A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 9.1 m/s. After the collision, the 0.20-kg puck has a speed of 5.5 m/s at an angle of θ = 53° to the positive x-axis. (a) Determine the velocity of the 0.30-kg puck after the collision, (b) Find the fraction of kinetic energy lost in the collision.

Answer:

B. Movement of regolith down a slope under the force of gravity

It is also known as slope movement or mass movement.

Answer:

Explanation:

(a) Work done, W = 1.82 x 10^4 J

(b) internal energy, U = - 4.07 x 10^4 J ( as it decreases)

(c) According to the first law of thermodynamics  

Q = W + U

Q = 1.82 x 10^4 - 4.07 x 10^4

Q = - 2.25 x 10^4 J

The work done (W), the change in internal energy (ΔU), and heat transferred (Q) are 1.82 x 10^4 J, -4.07 x 10^4 J, and -2.25 x 10^4 J respectively as per the First Law of Thermodynamics.

The values listed in the question are for the work done (W), and the change in internal energy (U). In physics, these concepts are investigated under a principle called the First Law of Thermodynamics, which is basically a version of the law of conservation of energy as applied for thermal processes. As per this law, the change in internal energy (U) of a system is equal to the heat added to the system (Q) minus the work done by the system (W).

Hence, the formula can be given as: ΔU = Q - W.

If we substitute the given values:
-W = Work done = 1.82 x 10^4 J and ΔU = internal energy decreases = -4.07 x 10^4 J

Now, the formula can be rewritten as:
:-Q = ΔU + W = -4.07 x 10^4 J + 1.82 x 10^4 J = -2.25 x 10^4 J

Values:
(a) W = Work done = 1.82 x 10^4 J (Work done by the system is positive.)
(b) ΔU = Change in internal energy = -4.07 x 10^4 J (Decrease in internal energy is negative.)
(c) Q = Heat transferred = -2.25 x 10^4 J (As per the convention, heat added to the system is positive. Here, it's negative, which means heat is lost from the system.)

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Answer:

(a)1.57 kg

(b) 281.17 N/m

(c) 201 micrometer

(d) [tex]2.69\times 10^{-3}m/sec[/tex]

Explanation:

We have given that value of inductor L = 1.57 Henry

Inductive energy [tex]E=5.76\mu j=5.76\times 10^{-6}J[/tex]

Maximum charge [tex]Q=201\mu C=201\times 10^{-6}C[/tex]

(A) In electrical mechanical system mass corresponds to inductance

So mass will be m = 1.57 kg

(B) We have given energy [tex]E=\frac{Q^2}{2C}[/tex]

[tex]C=\frac{Q^2}{2E}=\frac{(201\times 10^{-6})^2}{2\times 5.7\times 10^{-6}}=3543.94\times 10^{-6}[/tex]

In electrical mechanical system spring constant is equivalent to [tex]\frac{1}{C}[/tex]

So spring constant [tex]k=\frac{1}{C}=\frac{1}{3543.94\times 10^{-6}}=282.17N/m[/tex]

(c) Displacement is equivalent to maximum charge

So displacement will be [tex]x=201\mu m[/tex]

(d) Maximum speed is correspond to maximum current

As maximum current [tex]i_m=\frac{Q}{\sqrt{LC}}=\frac{201\times 10^{-6}}{\sqrt{1.57\times 3543.94\times 10^{-6}}}=2.69\times 10^{-3}A=2.69\times 10^{-3}m/sec[/tex]

Answer:

[tex]v=28m/s[/tex]

Explanation:

The vertical component of the normal force must cancel out with the weight of the car taking the curve:

[tex]N_y=W[/tex]

[tex]Ncos\theta=mg[/tex]

(Notice it has to be cos and not sin, because the angle [tex]\theta[/tex] is the slope, for null slope [tex]N_y=Ncos(0)=N[/tex], as it should be).

The horizontal component of the normal force must be the centripetal force, that is:

[tex]N_x=F_{cp}[/tex]

[tex]Nsen\theta=ma_{cp}[/tex]

[tex](\frac{mg}{cos\theta})sin\theta=m\frac{v^2}{r}[/tex]

[tex]gtan\theta=\frac{v^2}{r}[/tex]

[tex]v=\sqrt{grtan\theta}=\sqrt{(9.8m/s^2)(80m)tan(45^{\circ})}=28m/s[/tex]

To find the safe speed on a banked curve without sliding up or down, calculate the square root of a specific formula involving the radius, acceleration due to gravity, and the tangent of the angle.

To find the safe speed with which to take the curve without sliding up or down, we need to consider the force components acting on the vehicle. The gravitational force can be split into two components: the vertical component and the horizontal component. The vertical component helps keep the vehicle on the road, while the horizontal component provides the centripetal force needed to keep the vehicle moving in a curved path. By setting up an equation involving the gravitational force, the normal force, and the friction force, we can calculate the safe speed as the square root of the product of the radius of the curve, the acceleration due to gravity, and the tangent of the banking angle.

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Answer: 28

Explanation:

According to Newton's law of Gravitation the gravitational force on the surface of the Sun  [tex]Fg_{s}[/tex] is:

[tex]Fg_{s}=mg_{sun}=G\frac{Mm}{r^{2}}[/tex] (1)

Where:

[tex]m[/tex] is your mass

[tex]g_{sun}[/tex] is the acceleration due gravity on the surface of the Sun

[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the gravitational constant

[tex]M=1.99(10)^{30} kg[/tex] is the mass of the Sun

[tex]r=6.96(10)^{8} m[/tex] is the radius of the Sun

Simplifying:

[tex]g_{sun}=G\frac{M}{r^{2}}[/tex] (2)

[tex]g_{sun}=6.674(10)^{-11} \frac{m^{3}}{kgs^{2}} \frac{1.99(10)^{30} kg}{(6.96(10)^{8} m)^{2}}[/tex] (3)

[tex]g_{sun}=274 m/s^{2}[/tex] (4)

Since the acceleration due gravity on Earth is [tex]g_{E}=9.8 m/s^{2}[/tex], the relation is:

[tex]\frac{g_{sun}}{g_{E}}=\frac{274 m/s^{2}}{9.8 m/s^{2}}[/tex]

[tex]\frac{g_{sun}}{g_{E}}=27.9 \approx 28[/tex]

Hence, your weight would increase by a factor of 28.

The acceleration of gravity on the surface of the Sun is approximately 274 m/s^2. If you could stand on the Sun, your weight would increase by a factor of approximately 274.

To calculate the acceleration of gravity on the surface of the Sun, we can use the formula for gravitational acceleration: g = GM/R^2. Given that the mass of the Sun (M) is 1.99 ✕ 10^30 kg, the radius of the Sun (R) is 6.96 ✕ 10^8 m, and the gravitational constant (G) is 6.67 ✕ 10^-11 N · m^2/kg^2, we can substitute these values into the formula to find the acceleration of gravity on the surface of the Sun. Using the formula, the acceleration of gravity on the surface of the Sun is approximately 274 m/s^2.

To calculate the factor by which your weight would increase if you could stand on the Sun, we can use the formula W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. On Earth, the acceleration due to gravity is approximately 9.8 m/s^2 and the weight of a 1.0-kg object is 9.8 N. Since the acceleration of gravity on the surface of the Sun is 274 m/s^2, we can calculate the weight on the Sun by multiplying the mass by the acceleration of gravity on the Sun. Dividing the weight on the Sun by the weight on Earth will give us the factor by which your weight would increase if you could stand on the Sun. Using this calculation, the weight on the Sun would be approximately 274 times greater than the weight on Earth.

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Answer:A) The wavelengths are:8cm for first harmonic

      5.3cm for second harmonic

      4cm for third harmonic

      3.2cm for fourth harmonic

B) The frequencies are:

44.25Hz for first harmonic

66.38Hz for second harmonic

88.5Hz for third harmonic

110.63Hz for fourth harmonic

C) The wavelengths are:

10.66Cm for first harmonic

6.4 cm for second harmonic

4.57cm for third harmonic

3.5cm for fourth harmonic

D) The frequencies are:

33.19Hz for first harmonic

55.31 Hz for second harmonic

77.44Hz for third harmonic

99.56Hz for fourth harmonic

Explanation: since the mouth is being modelled as an organ pipe, then we can say that an open mouth is a model of an organ pipe open at both ends.

For an open pipe all harmonics are possible but only odd harmonic's are possible in closed pipes. A closed mouth is a model of a closed pipe, since closed pipes are open at one end.

Answer:

[tex]I_s=5.8A[/tex]

Explanation:

Not considering any type of losses in the transformer, the input power in the primary is equal to the output power in the secondary:

[tex]P_p=P_s[/tex]

So:

[tex]V_p*I_p=V_s*I_s[/tex]

Where:

[tex]V_p=Voltage\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\V_s=Voltage\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil\\I_p=Current\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\I_s=Current\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil[/tex]

Solving for [tex]I_s[/tex]

[tex]I_s=\frac{V_p*I_p}{V_s}[/tex]

Replacing the data provided:

[tex]I_s=\frac{110*29}{550} =5.8A[/tex]

Answer:

d= 7.32 mm

Explanation:

Given that

E= 110 GPa

σ = 240 MPa

P= 6640 N

L= 370 mm

ΔL = 0.53

Area A= πr²

We know that  elongation due to load given as

[tex]\Delta L=\dfrac{PL}{AE}[/tex]

[tex]A=\dfrac{PL}{\Delta LE}[/tex]

[tex]A=\dfrac{6640\times 370}{0.53\times 110\times 10^3}[/tex]

A= 42.14 mm²

πr² = 42.14 mm²

r=3.66 mm

diameter ,d= 2r

d= 7.32 mm

Answer:

[tex]d=7.32\ mm[/tex]

Explanation:

Given:

We know,

Stress:

[tex]\sigma=\frac{F}{A}[/tex]

where: A = cross sectional area

Strain:

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

& by Hooke's Law within the elastic limits:

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

[tex]\therefore 110\times 10^3=\frac{F}{A}\div \frac{\Delta l}{l}[/tex]

[tex]\therefore 110\times 10^3=\frac{6640\times 4}{\pi.d^2}\div \frac{0.53}{370}[/tex]

where: d = diameter of the copper rod

[tex]d=7.32\ mm[/tex]

To fix a grandfather clock that runs fast, you should lower the weight on the pendulum (option D), as this will increase the pendulum's period and slow down the clock's ticking rate.

If you have a grandfather clock at home that runs fast, you should make the pendulum swing slower to correct the time. To do this, you would lower the weight on the pendulum. The period of a pendulum, which determines the tick-tock rate of the clock, is directly affected by the length of the pendulum, not the weight, with a longer pendulum resulting in a slower tick-tock. Therefore, the correct adjustment would be option D: Lower the weight.

Adding or removing mass does not affect the period, so options C and E would not be effective. While decreasing the amplitude of swing might have a very small effect to speed up the pendulum (option B), this is not the standard method for adjusting the timekeeping of a clock and would not provide a consistent solution. Thus, the most accurate way to adjust the clock is by changing the length of the pendulum directly by lowering the weight.

Answer:0.00133 cm

Explanation:

Given

change in temperature [tex]\Delta T=90-20=70^{\circ}C[/tex]

Thermal Expansion coefficient of brass [tex]\alpha =0.000019 /^{\circ}C[/tex]

thickness t=1 cm

change in thickness is given by

[tex]\Delta t=t\cdot \alpha \cdot \Delta T[/tex]

[tex]\Delta t=1\times 0.000019\times 70[/tex]

[tex]\Delta t=0.00133\ cm[/tex]

Answer:u=66.67 m/s

Explanation:

Given

mass of meteor [tex]m=2.5 gm\approx 2.5\times 10^{-3} kg[/tex]

velocity of meteor [tex]v=40km/s \approx 40000 m/s[/tex]

Kinetic Energy of Meteor

[tex]K.E.=\frac{mv^2}{2}[/tex]

[tex]K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}[/tex]

[tex]K.E.=2\times 10^6 J[/tex]

Kinetic Energy of Car

[tex]=\frac{1}{2}\times Mu^2[/tex]

[tex]=\frac{1}{2}\times 900\times u^2[/tex]

[tex]\frac{1}{2}\times 900\times u^2=2\times 10^6 [/tex]

[tex]900\times u^2=4\times 10^6[/tex]

[tex]u^2=\frac{4}{9}\times 10^4[/tex]

[tex]u=\frac{2}{3}\times 10^2[/tex]

[tex]u=66.67 m/s[/tex]

Answer:

v = 67 m/s

Explanation:

The meteor has a mass (m) of 2.5 g and a speed (v) of 40 km/s. In SI units:

2.5 g × (1 kg / 10³ g) = 2.5 × 10⁻³ kg

40 km/s × (10³ m / 1 km) = 4.0 × 10⁴ m/s

The kinetic energy (KE) is:

KE = 1/2 × m × v² = 1/2 × (2.5 × 10⁻³ kg) × (4.0 × 10⁴ m/s)² = 2.0 × 10⁶ J

A 900 kg compact car, with the same kinetic energy, must have the following speed.

KE = 1/2 × m × v²

2.0 × 10⁶ J = 1/2 × 900 kg × v²

v = 67 m/s

The average induced emf in the coil, when it is rotated in Earth's Magnetic field, is 15 Volts.

The question is about the change in magnetic flux, which induces an electromotive force (emf) in a loop according to Faraday's law. The formula to calculate the change in magnetic flux is ΔФ = B × A × N, where B is the magnetic field, A is the area, and N is the number of coil turns.

Given B = 6.0×10⁻⁵T, A = 25cm² = 2.5 × 10⁻³ m² (since 1 m² = 10,000 cm²), and N=1000 turns, the change in magnetic flux equals to 0.15 Wb. According to Faraday's law, which is |emf| = |dФ/dt|, and dФ/dt is the rate of change of magnetic flux, thus |emf| = |0.15 / 0.010|= 15 V. Therefore, the average emf induced in the coil is 15 Volts.

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Final answer:

The gravitational force between two objects can be calculated using Newton's Law of Universal Gravitation, and for a 35.0 kg object to experience a net force of zero, it must be at a point where gravitational pulls from surrounding masses are balanced.

Explanation:

To solve the student’s question regarding the gravitational forces, we will use Newton's Law of Universal Gravitation which states the gravitational force (F) between two masses (m1 and m2) is directly proportional to the product of their masses and inversely proportional to the square of the distance (r) between their centers. It is mathematically expressed as F = G * (m1 * m2) / r², where G is the gravitational constant (6.674 × 10-11 N·m²/kg²).

(a) To find the net gravitational force exerted on the 35.0 kg object placed midway between the two objects with masses of 130 kg and 430 kg and separated by 0.300 m, we calculate the gravitational force from each mass separately and then find the vector sum which in this case will just be the difference, as the object is placed in the middle.

(b) For the 35.0 kg object to experience a net force of zero, it has to be placed at a point where the gravitational force due to both masses equals each other. This can be found by setting the gravitational forces from each mass to the 35.0 kg object equal and solving for the distance from one of the masses. There is only one point between the two masses where this can occur as per the symmetry of the system.

To solve the problem it is necessary to apply the concepts related to Chvorinov's Law, which states that

[tex]\frac{T_r}{T_c} = (\frac{V_r}{V_c}*\frac{SA_c}{SA_r})^2[/tex]

Where,

[tex]V_c[/tex] = Volume cube

[tex]SA_c[/tex] = Superficial Area from Cube

[tex]V_r[/tex] = Volume Rectangle

[tex]SA_r[/tex]= Superficial Area from Rectangle

Our values are given as (I will try to develop the problem in English units for ease of calculations),

[tex]V_c = 4^3 = 64in^3[/tex]

[tex]SA_c = 6*4^2=96in^2[/tex]

[tex]SA_r = 2*(1*16+4*1+16*4)=168in^2[/tex]

[tex]V_r = 4*1*16 = 64in^3[/tex]

Applying the Chvorinov equation we have to,

[tex]\frac{T_r}{T_c} = (\frac{V_r}{V_c}*\frac{SA_c}{SA_r})^2[/tex]

[tex]\frac{T_r}{T_c} = (\frac{64}{64}*\frac{96}{168})^2[/tex]

[tex]\frac{T_r}{T_c} = (\frac{96}{168})^2[/tex]

[tex]\frac{T_r}{T_c} = 0.3265[/tex]

The stipulated time for the cube is 14.5 then,

[tex]T_r = 0.3265*14.5[/tex]

[tex]T_r = 4.735min[/tex]

To solve this problem it is necessary to apply the concepts related to Hooke's Law as well as Newton's second law.

By definition we know that Newton's second law is defined as

[tex]F = ma[/tex]

m = mass

a = Acceleration

By Hooke's law force is described as

[tex]F = k\Delta x[/tex]

Here,

k = Gravitational constant

x = Displacement

To develop this problem it is necessary to consider the two cases that give us concerning the elongation of the body.

The force to keep in balance must be preserved, so the force by the weight stipulated in Newton's second law and the force by Hooke's elongation are equal, so

[tex]k\Delta x = mg[/tex]

So for state 1 we have that with 0.2kg there is an elongation of 9.5cm

[tex]k (9.5-l)=0.2*g[/tex]

[tex]k (9.5-l)=0.2*9.8[/tex]

For state 2 we have that with 1Kg there is an elongation of 12cm

[tex]k (12-l)= 1*g[/tex]

[tex]k (12-l)= 1*9.8[/tex]

We have two equations with two unknowns therefore solving for both,

[tex]k = 3.136N/cm[/tex]

[tex]l = 8.877cm[/tex]

In this way converting the units,

[tex]k = 3.136N/cm(\frac{100cm}{1m})[/tex]

[tex]k = 313.6N/m[/tex]

Therefore the spring constant is 313.6N/m

Answer:

The work done on the package by the machine's force is 676.94 J.

Explanation:

Given that,

Mass of package = 33.6 kg

Initial position [tex]r_{0}=(0.502i+0.751j+0.207k)\ m[/tex]

Final position [tex]r_{1}=(7.82i+2.17j+7.44k)\ m[/tex]

Final time = 11.9 s

Force [tex]F=(21.5i+42.5j+63.5k)[/tex]

Suppose we need to find the work done on the package by the machine's force

We need to calculate the displacement

Using formula of displacement

[tex]d=r_{1}-r_{0}[/tex]

Put the value into the formula

[tex]d=(7.82i+2.17j+7.44k)-(0.502i+0.751j+0.207k)[/tex]

[tex]d=7.318i+1.419j+7.233k[/tex]

We need to calculate the work done

Using formula of work done

[tex]W=F\dotc d[/tex]

[tex]W=(21.5i+42.5j+63.5k)\dotc(7.318i+1.419j+7.233k)[/tex]

[tex]W=676.94\ J[/tex]

Hence, The work done on the package by the machine's force is 676.94 J.

The measurement will be significantly affected.

Recall that the relationship between linear velocity and angular velocity is subject to the formula

[tex]v = \omega r[/tex],

Where r indicates the radius and [tex]\omega[/tex] the angular velocity.

As the radius increases, it is possible that the calibration is delayed and a higher linear velocity is indicated, that to the extent that the velocity is directly proportional to the radius of the tires.

Mounting larger-diameter tires on a truck leads to a systematic error in the speedometer reading, whereby the speedometer will underreport the truck's true linear speed due to an increase in linear distance covered per tire rotation at a constant angular velocity.

If larger-diameter tires are mounted on a truck, the speedometer reading will not be correct. This is due to the relationship between the angular velocity of the tires and the linear speed of the truck. Since the speedometer measures the angular speed of the tires to determine the truck's linear speed, changing to larger tires will cause a systematic error in the reading. For a fixed angular velocity, larger tires will cover a greater linear distance because of their larger circumference. Thus, the speedometer will display a speed lower than the truck's true linear speed. This affects the accuracy of the speedometer.

To further illustrate, consider a truck tire rotating with an angular velocity α. The linear (tangential) velocity v at the surface of the tire is given by v = rα, where r is the tire radius. With larger tires, for the same angular velocity α, r is bigger, hence v is larger. Therefore, when the tires are larger, the speedometer, which is calibrated for the standard tire size, underestimates the truck's speed because it is based on an incorrect assumption about the tire circumference.

Answer:

(C) Only if it starts moving

Explanation:

We know that work done is given by

[tex]W=F.d=Fdcos\Theta[/tex]

So there are two case in which work done is zero

First case is that when force and displacement are perpendicular to each other

And other case is that when there is no displacement

So for work to be done there must have displacement, if there is no displacement then there is no work done

So option (c) will be the correct option

Work in a physical sense is done on a desk only if the desk moves when you push it. Simply exerting force is not considered work without displacement in the direction of the force. So the correct option is C.

When considering whether you do work on a heavy desk by pushing it, it's important to understand the scientific definition of work. In physics, work is defined as the transfer of energy that occurs when a force is applied over a distance. Therefore, you do work on an object if, and only if, the object moves in the direction of the force. So the correct answer to the question is:

c. only if it starts moving.

Simply exerting a force on an object does not constitute work unless the object is displaced. For example, if you continue to push against a wall and it does not move, despite your effort and energy consumption, physically no work is done because there is no displacement.

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Answer:

1327.43362 rev/s

Explanation:

[tex]I_i[/tex] = Initial moment of inertia = 2.4 kgm² (arms outstretched)

[tex]I_f[/tex] = Final moment of inertia = 0.904 kgm² (arms close)

[tex]\omega_f[/tex] = Final angular velocity

[tex]\omega_i[/tex] = Initial angular velocity = 0.5 rev/s

Here the angular momentum is conserved

[tex]I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\frac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\frac{2.4\times 500}{0.904}\\\Rightarrow \omega_f=1327.43362\ rev/s[/tex]

The new rate of rotation is 1327.43362 rev/s

Answer:

The pressure inside the bubble is 666.67 [tex]N/m^{2}[/tex]

Solution:

As per the question:

Radius, R = [tex]2.22\times 10^{- 4}\ m[/tex]

Now,

Given that the surface tension of the wall is the same as that of soapy water.

The air trapped inside the bubble exerts pressure on the soap bubble which is given by:

Gauge Pressure, P = [tex]\frac{4T}{r}[/tex]

Also, the surface tension of the soapy water, [tex]T_{s} = 0.0370\ N/m^{2}[/tex]

To calculate the pressure inside the alveolus:

[tex]P_{i} = \frac{4T}{R} = \frac{4T_{s}}{R}[/tex]

[tex]P_{i} = \frac{4\times 0.0370}{2.22\times 10^{- 4}} = 666.67\ N/m^{2}[/tex]

Answer:

U = -3978.8 J

Explanation:

The work of the gravitational force U just depends of the heigth and is calculated as:

U = -mgh

Where m is the mass, g is the gravitational acceleration and h the alture.

for calculate the alture we will use the following equation:

h = L-Lcos(θ)

Where L is the large of the rope and θ is the angle.

Replacing data:

h = 12.2-12.2cos(58.4)

h = 5.8 m

Finally U is equal to:

U = -70(9.8)(5.8)

U = -3,978.8 J

Answer:

a. 4

Explanation:

Hi there!

The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the car.

v = speed of the car.

Let´s see how would be the equation if the velocity is doubled (2 · v)

KE2 = 1/2 · m · (2 · v)²

Distributing the exponent:

KE2 = 1/2 · m · 2² · v²

KE2 = 1/2 · m · 4 · v²

KE2 = 4 (1/2 · m · v²)      

KE2 = 4KE

Doubling the velocity increased the kinetic energy by 4.

Final answer:

The kinetic energy of a car increases by a factor of four when its speed is doubled due to the relationship KE = 1/2 m v^2, where kinetic energy is proportional to the square of the velocity.

Explanation:

When the speed of your car is doubled, the kinetic energy increases by a factor of four. This is because kinetic energy is proportional to the square of the velocity. The equation for kinetic energy (KE) is KE = 1/2 m v^2 where m is the mass and v is the velocity. If you double the velocity (v), the new kinetic energy will be 1/2 m (2v)^2 = 1/2 m (4v^2) = 4 times the original kinetic energy. Therefore, the correct answer to the question is a. 4.

Answer:

427.392 kJ

Explanation:

m = Mass of gas = 4.5 kg

Initial temperature = 200 K

Final temperature = 360 K

R = Mass specific gas constant = 296.8 J/kgK

[tex]\gamma[/tex] = Specific heat ratio = 1.5

Work done for a polytropic process is given by

[tex]W=\frac{mR\Delta T}{1-\gamma}\\\Rightarrow W=\frac{4.5\times 296.8(360-200)}{1-1.5}\\\Rightarrow W=-427392\ J\\\Rightarrow W=-427.392\ kJ[/tex]

The work input during the process is -427.392 kJ

The blood will spurt to a height of 0.077 meters or 7.7 centimeters.

The height to which blood spurts from a severed artery can be calculated using Torricelli's law, which is based on the principles of fluid dynamics.

The formula for the height h that blood spouts from the artery can be derived from Torricelli's law:

h = (2 * ΔP) / (ρ * g),

where:

To find ΔP, we can calculate the pressure difference between the cut artery and the atmosphere using the following formula:

ΔP = ρ * g * h.

Given that the specific gravity of blood is 1.050 g cm⁻³, we can find its density:

ρ_blood = [tex](1.050 g cm^{-3}) * (1 g / cm^3) = 1.050 g / cm^3.[/tex]

The specific gravity of mercury is 13.6 g[tex]cm^{-3[/tex], which corresponds to its density:

ρ_mercury = 13.6 g / [tex]cm^3[/tex].

Now, we can find the pressure difference:

ΔP = [tex](1.050 g / cm^3) * (9.81 m/s^2) * h[/tex],

where g is the acceleration due to gravity (approximately 9.81 m/s²).

We'll assume that the height (h) is in meters. We want to find h, so we'll solve for it:

h = ΔP / ((1.050 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex])).

Now, plug in the values:

h = ΔP / ((1.050 g / cm³) * (9.81 m/s²)).

h = (ρ_mercury * g * h) / ((1.050 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex])).

Now, solve for h:

h = (13.6 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex]) * h / ((1.050 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex])).

h = (13.6 / 1.050) * h.

h = 12.952 * h.

h = h / 12.952.

1 = 1 / 12.952.

h = 1 / 12.952.

h ≈ 0.077 m.

So, the blood will spurt to a height of approximately 0.077 meters or 7.7 centimeters.

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The blood spurt from a deep cut is determined by the blood pressure, which needs to exceed the vein pressure for fluid entry, and by the pressure gradient from the heart to smaller vessels. These factors, combined with the density disparity between blood and mercury, influence the height of the blood spurt.

To determine how high the blood will spurt when an individual experiences a deep cut that severs the radial artery near the elbow, we first need to understand the pressure at which the fluid, in this case blood, enters the vascular system. It must exceed the blood pressure in the vein, which is approximately 18 mm Hg above atmospheric pressure.

The blood pressure of a young adult is represented by 120 mm Hg at systolic (maximum output of the heart) and 80 mm Hg at diastolic (due to the elasticity of arteries maintaining pressure between heartbeats). Thus, the density of mercury which raises the mercury fluid in the manometer is 13.6 times greater than water, meaning the height of the fluid will be 1/13.6 of that in a water manometer.

The blood leaves the heart with a pressure of about 120 mm Hg but its pressure continues to decrease as it goes from the aorta to smaller arteries to small veins. The pressure differences in the circulation system are due to the blood flow as well as the position of the person.

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Answer:

A)[tex]T=209.94N[/tex]

B) [tex]F=75.24N[/tex]

Explanation:

Using the free body diagram and according to Newton's first law, we have:

[tex]\sum F_y=Tcos(21^\circ)-mg=0(1)\\\sum F_x=F-Tsin(21^\circ)=0(2)[/tex]

A) Solving (1) for T:

[tex]T=\frac{mg}{cos(21^\circ)}\\T=\frac{20kg(9.8\frac{m}{s^2})}{cos(21^\circ)}\\T=209.94N[/tex]

B) Solving (2) for F:

[tex]F=Tsin(21^\circ)\\F=(209.94N)sin(21^\circ)\\F=75.24N[/tex]

Answer:

s = 0.9689 m

Explanation:

given,

Height of fall of paratroopers = 362 m

speed of impact = 52 m/s

mass of paratrooper = 86 Kg

From from snow on him = 1.2 ✕ 10⁵ N

now using formula

F = m a

a = F/m

[tex]a = \dfrac{1.2 \times 10^5}{86}[/tex]

[tex]a =1395.35\ m/s^2[/tex]

Using equation of motion

v² = u² + 2 a s

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

[tex]s =\dfrac{52^2}{2\times 1395.35}[/tex]

s = 0.9689 m

The minimum depth of snow that would have stooped him is  s = 0.9689 m

The answer is option d. slows down.

The answer to the question is option d. slows down.

When the milk begins to leak out of the bottom of the railroad tank car, the car will experience a decrease in its mass as the milk is being lost. According to the Law of Conservation of Momentum, if the mass of an object decreases without any external forces acting on it, the object will slow down.

Therefore, as the milk leaks out of the tank car, its speed will slow down over time.

To solve this problem it is necessary to apply the concepts related to Gibbs free energy and spontaneity

At constant temperature and pressure, the change in Gibbs free energy is defined as

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

Where,

H = Entalpy

T = Temperature

S = Entropy

When the temperature is less than that number it is negative meaning it is a spontaneous reaction. [tex]\Delta  G[/tex] is also always 0 when using single element reactions. In numerical that implies [tex]\Delta G = 0[/tex]

At the equation then,

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

[tex]0 = \Delta H - T\Delta S[/tex]

[tex]\Delta H = T\Delta S[/tex]

[tex]T = \frac{\Delta H}{\Delta S}[/tex]

[tex]T = \frac{-93.8kJ}{-156.1J/K}[/tex]

[tex]T = \frac{-93.8*10^3J}{-156.1J/K}[/tex]

[tex]T = 600.89K[/tex]}

Therefore the temperature changes the reaction from non-spontaneous to spontaneous is 600.89K

A particular reaction, with ΔH° = − 93.8 kJ and ΔS° = − 156.1 J/K, will change from nonspontaneous to spontaneous at 601 K.

We want to know at what temperature a reaction changes from nonspontaneous to spontaneous, that is, at what temperature ΔG° = 0.

Given the standard enthalpy and entropy of the reaction, we can calculate that temperature using the following expression.

ΔG° = ΔH° - T . ΔS°

0 = -93.8 kJ - T . (-156.1 J/K)

T = 601 K

A particular reaction, with ΔH° = − 93.8 kJ and ΔS° = − 156.1 J/K, will change from nonspontaneous to spontaneous at 601 K.

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To solve this problem it is necessary to apply the concepts related to simple harmonic movement.

The maximum speed from the simple harmonic motion is given as

[tex]V = A\sqrt{\frac{K}{m}}[/tex]

Where,

K = Spring constant

m = mass

At this case m is a constant then

[tex]V \propto \sqrt{K}[/tex]

then the ratio is given by

[tex]\frac{v_2}{v_1}=\sqrt{\frac{K_1}{K_2}}[/tex]

According the statement,

[tex]v_2 = \sqrt{\frac{K_1}{K_2}}v_1[/tex]

[tex]v_2 = 2v_1[/tex]

Therefore the maximum speed becomes double: 2) It would increase by a factor of 2.