A woman is 1.6 m tall and has a mass of 50 kg. She moves past an observer with the direction of the motion parallel to her height. The observer measures her relativistic momentum to have a magnitude of 2.1 × 1010 kg·m/s. What does the observer measure for her height?

Answer:

Tracy’s constant acceleration during her time of contact with Tom is 1.25 m/s².

Explanation:

Given that,

Mass of Tracy = 49 kg

Mass of Tom = 77 kg

Distance = 0.4 m

Time = 0.8 s

We need to calculate Tracy’s constant acceleration during her time of contact with Tom

Using equation of motion

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

Where, s = distance

a = acceleration

t = time

Put the value into the formula

[tex]0.4=0+\dfrac{1}{2}a\times(0.8)^2[/tex]

[tex]a=\dfrac{0.4\times2}{(0.8)^2}[/tex]

[tex]a=1.25\ m/s^2[/tex]

Hence, Tracy’s constant acceleration during her time of contact with Tom is 1.25 m/s².

Tracy’s constant acceleration during her time of contact with Tom is 1.25 m/s².

Here we applied the motion equation i.e. shown below:

[tex]S = ut + \frac{1}{2}at^2[/tex]

Here s = distance

a = acceleration

t = time

So,

[tex]0.4 = \frac{1}{2} a \times (0.8)^2\\\\a = \frac{0.4\times 2}{(0.8)^2}[/tex]

=  1.25 m/s²

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Answer:

178378378.37 Pa

0.01897

Explanation:

F = Force = [tex]6.6\times 10^4\ N[/tex]

A = Area = [tex]3.7\times 10^{-4}\ m^2[/tex]

Y = Young's modulus of bone under compression = [tex]9.4\times 10^{9}\ Pa[/tex]

[tex]\varepsilon[/tex] = Strain

Stress is given by

[tex]\sigma=\frac{F}{A}\\\Rightarrow \sigma=\frac{6.6\times 10^4}{3.7\times 10^{-4}}\\\Rightarrow \sigma=178378378.37\ Pa[/tex]

The maximum stress that this bone can withstand is 178378378.37 Pa

Compression force is given by

[tex]F=Y\varepsilon A\\\Rightarrow \varepsilon=\frac{F}{YA}\\\Rightarrow \varepsilon=\frac{6.6\times 10^4}{9.4\times 10^{9}\times 3.7\times 10^{-4}} \\\Rightarrow \varepsilon=0.01897[/tex]

The strain that exists under a maximum-stress condition is 0.01897

The maximum stress that the femur can withstand is 1.78 * 10^8 Pascals, and the strain under this maximum stress condition is approximately 0.012 or 1.2%.

The subject of the question is the stress and strain on the femur bone, which is integral to the field of Physics. To address part (a) of the question, we must find the maximum stress that the femur can withstand. Stress is defined as the average force per unit area, or the force divided by the area. In this case, we are given a compressional force of 6.60 * 10^4 Newtons and a cross-sectional area of the femur as 3.70 * 10^-4 m2. We divide the force by the area to find the stress:

Stress = Force / Area = (6.60 * 10^4 N) / (3.70 * 10^-4 m2) = 1.78 * 10^8 Pa (Pascals).

For part (b) of the question, the strain under a maximum-stress condition can be calculated by dividing the stress by Young's modulus. For bone, the modulus can vary, and bones are brittle with the elastic region small. However, if we take an average value, say 1.5 * 10^10 Pa, then, the strain will be:

Strain = Stress / Young's modulus = (1.78 * 10^8 Pa) / (1.5 * 10^10 Pa) = 0.012.

So under maximum stress, the strain is 0.012 (dimensionless) or 1.2%. Remember, this is an approximation as the actual modulus can vary based on numerous factors such as age, diet and lifestyle of the individual.

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Answer:

[tex]T_2=630^{\circ}C[/tex]'

Explanation:

Original temperature of the gas, [tex]T_1=28^{\circ}C=301\ K[/tex]

From the ideal gas equation,

[tex]P_1V_1=nRT_1[/tex]

Since,

[tex]P_2=3P_1[/tex]

[tex]nRT_2=3(nRT_1)[/tex]

[tex]T_2=3T_1[/tex]

[tex]T_2=3\times 301[/tex]

[tex]T_2=903\ K[/tex]

or

[tex]T_2=630^{\circ}C[/tex]

So, the new temperature of the gas is 630 degree Celsius. Hence, this is the required solution.

Answer:

First bright fringe is at 1.82 mm

First dark fringe is at 2.83 mm

Solution:

As per the question:

Slit width, d = 1.19 mm = [tex]1.19\times 10^{- 3}\ m[/tex]

Distance from the screen, x = 3.53 m

Wavelength of the light, [tex]\lambda = 635\ nm = 635\times 10^{- 9}\ m[/tex]

Now,

We know that the 1st bright fringe from the central fringe is given by:

[tex]y = \frac{n\lambda x}{d}[/tex]

where

n = 1

[tex]y = \frac{1\times 635\times 10^{- 9}\times 3.53}{1.19\times 10^{- 3}} = 1.88\ mm[/tex]

Also, we know that the 1st dark fringe from the central fringe is given by:

[tex]y = \frac{(n + \frac{1}{2})\lambda x}{d}[/tex]

[tex]y = \frac{(1 + \frac{1}{2})\times 635\times 10^{- 9}\times 3.53}{1.19\times 10^{- 3}} = 2.83\ mm[/tex]

Answer:

3. One Newton is the force that gives a 1 kg body an acceleration of 1 m/s².

Explanation:

The force is a vector magnitude that represents every cause capable of modifying the state of motion or rest of a body or of producing a deformation in it.

Its unit in the International System is the Newton (N). A Newton is the force that when applied on a mass of 1 Kg causes an acceleration of 1 m/s².

1 N = kg*(m/s²)

Answer:

  y = 2.74 m

Explanation:

The linear thermal expansion processes are described by the expression

         ΔL = α L ΔT

Where α the thermal dilation constant for concrete is 12 10⁻⁶ºC⁻¹, ΔL is the length variation and ΔT the temperature variation in this case 20ªc

If the bridge is 250 m long and is covered by two sections each of them must be L = 125 m, let's calculate the variation in length

        ΔL = 12 10⁻⁶ 125 20

        ΔL = 3.0 10⁻² m

Let's use trigonometry to find the height

The hypotenuse     Lf = 125 + 0.03 = 125.03 m

Adjacent leg           L₀ = 125 m

       cos θ = L₀ / Lf

       θ = cos⁻¹ (L₀ / Lf)

       θ = cos⁻¹ (125 / 125.03)

       θ = 1,255º

We calculate the height

       tan 1,255 = y / x

       y = x tan 1,255

       y = 125 tan 1,255

       y = 2.74 m

When a bridge is subjected to a temperature increase, the concrete spans rise or buckle due to thermal expansion. The height to which they rise can be calculated using the formula for change in length. In this case, the height is 0.42 m.

When a bridge is subjected to a temperature increase, it expands due to thermal expansion. In this case, the two concrete spans of the bridge are placed end to end without allowing any room for expansion. Therefore, when the temperature increases by 20.0°C, the spans rise or buckle. The height to which they rise can be calculated using the formula:

change in length = coefficient of linear expansion x original length x change in temperature

In this case, the change in length is 0.84 m. To calculate the height y to which the spans rise, divide this change in length by the number of spans, which is 2. Therefore, the height is 0.42 m.

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Answer:

[tex]v=2.28m/s[/tex]

Explanation:

For the first mass we have [tex]m_1=3.5kg[/tex], and for the second [tex]m_2=6.0kg[/tex]. The pulley has a moment of intertia [tex]I_p=0.0352kgm^2[/tex] and a radius [tex]r_p=0.125m[/tex].

We solve this with conservation of energy.  The initial and final states in this case, where no mechanical energy is lost, must comply that:

[tex]K_i+U_i=K_f+U_f[/tex]

Where K is the kinetic energy and U the gravitational potential energy.

We can write this as:

[tex]K_f+U_f-(K_i+U_i)=(K_f-K_i)+(U_f-U_i)=0J[/tex]

Initially we depart from rest so [tex]K_i=0J[/tex], while in the final state we will have both masses moving at velocity v and the tangential velocity of the pully will be also v since it's all connected by the string, so we have:

[tex]K_f=\frac{m_1v^2}{2}+\frac{m_2v^2}{2}+\frac{I_p\omega_p^2}{2}=(m_1+m_2+\frac{I_p}{r_p^2})\frac{v^2}{2}[/tex]

where we have used the rotational kinetic energy formula and that [tex]v=r\omega[/tex]

For the gravitational potential energy part we will have:

[tex]U_f-U_i=m_1gh_{1f}+m_2gh_{2f}-(m_1gh_{1i}+m_2gh_{2i})=m_1g(h_{1f}-h_{1i})+m_2g(h_{2f}-h_{2i})[/tex]

We don't know the final and initial heights of the masses, but since the heavier, [tex]m_2[/tex], will go down and the lighter, [tex]m_1[/tex], up, both by the same magnitude h=1.25m (since they are connected) we know that [tex]h_{1f}-h_{1i}=h[/tex] and [tex]h_{2f}-h_{2i}=-h[/tex], so we can write:

[tex]U_f-U_i=m_1gh-m_2gh=gh(m_1-m_2)[/tex]

Putting all together we have:

[tex](K_f-K_i)+(U_f-U_i)=(m_1+m_2+\frac{I_p}{r_p^2})\frac{v^2}{2}+gh(m_1-m_2)=0J[/tex]

Which means:

[tex]v=\sqrt{\frac{2gh(m_2-m_1)}{m_1+m_2+\frac{I_p}{r_p^2}}}=\sqrt{\frac{2(9.8m/s^2)(1.25m)(6.0kg-3.5kg)}{3.5kg+6.0kg+\frac{0.0352kgm^2}{(0.125m)^2}}}=2.28m/s[/tex]

Final answer:

To solve this problem, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system (kinetic energy + potential energy) remains constant if no external forces are acting on the system.

Explanation:

Let's denote the initial position of the masses as position 1 and the final position as position 2. At position 1, the 3.5 kg mass is at a height h above the ground, and the 6.0 kg mass is at a height 1.25 m - h above the ground. At position 2, both masses are at a height 1.25 m above the ground.

The potential energy (PE) of the system at position 1 is:

PE1 = m1 * g * h + m2 * g * (1.25 m - h)

The potential energy (PE) of the system at position 2 is:

PE2 = (m1 + m2) * g * 1.25 m

Since the system is released from rest, the initial kinetic energy (KE) of the system is zero. The final kinetic energy (KE) of the system is:

KE2 = (1/2) * (m1 + m2) * v^2

where v is the speed of the masses after they have moved through 1.25 m.

According to the principle of conservation of mechanical energy, the total mechanical energy at position 1 is equal to the total mechanical energy at position 2:

PE1 + KE1 = PE2 + KE2

Substituting the expressions for PE1, PE2, and KE2, we get:

m1 * g * h + m2 * g * (1.25 m - h) = (m1 + m2) * g * 1.25 m + (1/2) * (m1 + m2) * v^2

Solving for v^2, we get:

v^2 = 2 * g * (m1 * h + m2 * (1.25 m - h) - (m1 + m2) * 1.25 m)

Substituting the given values for g, m1, m2, and h, we get:

v^2 = 2 * 9.8 m/s^2 * (3.5 kg * 1.25 m + 6.0 kg * (1.25 m - 1.25 m) - (3.5 kg + 6.0 kg) * 1.25 m)

Solving for v, we get:

v ≈ 2.3 m/s

So, the speed of the masses after they have moved through 1.25 m is approximately 2.3 m/s.

The sphere's angular acceleration is [tex]2.064 rad/s^2[/tex], and it will take approximately 10.17 seconds to decrease its rotational speed by 21.0 rad/s when experiencing a constant friction force.

Part A: Calculating Angular Acceleration

To find the angular acceleration, we must first calculate the torque. The torque ( au) caused by the friction force (F) is the product of the force and the radius (r) at which the force is applied, and torque is given by [tex]\tau = r \times F[/tex]. Using the diameter (d) to find the radius, we have r = d / 2 = 4.50 cm / 2 = 2.25 cm = 0.0225 m. The given friction force is 0.0200 N. Therefore, [tex]\tau = 0.0225 m \times 0.0200 N = 0.00045 Nm[/tex].

Next, we use the moment of inertia (I) for a sphere, [tex]I=\frac{2}{5} mr^2[/tex], to calculate the angular acceleration ([tex]\alpha[/tex]). The mass (m) is 220 g which is 0.220 kg. So, [tex]I = \frac{2}{5} \times 0.220 kg \times (0.0225 m)^2 = 0.000218025 kg m^2[/tex]. Angular acceleration, [tex]\alpha = \frac{\tau}{I} = \frac{0.00045 Nm}{0.000218025 kgm^2} \longrightarrow \alpha= 2.064 rad/s^2[/tex].

Part B: Time to Decrease Rotational Speed

To find the time (t) to decrease the rotational speed by 21.0 rad/s, we can use the equation [tex]\omega = \omega_0 + \alpha \times t[/tex], where [tex]\omega_0[/tex] is the initial angular velocity and [tex]\omega[/tex] is the final angular velocity. Since the sphere is slowing down due to friction, the angular acceleration will be negative, [tex]\alpha = -2.064 rad/s^2[/tex]. We are looking for the time it takes for the speed to decrease by 21.0 rad/s, so if the initial speed is [tex]\omega_0[/tex] and the final is [tex]\omega_0 - 21.0 rad/s[/tex], then [tex]0 = \omega_0 - 21.0 rad/s + \alpha \times t[/tex]. Solving for t gives [tex]t = \frac{21.0 rad/s}{2.064 rad/s^2} \approx t = 10.17 s[/tex].

The bungee jumper experiences a downward acceleration of 4.09 m/s^2 at the point when the bungee cord has lengthened to 25m.

The subject of this problem pertains to physics, specifically, the principles of dynamics and Hooke's Law. At the instant when the bungee cord has stretched to a total length of 25m, it has undergone an extension of 5m since its natural length is 20m. According to Hooke’s Law, this extension causes a restoring force F = kx, where k is the spring constant and x is the change in length.

In this case, F = 80 N/m × 5 m = 400 N which is the upward force exerted by the cord. The weight of the 70 kg person is mg, or 70 kg × 9.8 m/s² = 686 N downward. The net force acting on the jumper is the difference between the weight and the restoring force. This gives a net force of 686 N – 400 N = 286 N downward.

From Newton’s second law, force equals mass times acceleration, F = ma. Therefore, the acceleration of the jumper at the instant would be a = F/m, or 286 N ÷ 70 kg = 4.09 m/s² downward to 2 significant figures.

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Answer:

  v₂ = v/1.5= 0.667 v

Explanation:

For this exercise we will use the conservation of the moment, for this we will define a system formed by the two students and the cars, for this isolated system the forces during the contact are internal, therefore the moment conserves.

Initial moment before pushing

    p₀ = 0

Final moment after they have been pushed

    [tex]p_{f}[/tex] = m₁ v₁ + m₂ v₂

   p₀ =  [tex]p_{f}[/tex]

   0 = m₁ v₁ + m₂ v₂

   m₁ v₁ = - m₂ v₂

Let's replace

   M (-v) = -1.5M v₂

   v₂ = v / 1.5

  v₂ = 0.667 v

Conservation of momentum dictates that the total momentum before and after the push must be the same since there are no external forces. Therefore, student 2 and the cart will move in the opposite direction to student 1's movement, with a velocity of two-thirds of student 1's velocity.

This question relies on the concept of Conservation of Momentum. In this scenario, the frictionless carts allow us to use the principle of momentum conservation, because if there is no external force, the total momentum of the system remains constant.

Second Cart's Velocity Calculation:

Applying the principle, if student 1 and the cart (both with mass M) are initially at rest, the initial momentum of the system is 0. When student 1 pushes and goes in the opposite direction with velocity -v, the momentum of the system is still 0 (since no external forces are present). So, if student 2 (with mass 1.5M) is pushed in the opposite direction, for the total momentum to remain zero, the velocity (V2) of student 2 and the cart must be such that:
-M * v (student 1's momentum) + 1.5M * V2 (student 2's momentum) = 0.
Solving for V2 gives us V2 = -(-M * v) / 1.5M = 2v/3.

So, the velocity of student 2 and the cart is 2v/3 in the opposite direction to student 1's movement.

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Since there is a positive work on an object, this directly indicates that there is an increase in Energy on the body.

Work can be expressed in terms of Force and distance as

W = Fd

Where,

F = Force

d = Distance

The force in turn is a description of how mass accelerates according to Newton's second law, that is

F = ma

The mass (m) remains constant, but the acceleration (a) is constant, which implies directly that there is a gradual change in the velocity of the object.

Therefore the correct answer is: The object is speeding up.

The work is positive so the energy of the object is  increasing so the object is speeding up

As we know that the work is the product of the force and the displacement

[tex]W=F\times D[/tex]

Where,

F = Force

D= Distance

And from newtons second law we can see that

[tex]F=m\times a[/tex]

Since here mass will be constant to there will be a change in the velocity that is acceleration in the body so the energy of the body will change

Thus work is positive so the energy of the object is  increasing so the object is speeding up

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Answer:

correct option is (b) 2.7cm

Explanation:

given data

n = 3/2 = 1.5

sides of length  = 8 cm

to find out

How far from one of the faces of the block that what is the image distance of the picture from the face

solution

we get here distance that is

distance d = [tex]\frac{d_o}{n}[/tex]      ......................1

put here value as do is [tex]\frac{8}{2}[/tex]  = 4 cm

so

distance d = [tex]\frac{d_o}{n}[/tex]  

distance d = [tex]\frac{4}{1.5}[/tex]  

distance d = 2.6666666 = 2.67

so correct option is (b) 2.7cm

As the image distance of the picture from the face is 2.7 cm, the correct answer is (b) 2.7cm.

The problem is related to optics, specifically the refraction of light through a medium with a given refractive index, here Lucite, which has a refractive index of n = 1.5.

The picture is physically at the center of the block. Since the block has sides of length 8 cm, the center is 4 cm from any face.

To find the apparent distance due to refraction, we use the formula for the apparent depth (d') from the real depth (d) and the refractive index (n) given as:

[tex]d' = \frac{d}{n}[/tex]

Substituting the given values, we get:

d' = [tex]\frac{4 cm}{1.5}[/tex] ≈ 2.67 cm.

Rounding to the closest option, the apparent distance is 2.7 cm from one of the faces of the block.

Answer:

Q (reaction) = -69.7 kJ

Explanation:

Octane reacts with oxygen to give carbon dioxide and water.

C₈H₁₈ + 25 O₂ ---> 16 CO₂ +18 H₂O

This reaction is exothermic in nature. Therefore, the energy is released into the atmosphere. This reaction took place in a calorimeter, there the temperature (T) increases by 10 C. The heat capacity of the calorimeter is 6.97 kJ/C

The heat (q) of the reaction is calculated as follows:

Q= -cT, where c is the heat capacity of the calorimeter and T is the increase in temperature

q = -(6.97) x (10) = -69.7kJ

Since the heat capacity is given in kilo -joule per degree Celsius, therefore, the mass of octane is not required

Answer:

The mass of mars is [tex]6.61\times10^{23}\ kg[/tex]

Explanation:

Given that,

Time = 1.02 s

Height = 2.00 m

Radius of mars [tex]r= 3.39\times10^{6}\ m[/tex]

We need to calculate the acceleration due to gravity on mars

Using equation of motion

[tex]h = ut+\dfrac{1}{2}gt^2[/tex]

[tex]h = \dfrac{1}{2}gt^2[/tex]

[tex]g=\dfrac{2h}{t^2}[/tex]

Where, u = initial velocity

t = time

h = height

Put the value into the formula

[tex]g=\dfrac{2\times2.00}{(1.02)^2}[/tex]

[tex]g=3.84\ m/s^2[/tex]

We need to calculate the mass of mars

Using formula of gravity

[tex]g=\dfrac{Gm}{r^2}[/tex]

Put the value into the formula

[tex]3.84=\dfrac{6.67\times10^{-11}\times m}{(3.39\times10^{6})^2}[/tex]

[tex]m=\dfrac{3.84\times(3.39\times10^{6})^2}{6.67\times10^{-11}}[/tex]

[tex]m=6.61\times10^{23}\ kg[/tex]

Hence, The mass of mars is [tex]6.61\times10^{23}\ kg[/tex]

In order to find the mass of Mars, we can use a physics formula for gravity. The given values, such as the drop distance, time it took, the radius of Mars, and the universal gravitational constant, can be substituted into this formula. This, then, gives a theoretical value for the mass of Mars.

To calculate the mass of Mars, we use the formula to find gravity, which is defined as g = GM/r², where G is the universal gravitational constant. We can set this equal to something else: 2d / t², where d is the drop distance and t is the time it took for the marble to drop. We then can isolate M (the mass of Mars) on one side.

By substituting the given values into the formula, we get: M = (g * r²) / G = ((2 * 2.00 m) / (1.02 s)² * (3.39 × 10⁶ m)²) / (6.67 x 10⁻¹¹ N·m²/kg²).

This Netwonian calculation gives the mass of Mars, theoretically, based on the observed falling time of the marble.

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Answer:

The mass of [tex]_{53}^{131}I[/tex] is [tex]8.09\times10^{-9}\ g[/tex].

Explanation:

Given that,

Half life [tex]t_{\frac{1}{2}}=8.08\ days[/tex]

Sample emitting radiation = 1.00 mCi = [tex]3.7\times10^{7}\ dps[/tex]

We need to calculate the rate constant

Using formula of rate constant

[tex]\lambda=\dfrac{0.693}{t_{\frac{1}{2}}}[/tex]

[tex]\lambda=\dfrac{0.693}{8.08\times24\times60\times60}[/tex]

[tex]\lambda=9.92\times10^{-7}\ s^{-1}[/tex]

We need to calculate the numbers of atoms

Using formula of numbers of atoms

[tex]N_{0}=\dfrac{N}{\lambda}[/tex]

[tex]N_{0} =\dfrac{3.7\times10^{7}}{9.92\times10^{-7}}[/tex]

[tex]N_{0}=3.72\times10^{13}\ atoms[/tex]

We need to calculate the mass of [tex]_{53}^{131}I[/tex]

Using formula for mass

[tex]m=\dfrac{131\times3.72\times10^{13}}{6.023\times10^{23}}[/tex]

[tex]m=8.09\times10^{-9}\ g[/tex]

Hence, The mass of [tex]_{53}^{131}I[/tex] is [tex]8.09\times10^{-9}\ g[/tex].

The problem is related to the concept of radioactive decay, specifically of the isotope Iodine-131. The activity of the sample is calculated to be 37,000,000 decays/second. Through a series of calculations using formulas for decay constant, the number of atoms and finally the sample mass, we find the mass to be 9.3x10^-16 g/day.

The problem given relates to nuclear physics and utilizes the concept of radioactive decay. It specifically involves the isotope Iodine-131 (131 53I), which decays with a half-life of 8.08 days.

Assuming the activity of the sample is 1 mCi, which is equivalent to 37,000,000 decays/second (since 1 Ci=3.7x10^10 decays/sec, 1 mCi = 1/1000 Ci), we can use the activity formula which is given by A = λN, where A is activity, λ is the decay constant, and N is the number of atoms in the sample. We first need to calculate λ, which we can find using the formula λ = Ln(2)/T(1/2), where T(1/2) is the half-life.

λ = Ln(2)/8.08 days = 0.086/day. Therefore, N = A/ λ = 37,000,000 / 0.086 = 4.3x10^8 atoms/day.

The remaining calculation is the mass of the sample. The atomic mass of Iodine-131 is approximately 131 g/mol. Using Avogadro's number (6.02x10^23 atoms/mol), we can calculate the mass: M = N * (131 g/mol) / (6.02x10^23 atoms/mol) = 9.3x10^-16 g/day.

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Answer:

The average gauge pressure inside the vein is 110270.58 Pa

Explanation:

This question can be solved using the Bernoulli's Equation. First, in order to determine the outlet pressure of the needle, we need to find the total pressure exerted by the atmosphere and the fluid.

[tex]P_f: fluid's\ pressure\\P_f= \rho g h=1025\frac{kg}{m^3} \times 9.8 \frac{m}{s^2} \times 0.9 m=9040.5 Pa \\P_T: total\ pressure\\P_T=P_{atm}+P_f\\P_T=101325 Pa + 9040.5 Pa=110275.5 Pa\\[/tex]

Then, we have to find the fluid's outlet velocity with the transversal area of the needle, as follows:

[tex]S: transversal\ area \\S= \pi r^2=\pi (0.200 \times 10^{-3})^2=5.65 \times 10^{-7} m^2\\v=\frac{F}{S}=\frac{5.55 \times 10^{-8} \frac{m^3}{s}}{5.65 \times 10^{-7} m^2}=0.98\times 10^{-1} \frac{m}{s}[/tex]

As we have all the information, we can complete the Bernoulli's expression and solve to find the outlet pressure as follows:

[tex]P_T-P_{out}=\frac{1}{2} \rho v^2\\P_{out}=P_T-\frac{1}{2} \rho v^2=110275.5 Pa-\frac{1}{2} 1025\frac{kg}{m^3} (0.98\times 10^{-1} \frac{m}{s})^2=110275.5 Pa-4.92 Pa =110270.58 Pa[/tex]

Answer:

48.6 N

Explanation:

rate of mass per second, dm/dt = 5 kg/s

Velocity, v = 35 km/hr = 9.72 m/s

Force acting on the plate

F = v x dm/dt

F = 9.72 x 5 = 48.6 N

Thus, the force acting on the plate is 48.6 N.

To prevent the plate from moving horizontally, a force of 48.6 N is required. The initial horizontal momentum of the water is 48.6 kg*m/s.

To calculate the force needed to prevent the plate from moving horizontally when a horizontal water jet strikes a vertical plate, we first need to determine the change in momentum of the water. The water strikes the plate at a rate of 5 kg/s with a velocity of 35 km/hr. Let's convert the velocity to meters per second:

35 km/hr = (35 * 1000) / 3600 = 9.72 m/s.

Next, we calculate the initial horizontal momentum of the water jet:

Initial momentum = mass flow rate * velocity = 5 kg/s * 9.72 m/s = 48.6 kg*m/s.

The water stream moves in the vertical direction after the strike, so its horizontal momentum is reduced to zero. The change in momentum is:

Change in momentum = Initial momentum - Final momentum = 48.6 kg*m/s - 0 kg*m/s = 48.6 kg*m/s.

We use the change in momentum (impulse) to find the force since force is the rate of change of momentum:

Force = Change in momentum / Time

The mass flow rate gives the change in momentum per second, which means the force can be directly calculated as:

Force = 48.6 N

Thus the force required to prevent the plate from moving horizontally, a force of 48.6 N

Answer:

Explanation:

mass, m = 86 g

Amplitude, A = 1.3 mm

Maximum acceleration, a = 6.8 x 10^3 m/s^2

(a) Let T be the period of motion and ω be the angular frequency.

maximum acceleration, a = ω²A

6.8 x 10^3 = ω² x 1.3 x 10^-3

ω = 2287.0875 rad/s

T = 2π/ω

T = ( 2 x 3.14) / 2287.0875

T = 2.75 x 10^-3 second

T = 2.75 milli second

(b) maximum speed = ωA

v = 2287.0875 x 1.3 x 10^-3

v = 2.97 m/s

(c) Total mechanical energy

T = 1/2 mω²A²

T = 0.5 x 86 x 10^-3 x 2287.0875 x 2287.0875 x 1.3 x 1.3 x 10^-6

T = 0.38 J

(d) At maximum displacement, the acceleration is maximum

F = m x a = 86 x 10^-3 x 6.8 x 1000

F = 541.8 N

(e) acceleration a = ω² y = ω² x A / 2

a =  2287.0875 x 2287.0875 x 1.3 x 10^-3 / 2

a = 3400 m/s^2

Force, F = 86 x 10^-3 x 3400

F = 292.4 N

The required velocity of the larger object for the smaller object to reach a maximum height of 3m in an elastic collision can be calculated using the equation v0 = sqrt((2*m*g*h_max)/M). This concept relies on the conservation of energy and momentum in an elastic collision.

The central physics concept in this question is the conservation of energy and momentum in an elastic collision. In an elastic collision, both momentum and kinetic energy are conserved. According to the conservation of energy, the kinetic energy of the larger mass before the collision is converted into the potential energy of the smaller mass at its maximum height.

Expressing the conservation of energy mathematically, we have:

1/2*M*v0² = m*g*h_max

Solving for v0, we have:

v0 = sqrt((2*m*g*h_max)/M)

where v0 is the velocity with which the player should release the larger object, m is the mass of the smaller object (1.68 kg), g is the acceleration due to gravity (9.8 m/s²), h_max is the maximum height the smaller object will reach (3.0 m), and M is the mass of the larger object (5.41 kg).

Please do remember to plug in the correct values and units when solving the equation.

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The player must release the larger object with a velocity of 1.95 m/s so that the smaller object just reaches the target maximum height of 3.0 m.

Since the collision of the two masses is elastic, the total kinetic energy of the system before the collision is equal to the total kinetic energy of the system after the collision.

We can use the principle of conservation of mechanical energy to solve this problem. The mechanical energy of a system is the sum of its potential energy and kinetic energy. At the beginning of the collision, the larger object has only potential energy due to its height, while the smaller object is at rest on the ground and has no energy. After the collision, the larger object has some kinetic energy and the smaller object has both potential energy and kinetic energy.

Let's denote the velocity of the larger object before the collision as v_0 and the velocity of the smaller object after the collision as v.

The potential energy of the larger object before the collision is:

PE_initial = Mgh

where:

M is the mass of the larger object (kg)

g is the acceleration due to gravity (m/s^2)

h is the height of the larger object above the ground (m)

The kinetic energy of the larger object before the collision is:

KE_initial = (1/2)Mv_0^2

The total mechanical energy of the system before the collision is:

E_initial = PE_initial + KE_initial = Mgh + (1/2)Mv_0^2

After the collision, the potential energy of the smaller object is:

PE_final = mgh_{max}

where:

m is the mass of the smaller object (kg)

g is the acceleration due to gravity (m/s^2)

h_{max} is the maximum height reached by the smaller object (m)

The kinetic energy of the smaller object after the collision is:

KE_final = (1/2)mv^2

The total mechanical energy of the system after the collision is:

E_final = PE_final + KE_final = mgh_{max} + (1/2)mv^2

Since the collision is elastic, we know that E_initial = E_final. Therefore, we can set the two equations equal to each other and solve for v_0.

Mgh + (1/2)Mv_0^2 = mgh_{max} + (1/2)mv^2

Solving for v_0, we get:

v_0 = sqrt((2mg(h - h_{max}))/(M + m))

Plugging in the values, we get:

v_0 = sqrt((2(5.41 kg)(9.81 m/s^2)(0.3 m))/(5.41 kg + 1.68 kg)) = 1.95 m/s

Therefore, the player must release the larger object with a velocity of 1.95 m/s so that the smaller object just reaches the target maximum height of 3.0 m.

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Answer:

The acceleration is 3.62 m/s²

Explanation:

Step 1: Data given

mass of the shell = 1.65 kg

angle = 38.0 °

Step 2: Calculate the acceleration

We have 2 forces working on the line of motion:

⇒ gravity down the slope = m*g*sinα

       ⇒ provides the linear acceleration

⇒ friction up the slope = F

      ⇒ provides the linear acceleration and also the torque about the CoM.

∑F = m*a = m*g*sin(α) - F

I*dω/dt = F*R

The spherical shell with mass m has moment of inertia I=2/3*m*R² Furthermore a pure rolling relates  dω/dt and a through a = R dω/dt. So the two equations become

m*a = m*g sin(α) - F

2/3*m*a = F  

IF we combine both:

m*a = m*g*sin(α) - 2/3*m*a

1.65a = 1.65*9.81 * sin(38.0) - 2/3 *1.65a

1.65a + 1.1a = 9.9654

2.75a = 9.9654

a = 3.62 m/s²

The acceleration is 3.62 m/s²

The magnitude of the acceleration of the center of mass of the spherical shell is 3.62 m/s².

Newton’s second law of motion shows the relation between the force mass and acceleration of a body. It says, that the net force applied on the body is equal to the product of mass of the body and the acceleration of it.

It can be given as,

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

Here, (m) is the mass of the body and (a) is the acceleration.

A hollow spherical shell with mass 1.65 kg rolls without slipping down a slope that makes an angle of 38.0 ∘ with the horizontal. The free-fall acceleration g is 9.80 m/s2 .

The total friction force is equal to the force of gravity acting downward of the slope.

[tex]\sum F=mg\sin(\theta)-F\\[/tex]

For the force acting on the rotating spherical shell is,

[tex]F=\dfrac{2}{3}ma[/tex]

Put this value in above equation,

[tex]\sum F=(1.65)(9.80)\sin(38)-\dfrac{2}{3}(1.65)a\\ma=(1.65)(9.80)\sin(38)-\dfrac{2}{3}(1.65)a\\(1.65)a=(1.65)(9.80)\sin(38)-\dfrac{2}{3}(1.65)a\\a=(9.80)\sin(38)-\dfrac{2}{3}a\\a=3.62\rm \; m/s^2[/tex]

Thus, the magnitude of the acceleration of the center of mass of the spherical shell is 3.62 m/s².

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Answer:

given,

mass of uniform bar = 2.9 Kg

length of string = 3 m

mass of monkey = 1.45 Kg

monkey is sitting at the center

Weight of the beam will also consider to as the center

total downward force  = 2.9 g  + 1.45 g

now tension will balance the force

T₁ + T₂ = 2.9 g  + 1.45 g

T₁ + T₂ = 4.35 g

taking moment about T₂ string

now,

T₁ x (3-0.74) - 4.35 g x (1.5 - 0.74) = 0

T₁ x  2.26 = 32.3988

T₁ = 14.34 N

Answer:

  [tex]v_{f}[/tex] = 3,126 m / s

Explanation:

In a crash exercise the moment is conserved, for this a system formed by all the bodies before and after the crash is defined, so that the forces involved have been internalized.

the car has a mass of m = 1.50 kg a speed of v1 = 4.758 m / s and the mass of the train is M = 3.60 kg and its speed v2 = 2.45 m / s

Before the crash

    p₀ = m v₁₀ + M v₂₀

After the inelastic shock

    [tex]p_{f}[/tex]= m [tex]v_{1f}[/tex] + M [tex]v_{2f}[/tex]

    p₀ =  [tex]p_{f}[/tex]

    m v₀ + M v₂₀ = m [tex]v_{1f}[/tex]  + M [tex]v_{2f}[/tex]

We cleared the end of the train

     M [tex]v_{2f}[/tex] = m (v₁₀ - v1f) + M v₂₀

Let's calculate

     3.60 v2f = 1.50 (2.15-4.75) + 3.60 2.45

     [tex]v_{2f}[/tex]  = (-3.9 + 8.82) /3.60

      [tex]v_{2f}[/tex]  = 1.36 m / s

As we can see, this speed is lower than the speed of the car, so the two bodies are joined

set speed must be

      m v₁₀ + M v₂₀ = (m + M) [tex]v_{f}[/tex]

      [tex]v_{f}[/tex]  = (m v₁₀ + M v₂₀) / (m + M)

      [tex]v_{f}[/tex]  = (1.50 4.75 + 3.60 2.45) /(1.50 + 3.60)

      [tex]v_{f}[/tex] = 3,126 m / s

To develop this problem it is necessary to apply the concepts developed by Clausius - Claperyron.

This duet found the relationship between temperature and pressure expressed as,

[tex]ln P = constant - \frac{\Delta H}{RT}[/tex]

For the two states that we have then we could define the pressure and temperature in each of them as

[tex]ln(\frac{P_2}{P_1}) = \frac{-\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]

Where,

[tex]P_{1,2}[/tex]= Pressure at state 1 and 2

[tex]T_{1,2}[/tex]= Temperature at state 1 and 2

[tex]\Delta H[/tex]= Enthalpy of Vaporization of a substance

R = Gas constant (8.134J/mol.K)

Our values are given by,

[tex]P_1 = 1atm \\\Delta H = 38.56*10^{-3} J/mol \\R = 8.134J/mol.K\\T_1 = 78.4\°C = 351.55K\\T_2 =38.8\°C = 311.95K[/tex]

Therefore replacing we have that,

[tex]ln(\frac{P_2}{P_1}) = \frac{-\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]

[tex]ln(\frac{P_2}{1atm}) = \frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})[/tex]

[tex]ln(P_2) - Ln(1atm) = \frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})[/tex]

[tex]P_2 = e^{\frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})}[/tex]

[tex]P_2 = 0.187355atm[/tex]

Therefore the pressure of the Ethanol at 38.8°C is 0.187355atm

The work done to compress an ideal gas in a cylinder to 1/6 of its initial volume at constant temperature is -78.3 Joules.

This question involves the principle of work done in compressing a gas, a core concept in thermodynamics. The work done on an ideal gas at constant temperature (isothermal process) is given by the formula W = PΔV, where W is the work done, P is the atmospheric pressure, and ΔV is the change in volume.

Given that the initial volume (V1) is 9.33 * 10^-4 m³ and the final volume (V2) is 1/6 of the initial volume, so V2 = V1/6. Thus, the change in volume ΔV = V2 - V1 = V1/6 - V1 = -5V1/6.

Substituting the values into the formula and solving, we get W = PΔV = (1.013 * 10^5 Pa)(-5V1/6) = -5/6 * 1.013 * 10^5 Pa * 9.33 * 10^-4 m³ = -78.3 Joules.

Note that the work done is negative, which in this context means that work is done on the system (gas), not by it.

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Answer:

The unknown quantities are:

E and F

The final velocity of the proton is:

√(8/3) k e^2/(m*r)

Explanation:

Hello!

We can solve this problem using conservation of energy and momentum.

Since both particles are at rest at the beginning, the initial energy and momentum are:

Ei = k (q1q2)/r

pi = 0

where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)

and q1 = e and q2 = 2e

When the distance between the particles doubles, the energy and momentum are:

Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

pf = m1v1 + m2v2

with m1 = m,   m2 = 4m,    v1=vf_p,    v2 = vf_alpha

The conservation momentum states that:

pi = pf      

Therefore:

m1v1 + m2v2 = 0

That is:

v2 = (1/4) v1

The conservation of energy states that:

Ei = Ef

Therefore:

k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

Replacing

      m1 =  m, m2 = 4m, q1 = e, q2 = 2e

      and   v2 = (1/4)v1

We get:

(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 =  k e^2/r + (1/8)mv1^2

(3/8) mv1^2 = k e^2/r

v1^2 = (8/3) k e^2/(m*r)

To solve this problem it is necessary to apply the concepts related to electromotive force or induced voltage.

By definition we know that the induced emf in the loop is equal to the negative of the change in the magnetic field, that is,

[tex]\epsilon = -A \times \frac{\Delta B}{\Delta t}[/tex]

[tex]\epsilon = -A \times (\frac{B_f-B_i}{t_f-t_i})[/tex]

Where A is the area of the loop, B the magnetic field and t the time.

Replacing with our values we have that

[tex]\epsilon = -(\pi (1.5*10^{-2})^2)(\frac{0-23*10^{-6}}{7*10^{-3}-0})[/tex]

[tex]\epsilon = 2.3225*10^{-6}V[/tex]

Therefore the thermal energy produced is given by

[tex]E = P*t = \frac{\epsilon^2}{R}t[/tex]

[tex]E = \frac{(2.3225*10^{-6})^2}{8*10^{-6}}*(7*10^{-3})[/tex]

[tex]E = 4.719*10^{-9}J[/tex]

The thermal energy produced in the loop is [tex]4.719*10^{-9}J[/tex]

Answer:

Explanation:

Kinetic energy of ball

= .5 x .5 x 20²

= 100 J

Original kinetic energy of door = 0

Total kinetic energy before ball hitting the door

= 100 J

We shall apply law of conservation of momentum to calculate angular velocity of the door after ball hitting it.

change in angular  momentum of ball

= mvr - mur , u is initial velocity and v is final velocity of ball

= .5 ( 20 + 16 ) x .06

= 1.08

Change in angular momentum of door

= I x ω - 0

1/3 x 35 x .09² x ω

= .0945 x ω

so

.0945 x ω = 1.08

ω = 11.43

rotational K E of door after collision

= 1/2 I ω²

= .5 x .0945 x 11.43 ²

= 6.17 J  

Kinetic energy of ball after collision

= 1/2 x .5 x 16²

= 64

Total KE of door and ball

= 64 + 6.17

= 70.17 J

LOSS OF ENERGY

= 100 - 70.17 J

= 29.83 J

The minimum acceleration of the wagon required for the book not to slide down can be determined using the coefficient of static friction (μs) between the book and the vertical back of the wagon. The expression for the minimum acceleration is given by a = μs * g.

The minimum acceleration of the wagon required for the book not to slide down can be determined using the coefficient of static friction (μs) between the book and the vertical back of the wagon. The expression for the minimum acceleration is given by a = μs * g, where g is the acceleration due to gravity.

The mass of the book does not matter in this scenario. The coefficient of static friction is a property that depends on the nature of the surfaces in contact, and it determines the maximum force of static friction that can act between the book and the wagon before the book starts to slide down. The acceleration required to prevent sliding is independent of the mass of the book.

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Answer:

Explanation:

Given

mass of balls [tex]m= 5 kg[/tex]

[tex]N=45.6 rev/s[/tex]

angular velocity [tex]\omega =2\pi N=286.55 rad/s[/tex]

Length of Rod [tex]2L=1.1 m[/tex]

Tension in the Second half of rod

[tex]T_2=m\omega ^2(2L)=2m\omega ^2L[/tex]

[tex]T_2=5\times (286.55)^2\times 1.1[/tex]

[tex]T_2=451.609 kN[/tex]

For First Part

[tex]T_1-T_2=m\omega ^2L[/tex]

[tex]T_1=T_2+m\omega ^2L[/tex]

[tex]T_1=3 m\omega ^2L[/tex]

[tex]T_1=3\times 5\times (286.55)^2\times 0.55[/tex]

[tex]T_1=677.41 kN[/tex]

Final answer:

The question deals with the tension in a rotating rod with attached masses. Tension in the first half of the rod is dictated by the centripetal force for one mass, while the second half must account for two masses. It's a problem in the field of Physics, specifically rotational dynamics at the college level.

Explanation:

The student's question revolves around the concepts of rotational motion and tension in a system consisting of a massless rod with balls of equal mass attached at different points. Given the angular speed and the positions of the masses, we are to find the tension in two parts of the rod during rotation.

The tension for the first half of the rod can be calculated by considering the centripetal force required to keep the ball rotating in a circle of radius L, the middle of the rod. Since the ball at L is the only mass in the first segment, we only need to consider its centripetal force requirement.

For the second segment of the rod, from L to 2L, the tension must accommodate the centripetal force for both masses, one at the middle and the other at the end. The ball at the end experiences more tension because it is further from the pivot point and thus has a larger radius of rotation.

Note that actual equations and calculations are not provided here, as the question seems to request conceptual explanations rather than specific numerical solutions.

Answer:

α₂= 0.1  rad/s²

t= 62.8 s

Explanation:

Given that

For small wheel

r₁= 1.2 cm

α₁ = 3 rad/s²

For large wheel

r₂= 36 cm

Angular acceleration = α₂  rad/s²

The tangential acceleration for the both wheel will be same

a = α₁ r₁=α₂ r₂

Now by putting the values in the above equation

α₁ r₁=α₂ r₂

3 x 1.2 = 36 x α₂

α₂= 0.1  rad/s²

Given that

N = 60 rpm

Angular speed in rad/s ω

[tex]\omega = \dfrac{2\pi N}{60}[/tex]

[tex]\omega = \dfrac{2\pi \times 60}{60}[/tex]

ω = 6.28 rad/s

Time taken is t

ω = α₂ t

6.28 = 0.1 t

t= 62.8 s

The angular acceleration of the pottery wheel is 0.1 rad/s². It will rotate at 60 rpm after approximately 63 seconds.

The concept involved in the question relates to the principle of angular momentum conservation. The angular acceleration of the pottery wheel can be calculated from the given angular acceleration of the small wheel and the ratio of their radii. The equation that connects linear acceleration (a), angular acceleration (alpha), and radius (r) is a = alpha*r. Given that the small wheel has an angular acceleration, alpha1=3 rad/s² and radius r1=1.2 cm, while the pottery wheel has radius r2=36 cm, the linear acceleration of both is the same. Hence, the angular acceleration of the pottery wheel, alpha2 = a/r2 = (alpha1*r1)/r2 = (3 * 1.2 cm)/(36 cm) = 0.1 rad/s².

Now, for how long until the pottery wheel rotates at 60 rpm, first convert 60 rpm to rad/s. Note that 1 rev = 2π rad, and 1 min = 60 s. So, 60 rev/min = (60 * 2π rad)/(60 s) = 2π rad/s = final angular velocity, omega.

Then, use the formula for time in angular motion having constant angular acceleration: t = (omega - omega_initial) / alpha, and since the pottery wheel starts from rest, omega_initial = 0. Hence, t = omega/alpha2 = (2π rad/s)/(0.1 rad/s²) = 20π s, or approximately 63 seconds.

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