Assume that a pitcher throws a baseball so that it travels in a straight line parallel to the ground. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. Define the direction the pitcher originally throws the ball as the +x direction. The impulse on the ball caused by the bat will bo in the negative x direction. Part E Now assume that the pitcher in Part D throws a 0.145-kg baseball parallel to the ground with a speed of 32 m/s in the +x direction. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. What is the ball's velocity just after leaving the bat if the bat applies an impulse of -8.4 N s to the baseball? Enter your answer numerically in meters per second using two significant figures.

Answer:

[tex]T=172.37\ K[/tex]

[tex]P_s=11.1095\ Pa[/tex]

Explanation:

Triple point is that temperature and pressure for a substance at which the chemical substance coexists in liquid, solid and gaseous state with mutual equilibrium.

Given expressions of vapor pressure:

[tex]ln (P_s)=24.320 -\frac{3777  K}{T}[/tex] ........................................(1)

[tex]ln (P_l)=17.892 -\frac{2669 K}{T}[/tex] ..........................................(2)

Equating the vapor pressure of two phases of chlorine:

[tex]24.320 -\frac{3777  K}{T}=17.892 -\frac{2669 K}{T}[/tex]

[tex]T=172.37\ K[/tex]

The vapor pressure at this temperature:

[tex]ln (P_s)=24.320 -\frac{3777 }{172.37}[/tex]

[tex]P_s=11.1095\ Pa[/tex]

To find the triple point of chlorine, we set the vapor pressure equations of solid and liquid chlorine equal and solve for temperature, then use either equation to find the vapor pressure. However, numerical values for the vapor pressure equations are required to compute the specific temperature and pressure. The Clausius-Clapeyron equation can also be relevant in such calculations.

The triple point of chlorine is determined by finding the temperature and pressure at which the vapor pressures of solid and liquid chlorine are equal. Given the vapor pressure equations for solid and liquid chlorine, both can be set equal to each other and solved for temperature (T), as both will have the same pressure at the triple point.

Unfortunately, without the numerical expressions for ln(Ps) and ln(Pl), it is impossible to provide a numerical answer. Typically, this would involve algebraic manipulation to find T, and then either expression can be used to solve for the vapor pressure at the triple point.

For example, the Clausius-Clapeyron equation is used to relate the temperatures and pressures for phase changes, such as moving from solid to gas (sublimation) at the triple point. However, more information is necessary to calculate the temperature and pressure explicitly.

Answer:

0 V

Explanation:

given,                                        

Inductance  = L = 0.0345 H        

resistor of resistance = R = 30.5 Ω                          

connected in series with battery of voltage = 3.20 V

at t= 0 switch is closed                                                      

potential difference when switch is closed = ?                                      

The potential difference when the switch is just closed across the resistor will be equal to Zero.                                        

Because at t =0 the capacitor will act as a short circuit.

hence, the potential will be 0 V                                

Final answer:

The moment of inertia for each object can be calculated using specific formulas. The formulas are: For a thin hoop: I = mR², for a solid cylinder: I = (1/2)MR², for a solid sphere: I = (2/5)MR², and for a thin spherical shell: I = (2/3)MR².

Explanation:

The moment of inertia for each object as it rotates about its axis can be calculated using the formulas for moment of inertia for different objects:

1. Hoop: The moment of inertia for a thin hoop rotating about an axis perpendicular to its plane is given by the formula I = mR², where m is the mass and R is the radius of the hoop. In this case, the mass is 4.01 kg and the radius is 0.260 m, so the moment of inertia for the hoop is (4.01 kg)(0.260 m)².

2. Solid Cylinder: The moment of inertia for a solid cylinder rotating about its central axis is given by the formula I = (1/2)MR², where M is the mass and R is the radius of the cylinder. In this case, the mass is 4.01 kg and the radius is 0.260 m, so the moment of inertia for the solid cylinder is (1/2)(4.01 kg)(0.260 m)².

3. Solid Sphere: The moment of inertia for a solid sphere rotating about its central axis is given by the formula I = (2/5)MR², where M is the mass and R is the radius of the sphere. In this case, the mass is 4.01 kg and the radius is 0.260 m, so the moment of inertia for the solid sphere is (2/5)(4.01 kg)(0.260 m)².

4. Thin Spherical Shell: The moment of inertia for a thin spherical shell rotating about its central axis is given by the formula I = (2/3)MR², where M is the mass and R is the radius of the shell. In this case, the mass is 4.01 kg and the radius is 0.260 m, so the moment of inertia for the thin spherical shell is (2/3)(4.01 kg)(0.260 m)².

Answer:

a = - 0.248 m/s²

Explanation:

Frictional drag force

F = ½ *(ρ* v² * A * α)

ρ = density of air  , ρ = 1.295 kg/m^3

α = drag coef , α = 0.250

v = 100 km/h x 1000m / 3600s

v =  27.77 m/s

A = 2.20m^2

So replacing numeric in the initial equation

F = ½ (1.295kg/m^3)(27.77m/s)²(2.30m^2)(0.26)

F = 298.6 N

Now knowing the force can find the acceleration

a = - F / m

a = - 298.6 N / 1200 kg

a = - 0.248 m/s²

Final answer:

The question involves using aerodynamic drag force and Newton's second law to calculate the deceleration of a sports car shifted into neutral.

Explanation:

The question revolves around calculating the initial acceleration of a sports car weighing 1200 kg that has been shifted into neutral and allowed to coast, assuming the car was traveling at 100 km/hr, and given an aerodynamic drag coefficient of 0.250 and a frontal area of 2.20 m2. To calculate acceleration due to the forces acting on the car, we use the formula for aerodynamic drag force which is Fd = 0.5 × Cd × ρ × A × v2 where Cd is the drag coefficient, ρ is the air density (approximately 1.225 kg/m3 at sea level), A is the frontal area, and v is the velocity in meters per second. To find the acceleration, we can then apply Newton's second law, F = m × a, where F is the net force acting on the car, m is the mass, and a is the acceleration. In this case, the net force is opposite to the direction of the velocity vector due to drag force, thus providing the deceleration or negative acceleration of the car.

Answer:

The induced emf for this silver circle during this period of expansion is 0.0314 V.

Explanation:

Given that,

Initial radius = 0.075 m

Magnetic field = 1.5 T

Radius = 0.125 m

Time t =1.5 s

We need to calculate the induced emf for this silver circle

Using formula of emf

[tex]\epsilon=\dfrac{d(BA)}{dt}[/tex]

[tex]\epsilon=\dfrac{B(A_{f}-A_{i})}{t}[/tex]

[tex]\epsilon=\dfrac{B(\pi r_{f}^2-\pi r_{i}^2)}{t}[/tex]

Put the value ino the formula

[tex]\epsilon=\dfrac{1.5(\pi\times(0.125)^2-\pi(0.075)^2)}{1.5}[/tex]

[tex]\epsilon=0.0314\ V[/tex]

Hence, The induced emf for this silver circle during this period of expansion is 0.0314 V.

The induced emf for the silver circle during this period of expansion is approximately [tex]\(\boxed{0.0314 \, \text{V}}\).[/tex]

The induced emf  in the silver circle during the expansion can be calculated using Faraday's law of induction, which states that the induced emf in a closed loop is equal to the negative rate of change of the magnetic flux through the loop. Mathematically, this is expressed as:

[tex]\[ \varepsilon = -\frac{d\Phi_B}{dt} \][/tex]

 The initial area [tex]\( A_i \)[/tex] when the radius [tex]\( r_i \[/tex]) is 0.075 m is:

[tex]\[ A_i = \pi r_i^2 = \pi (0.075 \, \text{m})^2 \][/tex]

 The final area [tex]\( A_f \)[/tex] when the radius [tex]\( r_f \)[/tex] is 0.125 m is:

[tex]\[ A_f = \pi r_f^2 = \pi (0.125 \, \text{m})^2 \][/tex]

 The change in area [tex]\( \Delta A \)[/tex] is:

[tex]\[ \Delta A = A_f - A_i = \pi (0.125 \, \text{m})^2 - \pi (0.075 \, \text{m})^2 \][/tex]

 The time taken for this change [tex]\( \Delta t \)[/tex] is 1.5 s.

 The induced emf [tex]\( \varepsilon \)[/tex] is then:

[tex]\[ \varepsilon = -\frac{d\Phi_B}{dt} = -B \frac{\Delta A}{\Delta t} \][/tex]

[tex]\[ \varepsilon = -1.5 \, \text{T} \times \frac{\pi ((0.125 \, \text{m})^2 - (0.075 \, \text{m})^2)}{1.5 \, \text{s}} \][/tex]

 Now, let's calculate the numerical value:

[tex]\[ \varepsilon = -1.5 \times \pi \times \frac{(0.125^2 - 0.075^2)}{1.5} \][/tex]

[tex]\[ \varepsilon = -1.5 \times \pi \times \frac{(0.015625 - 0.005625)}{1.5} \][/tex]

[tex]\[ \varepsilon = -1.5 \times \pi \times \frac{0.01}{1.5} \][/tex]

[tex]\[ \varepsilon = -1.5 \times \pi \times 0.00666\ldots \][/tex]

[tex]\[ \varepsilon \approx -0.0314 \, \text{V} \][/tex]

The negative sign indicates the direction of the induced emf according to Lenz's law, which is opposite to the direction of the change in magnetic flux. However, the magnitude of the induced emf is approximately 0.0314 V.

Therefore, the induced emf for the silver circle during this period of expansion is approximately [tex]\(\boxed{0.0314 \, \text{V}}\).[/tex]

Answer:[tex]\frac{F}{2}[/tex]

Explanation:

Given

weight of Object is W

Buoyant Force [tex]F_b=F[/tex]

let the density of First Liquid is  [tex]\rho [/tex]

volume of object [tex]=\frac{W}{\rho _0g}[/tex]

where [tex]\rho _0[/tex]=object density

thus buoyant Force in first Liquid is

[tex]F=\rho \times \frac{W}{\rho _0g}\times g=\frac{W\rho }{\rho _0}[/tex]

For Second Liquid

[tex]F'=\frac{\rho }{2}\times \frac{W}{\rho _0g}\times g=\frac{W\rho }{2\rho _0}[/tex]

[tex]F'=\frac{F}{2}[/tex]

To develop this problem it is necessary to apply the concepts related to the calculation of the Force through density and volume as well as the ideal gas law.

By definition, force can be expelled as

F = ma

Where,

m = mass

a = Acceleration

At the same time  the mass can be defined as function of density and Volume

[tex]m = \rho V[/tex]

Therefore if we do a sum in the spherical balloon we have,

[tex]\sum F = 0[/tex]

[tex]F_w +F_h-F_b=0[/tex]

Where,

[tex]F_W[/tex]= Force by weight of balloon

[tex]F_h[/tex]= Force by weight of helium gas

[tex]F_b[/tex]= Buoyant force

[tex]mg + V \rho g - V\rho_a g = 0[/tex]

Re-arrange to find [tex]\rho,[/tex]

[tex]\rho = \rho_a - \frac{m}{V}[/tex]

Our values are given as,

[tex]r= 1.55m[/tex]

[tex]V = \frac{4}{3} \pi r^3[/tex]

[tex]V = \frac{4}{3} \pi (1.55)^3[/tex]

[tex]V = 15.59m^3[/tex]

Replacing the values we have,

[tex]\rho = 1.19kg/m^3 - \frac{2.7}{15.59}[/tex]

[tex]\rho = 1.0168kg/m^3[/tex]

Applying the ideal gas law we have finally that

[tex]P = \frac{\rho}{M_0} RT[/tex]

Where,

P = Pressure

[tex]\rho =[/tex] Density

M_0 Molar mass (0.004Kg/mol for helium)

R= Gas constant

T = Temperature

Substituting

[tex]P = \frac{1.0168}{0.004} *8.314*290[/tex]

[tex]P = 612891.452Pa[/tex]

[tex]P = 0.613Mpa[/tex]

Therefore the absolute pressure of the helium gas is [tex]0.613Mpa[/tex]

When an ice cube, which floats with 90% submerged in water, is placed in a liquid that is less dense than water, less than 90% of the cube will be submerged. This is due to the interplay between the densities of the ice cube and the liquid.

The extent to which an object is submerged in a fluid depends on the density of both the object and the fluid. In the case of an ice cube in water, the ice cube is less dense than the water, causing about 90% of it to be submerged under the surface. Now if you place the same ice cube in a liquid that is less dense than water, less than 90% of the ice cube will be submerged. This is because the ice cube is denser than this new liquid, and it displaces less liquid to balance its weight.

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Answer:

a. 4d.

If the car travels at a velocity of 2v, the minimum stopping distance will be 4d.

Explanation:

Hi there!

The equations of distance and velocity of the car are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x =  position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

Let´s find the time it takes the car to stop using the equation of velocity. When the car stops, its velocity is zero. Then:

velocity = v0 + a · t      v0 = v

0 = v + a · t

Solving for t:

-v/a = t

Since the acceleration is negative because the car is stopping:

v/a = t

Now replacing t = v/a in the equation of position:

x = x0 + v0 · t + 1/2 · a · t²     (let´s consider x0 = 0)

x = v · (v/a) + 1/2 · (-a) (v/a)²    

x = v²/a - 1/2 · v²/a

x = 1/2 v²/a

At a velocity of v, the stopping distance is 1/2 v²/a = d

Now, let´s do the same calculations with an initial velocity v0 = 2v:

Using the equation of velocity:

velocity = v0 + a · t

0 = 2v - a · t

-2v/-a = t

t = 2v/a

Replacing in the equation of position:

x1 = x0 + v0 · t + 1/2 · a · t²  

x1 = 2v · (2v/a) + 1/2 · (-a) · (2v/a)²

x1 = 4v²/a - 2v²/a

x1 = 2v²/a

x1 = 4(1/2 v²/a)

x1 = 4x

x1 = 4d

If the car travels at a velocity 2v, the minimum stopping distance will be 4d.

Answer:

4d

Explanation:

To solve this problem it is necessary to apply the concepts related to Kepler's second law and the conservation of angular momentum.

Kepler's second law tells us that the vector radius that unites a planet and the sun sweeps equal areas at equal times, that is, when the planet is farther from the sun, the speed at which it travels is less than when it is close to the sun.

The angular momentum is defined as

[tex]L = m*r*v[/tex]

Where,

m= mass

r = Radius

v = Velocity

For conservation of angular momentum

[tex]L_{apogee}=L_{perigee}[/tex]

[tex]mv_a*r_a = mv_p*r_p[/tex]

[tex]v_a*r_a= v_p*r_p[/tex]

[tex]r_a = \frac{v_p*r_p}{v_a}[/tex]

[tex]r_a = \frac{(4.46)(2.23*10^4)}{(3.54)}[/tex]

[tex]r_a = 2.81*10^4km[/tex]

Therefore the corresponding distance at apogee is [tex]2.81*10^4km[/tex]

To solve this problem we must resort to the Work Theorem, internal energy and Heat transfer. Summarized in the first law of thermodynamics.

[tex]dQ = dU + dW[/tex]

Where,

Q = Heat

U = Internal Energy

By reference system and nomenclature we know that the work done ON the system is taken negative and the heat extracted is also considered negative, therefore

[tex]W = -168J \righarrow[/tex]  Work is done ON the system

[tex]Q = -305.6J \rightarrow[/tex] Heat is extracted FROM the system

Therefore the value of the Work done on the system is -158.0J

Answer:

0.7274 m

Explanation:

X denotes the distance from the reference point

M denotes the mass of the point

The center of mass of a system is given by

[tex]X_{CM}=\sum_{n=1}^n\frac{X_n\times M_n}{M_n}[/tex]

Taking the position of Juliet as the reference point

[tex]X_{CM}=\frac{59\times 0+85\times 2.7+75\times 1.35}{59+85+75}\\\Rightarrow X_{CM}=1.51027\ m[/tex]

After the boat has moved the distance between the center of mass of the boat and the shore remains the same. The change is observed relative to the rear of the boat

[tex]X_{CM}=\frac{75\times 1.35+(85+59)\times 2.7}{59+85+75}\\\Rightarrow X_{CM}=2.23767\ m[/tex]

The displacement of the boat towards the shore is

[tex]2.23767-1.51027=0.7274\ m[/tex]

By applying the conservation of momentum for a system with no external forces, the distance the boat moves when Juliet goes to the rear to kiss Romeo is 2.124 meters toward the shore.

The question involves the concept of conservation of momentum, particularly in systems with no external forces. Here, we consider a scenario where Juliet transitions from the front to the rear of the boat, causing the boat to move toward shore. To solve for the movement of the boat toward the shore, we need to realize that the total momentum of the system (Juliet, Romeo, and the boat) before and after Juliet's movement must remain constant, as there is no external force in the horizontal direction.

Initially, both Juliet and the boat are at rest, so the total initial momentum is zero. When Juliet, having mass 59.0 kg, moves a distance of 2.70 m towards Romeo, the boat will move in the opposite direction to conserve momentum.

To find the distance the boat moves, we use the conservation of momentum equation: (mass of Juliet) x (distance Juliet moves) = (mass of boat) x (distance boat moves). Rearranging gives us the distance the boat moves: distance boat moves = (mass of Juliet x distance Juliet moves) / mass of boat.

By substituting the given values, we get:

distance boat moves = (59.0 kg x 2.70 m) / 75.0 kg
distance boat moves = 2.124 m

Therefore, the boat moves 2.124 meters toward the shore when Juliet moves to the rear to kiss Romeo.

Answer:

a) 2.1 × 10^-6 m b) 1.3 × 10^-4 m

Explanation:

The thermal expansion of super Invar 0.20 ×10^-6oC^-1 and the length of Super is 1.9 m

using the linear expansivity  formula

ΔL = αLoΔt where ΔL is the change in length meters, α is the linear expansivity of Super Invar in oC^-1 and Δt  is the change in temperature in oC. Substituting the value into the formula gives

ΔL = 0.2 × 10 ^-6 × 1.9 m ×5.5 = 2.1 × 10^-6 meters to two significant figure

b) the thermal expansion of steel is 1.3 × 10^-5 oC^-1 and the length of steel is 1.8 m

using the formula

ΔL of steel = 1.3 × 10^-5 × 1.8 ×  5.5 = 1.3 × 10^-4 m to two significant figure.

Answer:

Positive.

Explanation:

As a consequence of the photoelectric effect, electrons that will get hit by sufficiently energetic photons will abandon the metal surfaces exposed to bright sunlight. This decreases the negative charge of the surface, thus causing it to develop a positive net charge.

Answer:

2d

Explanation:

For any instance equivalent force acting on the body is

[tex]mg-kd= m\frac{d}{dt}\frac{dx}{dt}[/tex]

Where

m is the mass of the object

k is the force constant of the spring

d is the extension in the spring

and

d/dt(dx/dt)=  is the acceleration of the object

solving the above equation we get

[tex]x= Asin\omega t +d[/tex]

where

[tex]\omega= \sqrt{\frac{k}{m} } = \frac{2\pi}{T}[/tex]

A is the amplitude of oscillation from the mean position.

k= spring constant , T= time period

Here  we are assuming  that at t=T/4

x= 0   since, no extension in the spring

then

A=- d

Hence

x=- d sin wt + d

now, x is maximum when sin wt=- 1

Therefore,

x(maximum)=2d

Final answer:

When an object falls onto a spring, it will stretch the spring further than if it were slowly lowered due to kinetic energy turning into additional elastic potential energy. The maximum stretch can be calculated using energy conservation, resulting in a stretch that is √(2mgd/k) where d is the equilibrium stretch, m is the mass, g is gravity, and k is the spring constant.

Explanation:

Maximum Stretch of a Spring with a Falling Object

When an object of mass m is attached to a vertical spring and slowly lowered to its equilibrium position, the spring stretches by a distance d, producing potential energy in the spring equal to k·d²/2. Meanwhile, the work done by gravity equals mg·d. If instead the object is allowed to fall and attaches to the spring, the spring will stretch further due to the kinetic energy acquired during the fall, turning into additional elastic potential energy at maximum stretch.

If we assume no energy losses due to non-conservative forces, the energy conservation equation just before the object attaches to the spring would be:

Kinetic energy + gravitational potential energy = elastic potential energy at maximum extension.

Since the object starts from rest at a height d above the equilibrium position:

1/2·mv² + mgd = 1/2·k·x² (with x being the total distance the spring stretches).

At the point of maximum stretch, all kinetic energy has been converted into elastic potential energy, and the object is momentarily at rest, so:

mgd = 1/2·k·x²

In the special case where there are no other forms of energy conversion (like heat due to air resistance), the spring will stretch maximum distance x satisfying:

x = √(2·mg·d / k)

Using the initial condition that the spring stretches by d when the object is slowly lowered, we can deduce that the object in free fall will stretch the spring more than d due to extra energy from its fall. The exact factor would depend on the stiffness of the spring (expressed by k) and the mass of the object (m).

Answer:

3.2 C

Explanation:

e = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

Number of synapses = [tex]10^{16}[/tex]

Signals transferred by each synapse 1000 calcium ions [tex]Ca^{2+}[/tex]

Each calcium ion carries 2 charges

Number of charges would be [tex]2\times 1000=2000[/tex]

Total charges would be

[tex]N=10^{16}\times 2000=2\times 10^{19}[/tex]

Charge is given by

[tex]Q=Ne\\\Rightarrow Q=2\times 10^{19}\times 1.6\times 10^{-19}\\\Rightarrow Q=3.2\ C[/tex]

The total electric charge transfer in coulombs during that short time interval is 3.2 C

Final answer:

In a fictional scenario where 10¹⁶ synapses simultaneously transmit signals using 1000 calcium ions each, the total electric charge transferred is 3.20 coulombs, which is significant but still less than the 5 coulombs typically involved in a lightning flash.

Explanation:

Children have about 1016 synapses that can transfer signals between neurons in the brain and between neurons and muscle cells. If these synapses simultaneously transmit a signal, sending 1000 calcium ions (Ca²⁺) across the membrane at each synaptic ending, and given that the charge of each Ca²⁺ is twice that of the electron charge (e = 1.60 × 10⁻¹⁹ C), we can calculate the total electric charge transfer in coulombs during that time interval.

The total number of calcium ions transferred is 1016 synapses × 1000 ions/synapse = 10¹⁹ ions. Since each Ca²⁺ carries twice the charge of an electron, the charge per ion is 2 × 1.60 × 10⁻¹⁹ C = 3.20 × 10⁻¹⁹ C. Therefore, the total charge transferred is 1019 ions × 3.20 × 10⁻¹⁹ C/ion = 3.20 coulombs.

Comparing this to a lightning flash, which involves about 5 coulombs of charge transfer, we can see that the total charge transfer in this hypothetical scenario is significant, yet smaller than that of a lightning flash.

The angular acceleration of the two flywheels will be different due to their differing moments of inertia, which depend on both the mass and its distribution relative to the rotational axis. The removed hexagonal sections on flywheel A alters its moment of inertia.

The answer to your question is: The angular acceleration of the two flywheels is different but it is impossible to tell which is greater.

The reason for this is tied to the concept known as moment of inertia, an object's resistance to changes in its state of rotation. The moment of inertia depends not only on the mass of an object, but also on its distribution of mass relative to the axis around which it rotates. The hexagonal sections removed from flywheel A affect its distribution of mass, potentially lowering its moment of inertia.

Angular acceleration is the rate of change of angular velocity over time, and is given by the equation: torque = moment of inertia x angular acceleration. The torque on both flywheels is the same, since the same force is exerted on both through the identical masses falling from height h. Therefore, if the moments of inertia of the two flywheels are different, then their angular accelerations must also be different as they both satisfy the above equation.

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The angular acceleration of the two flywheels is different but it is impossible to tell which is greater.

The angular acceleration of the two flywheels is different but it is impossible to tell which is greater. The angular acceleration of a flywheel depends on both its mass and its moment of inertia, which takes into account the distribution of mass relative to the axis of rotation.

In this case, the larger mass of flywheel A would tend to result in a smaller angular acceleration, but the removal of material in hexagonal sections could change the moment of inertia and potentially increase the angular acceleration. Without specific information about the moment of inertia for each flywheel, it is impossible to determine which one would have a greater angular acceleration.

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Answer:

(B) You should catch the hall

Explanation:

B) You should catch the ball.

case 1 : when ball is catched.

Mass of person catching = M

mass of ball = m

velocity of the ball just before catching = v

velocity of ball just after catching = 0

[tex]\Delta P_{ball}[/tex] = change in momentum of the ball = m (0 - v) = - mv

since,

[tex]\Delta P_{person} =-\Delta P_{ball}[/tex]

[tex]\Delta P_{person} =mv[/tex]

case 2 : when the ball is deflected back

velocity of the ball just before catching = v

velocity of ball just after catching = - v

[tex]\Delta P_{ball}[/tex] = change in momentum of the ball = m (- v - v) = - 2mv

since,  

[tex]\Delta P_{person} =- \Delta P_{ball}[/tex]

[tex]\Delta P_{person} = 2mv[/tex]

clearly we can see that the change in momentum is minimum in case when the ball is catched and hence keeps the speed of skateboard minimum

Answer:

 ΔL = 1.43 10 -9 m

Explanation:

a) Let's start from Newton's second law, the force in a spring is elastic

    F = - k Δx

Let's divide the two sides by the area

    F / A = -k Δx / A

In general area is long by wide, the formulated pressure is

    P = F / A

    P = - k Δx / l x

    P = (-k / l) Δx/x

Call us at Young's constant module

    P = E Δx / x

Let's change x for L

   E = P / (ΔL/L)

b) we cleared

   ΔL = P L / E

Let's reduce the magnitudes to the SI system

   E = 70 GPa = 70 109 Pa

   L = 100mm (1m / 1000mm) = 0.100m

   P = 1 kN = 1 103 N

   ΔL = 1 103  0.100/ 70 109

   ΔL = 1.43 10 -9 m

Answer:

a. gender intensification

Explanation:

This theory states that in the adolescence both girls and boys are pressured to follow gender roles established culturally.

To solve the problem it is necessary to apply the concepts related to the calculation of discharge flow, Bernoulli equations and energy conservation in incompressible fluids.

PART A) For the calculation of the velocity we define the area and the flow, thus

[tex]A = \pi r^2[/tex]

[tex]A = pi (2*10^{-3})^2[/tex]

[tex]A = 12.56*10^{-6}m^2[/tex]

At the same time the rate of flow would be

[tex]Q = \frac{1L}{2s}[/tex]

[tex]Q = 0.5L/s = 0.5*10^{-3}m^3/s[/tex]

By definition the discharge is expressed as

[tex]Q = NAv[/tex]

Where,

A= Area

v = velocity

N = Number of exits

Q = NAv

Re-arrange to find v,

[tex]v = \frac{Q}{NA}[/tex]

[tex]v = \frac{0.5*10^{-3}}{44*12.56*10^{-6}}[/tex]

[tex]v = 0.9047m/s[/tex]

PART B) From the continuity equations formulated by Bernoulli we can calculate the speed of water in the pipe

[tex]P_1 + \frac{1}{2}\rho v_1^2+\rho gh_1 = P_2 +\frac{1}{2}\rho v^2_2 +\rho g h_2[/tex]

Replacing with our values we have that

[tex]1.5*10^5 + \frac{1}{2}(1000) v_1^2+(1000)(9.8)(0) = 10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)[/tex]

[tex]v_1^2 = 10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)-1.5*10^5 - \frac{1}{2}(1000)[/tex]

[tex]v_1 = \sqrt{10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)-1.5*10^5 - \frac{1}{2}(1000)}[/tex]

[tex]v_1 = 3.54097m/s[/tex]

PART C) Assuming that water is an incomprehensible fluid we have to,

[tex]Q_{pipe} = Q_{shower}[/tex]

[tex]v_{pipe}A_{pipe}=v_{shower}A_{shower}[/tex]

[tex]3.54097*A_{pipe}=0.9047*12.56*10^{-6}[/tex]

[tex]A_{pipe} = \frac{0.9047*12.56*10^{-6}}{3.54097}[/tex]

[tex]A_{pipe = 3.209*10^{-6}m^2[/tex]

The answer provides calculations for the speed of water from the shower head, the speed in the connected pipe, and the cross-sectional area of the pipe based on the given parameters.

This question asks for the rate of heat transfer from a copper fin, which requires the application of heat transfer theory, including the use of specific equations for fin efficiency and heat loss. However, a sample calculation provided deals with finding the initial temperature of a copper piece mixed with water, based on the conservation of energy and specific heat capacities.

The subject of this question involves the calculation of heat transfer from a fin made of copper. Given the material's thermal conductivity (k), the fin's dimensions, and the conditions surrounding it (base temperature, air temperature, and convection heat transfer coefficient h), we can find the rate of heat transfer. However, the actual calculation for the rate of heat transfer from this fin isn't provided directly in the context of the question. For such a problem, one would typically employ the fin equations from heat transfer theory, which might involve the use of Biot number, fin efficiency, and heat loss calculations. Instead, to answer a closely related question with given data, we can calculate the initial temperature of a 248-g piece of copper when dropped into 390 mL of water at 22.6°C, where the final temperature is measured as 39.9°C, assuming all heat transfer occurs between the copper and the water. This problem relies on the conservation of energy principle, requiring the application of the specific heat capacities of copper and water

Answer:

I will do more work against friction going around the floor.

ΔW = 5 J

Explanation:

Given

Ff₁ = 40 N

Ff₂ = 55 N

a = 3 m

b = 4 m

We have to get the distance d₁ as follows

d₁ = a + b = 3 m + 4 m = 7 m

And  d₂:

d₂ = √(a² + b²) = √((3 m)² + (4 m)²)

⇒   d₂ = 5 m

then we use the equations

W₁ = Ff₁*d₁ = (40 N)(7 m) = 280 J   (work against friction going around the floor)

W₂ = Ff₂*d₂ = (55 N)(5 m) = 275 J  (work against friction going across the rug)

since  W₂ < W₁  I will do more work against friction going around the floor.

Now, we apply the formula

ΔW = W₁ - W₂ = 280 J - 275 J = 5 J

When pushing the heavy box from one corner of the rug to the opposite corner, you will do more work against friction going across the rug than going around the floor. The extra work against friction across the rug is calculated to be 375 J.

When pushing the heavy box from one corner of the rug to the opposite corner, you will do more work against friction going across the rug. The force of friction on the rug is greater than the force of friction on the smooth concrete floor. To calculate the extra work, you can use the formula:

Extra work = (force of friction on the rug - force of friction on the floor) x distance

Extra work = (55 N - 40 N) x (diagonal distance of the rug)

Since the rug is 3 m by 4 m, the diagonal distance can be calculated using the Pythagorean theorem:

diagonal distance = √(3^2 + 4^2) = √(9 + 16) = √25 = 5 m

Therefore, the extra work against friction going across the rug is (55 N - 40 N) x 5 m = 75 N x 5 m = 375 J.

Answer:

  d. They both have the same change in magnitude of momentum because momentum is conserved.                

Explanation:

      We know that if there is no any external force on the system the linear     momentum of the system remain conserve.Or in the other other words change in the linear momentum is zero.

    Here the external force is zero ,that is why initial and the final momentum will be conserved.

     change in the momentum  for both car as well as truck will be same.Because momentum should be conserved.

Therefore the answer is -d

The statement that best describes the situation is that they both have the same change in magnitude of momentum because momentum is conserved. Option D is correct

Momentum has to do with the strength an object has when moving with speed.

For the truck and the small car given in the question, there was no external force acting on the system.

Since there was no force acting on the system, we can conclude that the linear momentum of the system is conserved that is there is no change in the linear momentum of the system.

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Answer:

The net torque about the pivot is and the answer is 'c'

c. [tex]T_{net}=8.58[/tex]

Explanation:

[tex]T=F*d[/tex]

The torque is the force apply in a distance so it is the moment so depends on the way to be put it the signs so:

[tex]T_1=F_1*d_1[/tex]

[tex]T_1=7.8N*1.6m=12.48N*m[/tex]

[tex]T_2=F_2*d_2[/tex]

[tex]T_2=2.60N**cos(30)*3.0m[/tex]

[tex]T_2= - 3.9 N*m[/tex]

Now to find the net Torque is the summation of both torques

[tex]T_{net}=T_1+T_2[/tex]

[tex]T_{net}=12.48N-3.9N=8.58N[/tex]

The net torque about the pivot is calculated by finding the individual torques caused by each force and then subtracting the clockwise torque from the counterclockwise torque, resulting in a positive net torque of 8.58 N·m. So the correct option is c.

The student is asking to calculate the net torque about the pivot on a rod being acted upon by two forces causing opposing torques. To find the net torque, we need to calculate each torque and then subtract the clockwise (cw) torque from the counterclockwise (ccw) torque. Torque (τ) can be calculated using the formula τ = rFsin(θ), where r is the distance from the pivot, F is the force applied, and θ is the angle at which the force is applied relative to the rod.

Step-by-step calculation:

Since the counterclockwise (ccw) torque is the positive direction, the net torque of 8.58 N·m is positive, giving us option c) 8.58 N·m as the correct answer.

Answer:

The number of available energy states per unit volume is [tex]4.01\times10^{48}[/tex]

Explanation:

Given that,

Average energy  [tex]E=2\times10^{-4}\ eV[/tex]

Photon = [tex]4\times10^{-5}\ eV[/tex]

We need to calculate the number of available energy states per unit volume

Using formula of energy

[tex]g(\epsilon)d\epsilon=\dfrac{8\pi E^2dE}{(hc)^3}[/tex]

Where, E = energy

h = Planck constant

c = speed of light

Put the value into the formula

[tex]g(\epsilon)d\epsilon=\dfrac{8\times\pi\times2\times10^{-4}\times4\times10^{-5}\times1.6\times10^{-19}}{(6.67\times10^{-34}\times3\times10^{8})^3}[/tex]

[tex]g(\epsilon)d\epsilon=4.01\times10^{48}[/tex]

Hence, The number of available energy states per unit volume is [tex]4.01\times10^{48}[/tex]

Answer:

the length of stretched spring  in cm is 22

Explanation:

given information:

spring length, x1 = 20 cm = 0.2 m

force, F = 100 N

the length of spring streches, x2 = 22 cm = 0.22 m

According to Hooke's law

F = - kΔx

k = F/*=(x2-x1)

  = 100/(0.22 - 0.20)

  = 5000 N/m

if the spring is now suspended from a hook and a 10.2-kg block is attached to the bottom end

m = 10.2 kg

W = m g

    = 10.2 x 9.8

    = 99.96 N

F = - k Δx

Δx = F / k

     = 99.96 / 5000

     = 0.02

Δx = x2- x1

x2 = Δx + x1

    = 0.20 + 0.02

    = 0.22 m

     = 22 cm

Answer:

a)Uo= 2 m/s

b)[tex]u_{avg}=1 \ m/s[/tex]

c)Q=7.54  x 10⁻⁴ m³/s

Explanation:

Given that

[tex]u(r)=2\left(1-\dfrac{r^2}{R^2}\right)[/tex]

Diameter ,D= 3.1 cm

Radius ,R= 1.55 cm

We know that in the pipe flow the general equation for laminar fully developed flow given as

[tex]u(r)=U_o\left(1-\dfrac{r^2}{R^2}\right)[/tex]

Uo=Maximum velocity

Therefore maximum velocity

Uo= 2 m/s

The average velocity

[tex]u_{avg}=\dfrac{U_o}{2}[/tex]

[tex]u_{avg}=\dfrac{2}{2}\ m/s[/tex]

[tex]u_{avg}=1 \ m/s[/tex]

The volume flow rate

[tex]Q=u_{avg}. A[/tex]

[tex]Q=\pi R^2\times u_{avg}\ m^3/s[/tex]

[tex]Q=\pi \times (1.55\times 10^{-2})^2\times 1\ m^3/s[/tex]

Q=0.000754 m³/s

Q=7.54  x 10⁻⁴ m³/s

Answer:

C = 3.77*10⁻¹⁰ F = 377 pF

Q = 1.13*10⁻⁵ C

Explanation:

Given

D = 8.0 cm = 0.08 m

d = 0.95 cm = 0.95*10⁻² m

k = 80.4  (dielectric constant of the milk)

V = 30000 V

C = ?

Q = ?

We can get the capacitance of the system applying the formula

C = k*ε₀*A / d

where

ε₀ = 8.854*10⁻¹² F/m

and   A = π*D²/4 = π*(0.08 m)²/4

⇒  A = 0.00502655 m²

then

C = (80.4)*(8.854*10⁻¹² F/m)*(0.00502655 m²) / (0.95*10⁻² m)

⇒  C = 3.77*10⁻¹⁰ F = 377 pF

Now, we use the following equation in order to obtain the charge on each plate when they are fully charged

Q = C*V

⇒  Q = (3.77*10⁻¹⁰ F)*(30000 V)

⇒  Q = 1.13*10⁻⁵ C

The capacitance of the system is calculated by applying the physics of parallel-plate capacitors, using characteristics of the plates and the milk as the dielectric constant. The charge on each plate when fully charged is then deduced from the calculated capacitance and the applied voltage.

The concept this question revolves around is the electrical mechanism in a parallel-plate capacitor. This system uses the property of the dielectric constant (the same as water for milk, in this case) of the substance between the plates.

The capacitance of such a system may be calculated as C = ε * (A/d), where ε denotes the permittivity of the substance (milk), A is the area of one of the plates, and d is the distance separating the plates. For a circular plate, the area can be computed as A = π * (D/2)^2 where D is the diameter of the plate.

Here, ε = 8.854*10^-12 F/m (for water), D = 8 cm = 0.08 m (converted into meters for unit consistency), and d = 0.95 cm = 0.0095 m. Substituting these values, we compute the capacitance C which should then be used to calculate the charge on each plate when fully charged using the formula, Q = C * V, where V is the voltage applied.

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To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

By definition kinetic energy is defined as

[tex]KE = \frac{1}{2} mv^2[/tex]

Where,

m = Mass

v = Velocity

On the other hand we have the conservation of the moment, which for this case would be defined as

[tex]m*V_i = m_1V_1+m_2V_2[/tex]

Here,

m = Total mass (8Kg at this case)

[tex]m_1=m_2 =[/tex] Mass each part

[tex]V_i =[/tex] Initial velocity

[tex]V_2 =[/tex] Final velocity particle 2

[tex]V_1 =[/tex] Final velocity particle 1

The initial kinetic energy would be given by,

[tex]KE_i=\frac{1}{2}mv^2[/tex]

[tex]KE_i = \frac{1}{2}8*5^2[/tex]

[tex]KE_i = 100J[/tex]

In the end the energy increased 100J, that is,

[tex]KE_f = KE_i KE_{increased}[/tex]

[tex]KE_f = 100+100 = 200J[/tex]

By conservation of the moment then,

[tex]m*V_i = m_1V_1+m_2V_2[/tex]

Replacing we have,

[tex](8)*5 = 4*V_1+4*V_2[/tex]

[tex]40 = 4(V_1+V_2)[/tex]

[tex]V_1+V_2 = 10[/tex]

[tex]V_2 = 10-V_1[/tex](1)

In the final point the cinematic energy of EACH particle would be given by

[tex]KE_f = \frac{1}{2}mv^2[/tex]

[tex]KE_f = \frac{1}{2}4*(V_1^2+V_2^2)[/tex]

[tex]200J=\frac{1}{2}4*(V_1^2+V_2^2)[/tex](2)

So we have a system of 2x2 equations

[tex]V_2 = 10-V_1[/tex]

[tex]200J=\frac{1}{2}4*(V_1^2+V_2^2)[/tex]

Replacing (1) in (2) and solving we have to,

200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)

PART A: [tex]V_1 = 10m/s[/tex]

Then replacing in (1) we have that

PART B: [tex]V_2 = 0m/s[/tex]