = 8,181.229 cm^3
Mass of gold sphere = volume × density = 8,181.229 cm^3 × 19.3 g/cm^3 = 157,897.52 g
The mass in scientific notation is 1.5789752 × 105 g.

(b) Volume of platinum cube = edge^3 = (0.049 cm)^3 = 1.17649 × 10^-4 cm^3
Mass of platinum cube = volume × density
= 1.17649 × 10^-4 cm^3 × 21.4 g/cm^3 = 2.517269 × 10^-3 g

(c) Mass of ethanol = volume × density
= 67.1 mL × 0.798 g/mL = 53.5458 g