A person who weighs 685 N steps onto a spring scale in the bathroom, and the spring compresses by 0.88 cm. (a) What is the spring constant? N/m (b) What is the weight of another person who compresses the spring by 0.38 cm?

Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

Mass of the truck = 7100 kg

Angle = 15°

velocity = 15m/s

Acceleration = 1.5 m/s²

Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinθ

⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes  m1*gsinθ .

Resultant force on truck is F = T – m1*gsinθ  

F causes the acceleration of the truck: F= m*a

This gives the equation:

T – m1*gsinθ = m1*a  

T = m1(a + gsinθ)

Combining both equations gives:

(m1+m2)*g*sinθ = m1*(a + gsinθ)

m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ

m2*g*sinθ = m1*a

Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:

m2*g*sinθ = (7100 – m2)*a

m2*g*sinθ = 7100a – m2a

m2*gsinθ + m2*a = 7100a

m2* (gsinθ + a) = 7100a

m2 = 7100a/(gsinθ  + a)

m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)

m2 = 2636.8 kg

The load has a mass of 2636.8 kg

The mass of the load is 2636.8 kg

Mass of the truck m = 7100 kg

Angle θ = 15°

velocity v = 15m/s

Acceleration a = 1.5 m/s²

Now let us assume that the mass of the truck is m₁ and the mass of the load is m₂. The thrust from the engine be T.

Initially, the thrust (T) balances the total weight:

[tex]T = (m_1+m_2) g sin\theta[/tex]

When the load falls off, the thrust (T) remains the same

so the net force on the truck

[tex]F = T -m_1gsin\theta[/tex]

[tex]T-m_1gsin\theta = m_1a \\\\ T = m_1(a + gsin\theta)[/tex]

comparing both the equation for T we get:

[tex](m_1+m_2)gsin\theta = m_1(a + gsin\theta)\\\\m_1gsin\theta + m_2gsin\theta =m_1a + m_1gsin\theta\\\\m_2gsin\theta = m_1a[/tex]

Now, m₁+m₂ = 7100kg

m₁= 7100 – m₂

[tex]m_2gsin\theta = (7100-m_2)a\\\\m_2gsin\theta = 7100a-m_2a\\\\m_2gsin\theta + m_2a = 7100a\\\\m_2 (gsin\theta + a) = 7100a\\\\m_2 = 7100a/(gsin\theta + a)\\\\m_2 = (7100 \times1.5) / (9.8sin(15) + 1.5)\\\\m_2 = 2636.8 kg[/tex]

The load has a mass of 2636.8 kg

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Answer: It is not gold. The density is 1.25 g/cm3, very little compared with gold.

Explanation:

According to Archimedes’ principle, any body submerged in a liquid, receives an upward force, called buoyant force, which magnitude is equal to the weight of the liquid that displaces.

This weight can be calcultated as follows:

Fb= δH2O . Vbody g

We are told that the volume of the liquid displaced (equal to the volume of the body as it is completely submerged in water) is 400 cm3.

If we know that the mass of the bracelet is 0.5 Kg, we can find out the density of the material, just applying the same relationship between mass, density and volume:

m = δbracelet. V  

δbracelet = m / V = 500 g / 400 cm3= 1.25 g/cm3

As the gold density is approximately 19.3 g/cm3, it is clear that the bracelet is not made from gold, but from something less dense.

The density of Vadim's bracelet is calculated to be 1.25 g/cm³, which is significantly lower than the density of pure gold (19.3 g/cm³). Thus, the bracelet is not made of gold and could be made from a lighter metal or alloy.

To determine if Vadim's bracelet is gold, we must calculate its density and compare it to the density of pure gold. The density of a substance is the mass-to-volume ratio, and for gold, it is 19.3 g/cm³. Given that the bracelet has a mass of 0.5 kg (or 500 g) and a volume displacement of 400 cm³ when submerged in water, we can calculate its density using the formula:

Density = Mass / Volume

density = 500 g / 400 cm³ = 1.25 g/cm³

The calculated density of 1.25 g/cm³ is significantly lower than the density of pure gold, which suggests that the bracelet is not made of gold. Since the density is lower than that of gold, the bracelet could be made of a metal or alloy with a lower density, such as aluminium or a mixture that includes materials with lighter densities.

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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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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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:

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]\frac{r}{r_0}=-4.4x10^{-5}[/tex]

Explanation:

Using the equation to find the fraction does the radius of the sphere change when it is exposed to Venusian atmosphere.

[tex]\frac{r}{r_0}=\frac{1}{3}*\frac{V}{V_0}[/tex]

Replacing to find the relation between the radius

[tex]\frac{V}{V_0}=-\frac{P}{B}[/tex]

Replacing numeric to find the relation

[tex]\frac{r}{r_0}=\frac{1}{3}*\frac{8.9x10^6pa}{6.7x10^{10}pa}[/tex]

[tex]\frac{r}{r_0}=\frac{1}{3}*-1.33x10^{-4}[/tex]

So the relation is

[tex]\frac{r}{r_0}=-4.4x10^{-5}[/tex]

Answer:

Frequency, f = 3.35 Hz

Explanation:

It is given that,

Mass of the object, m = 8 kg

Stretching in the spring, x = 2.2 cm

When the spring is hanged up in the Earth's gravitational field, its weight is balanced by the force in the spring. So,

[tex]mg=kx[/tex]

k is the spring constant

[tex]k=\dfrac{mg}{x}[/tex]

[tex]k=\dfrac{8\times 9.8}{2.2\times 10^{-2}}[/tex]            

k = 3563.63 N/m

Let f is the frequency of oscillation. Its expression is given by :

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{3563.63}{8}}[/tex]        

f = 3.35 Hz

So, the frequency of oscillation by the spring is 3.35 Hz. Hence, this is the required solution.                                                                                                                                                                                                            

Answer:

h=12.04m

Explanation:

1) Notation and some important concepts

[tex]\omega_i[/tex] represent the initial angular velocity

[tex]v_i[/tex] represent the initial velocity

[tex]h[/tex] represent the final height reached by the mass

[tex]M[/tex] represent the mass of the object

[tex]W_f[/tex] represent the work due the friction force (variable of interest)

[tex]KE_{rot}[/tex] represent the rotational energy

[tex]KE_{tra}[/tex] represent the transitional kinetic energy

[tex]PE=mgh[/tex] represent the potential energy

[tex]I=0.8MR^2[/tex] represent the rotational inertia

W= 395 N is the weight of the object

For this problem we can use the principle of energy conservation, this principle states that "the total energy of an isolated system remains constant; it is said to be conserved over time".

At the begin the wheel had rotational energy defined as "The kinetic energy due to rotational motion. Is a scalar quantity and have units of energy usually Joules". And this energy is represented by the following formula: [tex]KE_{rot}=\frac{1}{2}I\omega^2_i[/tex]

At the starting point the wheel also had kinetic energy defined as "The energy of mass in motion" and is given by the formula : [tex]KE_{tran}=\frac{1}{2}mv_i^2_i[/tex]

At the end of the movement we have potential energy since the mass is at height h the potential energy is defined as "The energy stored within an object, due to the object's position, arrangement or state" and is given by the formula [tex]PE=mgh[/tex].

Since we have friction acting we have a work related to the force of friction and we need to subtract this from the formula of conservation of energy

2) Formulas to use

The figure attached is an schematic draw for the problem

Using the principle of energy conservation we have:

[tex]KE_{rot}+KE_{tran}-W_f =PE[/tex]

Replacing the formulas for each energy w ehave:

[tex]\frac{1}{2}I\omega^2_i+\frac{1}{2}mv_i^2_i-W_f =Mgh[/tex]    (1)

We also know that [tex]v_i =\omega_i R[/tex] and [tex]I=0.8MR^2[/tex] so if we replace this into equation (1) we got:

[tex]0.8(\frac{1}{2})M(R\omega_i)^2 +\frac{1}{2}M(\omega_i R)^2-W_f=Mgh[/tex]    (2)

We also know that the weight is defined as [tex]W=mg[/tex] so then [tex]M=\frac{W}{g}[/tex], so if we replace this into equation (2) we have:

[tex]0.8(\frac{1}{2})\frac{W}{g}(R\omega_i)^2 +\frac{1}{2}\frac{W}{g}(\omega_i R)^2-W_f=Wh[/tex]    (3)

So then if we solve for h we got:

[tex]h=\frac{0.8(\frac{1}{2})\frac{W}{g}(R\omega_i)^2 +\frac{1}{2}\frac{W}{g}(\omega_i R)^2-W_f}{W}[/tex]    (4)

3) Solution for the problem

Now we can replace the values given into equation (4):

[tex]h=\frac{0.8(\frac{1}{2})\frac{395N}{9.8\frac{m}{s^2}}(0.652m(23.1\frac{rad}{s}))^2 +\frac{1}{2}\frac{395}{9.8\frac{m}{s^2}}(23.1\frac{rad}{s}(0.652m))^2-3470J}{395N}=12.04m[/tex]    (4)

So then our final answer would be h=12.04m

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

The television set moves sideways approximately 0.15 µm.

To calculate the distance the television set moves sideways, we need to use the equation for shear deformation. The equation is Ax = (F * L0 * L2) / (S * A), where Ax is the displacement, F is the force applied, L0 is the original length, L2 is the height of the rubber pads, S is the shear modulus, and A is the cross-sectional area. Substituting the known values, we have Ax = (200 N * 1.0 cm) / (2.6 x 10^6 N/m^2 * (3.14 x (0.0060 m)^2)). Solving for Ax, we find Ax ≈ 1.50 x 10^-7 m, or 0.15 µm.

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Answer:

3 cm

Explanation:

To answer this question the equation of continuity can be used

For us to use the equaition of continuity, we will make a few assumptions:

[tex]A_{1}v_{1} = A_{2} v_{2} \\\pi (3cm)^{2} * 5m/s = \pi r^{2} *20m/s\\\\r = 1.5 cm\\D = 2r = 2*1.5cm=3 cm[/tex]

Answer:

Halogen

0.85294

Explanation:

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

b = Wien's displacement constant = [tex]2.897\times 10^{-3}\ mK[/tex]

T = Temperature

From Wien's law we have

[tex]\lambda_m=\frac{b}{T}\\\Rightarrow \lambda_m=\frac{2.897\times 10^{-3}}{2900}\\\Rightarrow \lambda_m=9.98966\times 10^{-7}\ m[/tex]

Frequency is given by

[tex]\nu=\frac{c}{\lambda_m}\\\Rightarrow \nu=\frac{3\times 10^8}{9.98966\times 10^{-7}}\\\Rightarrow \nu=3.00311\times 10^{14}\ Hz[/tex]

For Halogen

[tex]\lambda_m=\frac{b}{T}\\\Rightarrow \lambda_m=\frac{2.897\times 10^{-3}}{3400}\\\Rightarrow \lambda_m=8.52059\times 10^{-7}\ m[/tex]

Frequency is given by

[tex]\nu=\frac{c}{\lambda_m}\\\Rightarrow \nu=\frac{3\times 10^8}{8.52059\times 10^{-7}}\\\Rightarrow \nu=3.52088\times 10^{14}\ Hz[/tex]

The maximum frequency is produced by Halogen bulbs which is closest to the value of [tex]5.5\times 10^{14}\ Hz[/tex]

Ratio

[tex]\frac{3.00311\times 10^{14}}{3.52088\times 10^{14}}=0.85294[/tex]

The ratio of Incandescent to halogen peak frequency is 0.85294

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

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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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

Answer

given,                                    

radius of tire = 25 cm = 0.25 m

initial velocity = 0 m/s                

final velocity = 30 m/s                  

time = 10 s                                        

a) acceleration with respect to center  

   v = u + a t                      

   30 = 0 + 10 a                      

    a = 3 m/s²                            

   radial acceleration            

  [tex]a_r = \dfrac{v^2}{r}[/tex]      

  [tex]a_r = \dfrac{2^2}{0.25}[/tex]

  [tex]a_r =16\ m/s^2[/tex]

  [tex]a=\sqrt{a_r^2 + a_t^2}[/tex]

  [tex]a=\sqrt{16^2 + 3^2}[/tex]

  [tex]a=16.28\ m/s[/tex]

acceleration w.r.t center of tire = [tex]a=16.28\ m/s[/tex]

b) acceleration with respect to road will be equal to 3 m/s²

The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.

The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. In order to determine the heat transfer and entropy change, we need to use the first and second laws of thermodynamics. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.

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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:

The radius of the second coil will be 7.5 cm

Explanation:

We have given that radius of the first coil [tex]r_1=5.1cm=0.051m[/tex]

Peak emf of the first generator [tex]e_1=1.8Volt[/tex]

Peak emf of the second generator [tex]e_2=3.9Volt[/tex]

We know that induced emf is rate of change of magnetic flux , that is equal to magnetic field multiply by change in area

So emf will be proportional to the change in area

So [tex]\frac{e_1}{e_2}=\frac{A_1}{A_2}[/tex]

[tex]\frac{1.8}{3.9}=\frac{0.051^2}{r_2^2}[/tex]

[tex]r_2=0.075m = 7.5cm[/tex]

So the radius of the second coil will be 7.5 cm

Answer:

275 MPa, -175 MPa

-0.63636

450 MPa

Explanation:

[tex]\sigma_{max}[/tex] = Maximum stress

[tex]\sigma_{min}[/tex] = Minimum stress

[tex]\sigma_m[/tex] = Mean stress = 50 MPa

[tex]\sigma_a[/tex] = Stress amplitude = 225 MPa

Mean stress is given by

[tex]\sigma_m=\frac{\sigma_{max}+\sigma_{min}}{2}\\\Rightarrow \sigma_{max}+\sigma_{min}=2\sigma_m\\\Rightarrow \sigma_{max}+\sigma_{min}=2\times 50\\\Rightarrow \sigma_{max}+\sigma_{min}=100\ MPa\\\Rightarrow \sigma_{max}=100-\sigma_{min}[/tex]

Stress amplitude is given by

[tex]\sigma_a=\frac{\sigma_{max}-\sigma_{min}}{2}\\\Rightarrow \sigma_{max}-\sigma_{min}=2\sigma_a\\\Rightarrow \sigma_{max}-\sigma_{min}=2\times 225\\\Rightarrow \sigma_{max}-\sigma_{min}=450\ MPa\\\Rightarrow 100-\sigma_{min}-\sigma_{min}=450\\\Rightarrow -2\sigma_{min}=350\\\Rightarrow \sigma_{min}=-175\ MPa[/tex]

[tex]\sigma_{max}=100-\sigma_{min}\\\Rightarrow \sigma_{max}=100-(-175)\\\Rightarrow \sigma_{max}=275\ MPa[/tex]

Maximum stress level is 275 MPa

Minimum stress level is -175 MPa

Stress ratio is given by

[tex]R=\frac{\sigma_{min}}{\sigma_{max}}\\\Rightarrow R=\frac{-175}{275}\\\Rightarrow R=-0.63636[/tex]

The stress ratio is -0.63636

Stress range is given by

[tex]\sigma_{max}-\sigma_{min}=450\ MPa[/tex]

Magnitude of the stress range is 450 MPa

The maximum and minimum stress levels could be calculated as 275 MPa and -175 MPa respectively. The stress ratio comes out to be -0.64, and the magnitude of the stress range is 450 MPa.

In material fatigue studies, the stress levels in a material are often evaluated. In your case:

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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:

7.85 m/s^2

Explanation:

linear or tangential acceleration= dv/dt

⇒[tex]a_t= \frac{12.5-10}{3}[/tex]

=0.83 m/s^2

radial acceleration is given by = [tex]\frac{v^2}{r}[/tex]

⇒[tex]a_r =\frac{12.5^2}{20}[/tex]

= 7.81 m/s^2

total acceleration

[tex]a_T= \sqrt{a_t^2+a_r^2}[/tex]

putting values we get

[tex]a_T= \sqrt{0.83^2+7.81^2}[/tex]

= 7.85 m/s^2

Answer:

Explanation:

Tangential acceleration = ( 12.5 - 10 )/ 3

a_t= .833 m /s²

radial acceleration

= v² / R

12.5² / 20 ( 12.5 m/s is the velocity when it leaves the curve )

a_r= 7.81 ms⁻²  

Total acceleration √( .833² + 7.81²)

=  √( 61.6939)

= 7.85 m/s⁻

Answer:

a) V = 465.9 m/s

b) θ = 70.529°

Explanation:

Let's first calculate angular velocity of earth:

[tex]\omega=\frac{2\pi}{23.9h}*1h/3600s[/tex]

Velocity of a person on Ecuador will be:

[tex]V_E = \omega*R[/tex]

[tex]V_E = 465.9 m/s[/tex]

For part b, since angular velocity is the same:

[tex]\frac{\omega*R}{3}=\omega*(R*cos\theta )[/tex]

Solving for θ:

[tex]\theta=acos(1/3)[/tex]

[tex]\theta=70.529\°[/tex]

(a)   speed at the equator: 465.9 m/s

(b) Latitude for 1/3 speed: 50.6°

(a) The tangential speed of a person living in Ecuador is equal to the circumference of the Earth at the equator divided by the period of rotation. The circumference of the Earth at the equator is 2πr, where r is the radius of the Earth. The period of rotation is 23.9 hours, which is equal to 23.9 × 3600 seconds. Therefore, the tangential speed of a person living in Ecuador is:

v = 2πr / T = 2π(6.38 ×[tex]10^6[/tex] m) / (23.9 × 3600 s) ≈ 465.9 m/s

(b) The tangential speed of a person living at a latitude of θ is equal to vr * cos(θ), where vr is the tangential speed of a person living in Ecuador and r is the radius of the Earth. Therefore, we can solve for θ:

cos(θ) = vr / v = 465.9 m/s / v

Solving for θ, we get:

θ = arccos(vr / v) ≈ 50.6°

Therefore, the latitude at which the tangential speed is one-third that of a person living in Ecuador is approximately 50.6°.

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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

given,

resistance = 0.05 Ω

internal resistance of battery = 0.01 Ω

electromotive force = 12 V

a) ohm's law

        V = IR

     and volage

   [tex]V = \epsilon - Ir[/tex]

now,

   [tex]IR = \epsilon - Ir[/tex]

   [tex]I(R+r) = \epsilon[/tex]

   [tex]I= \dfrac{\epsilon}{R+r}[/tex]

inserting the values

   [tex]I= \dfrac{12}{0.05+0.01}[/tex]

      I = 200 A

b) Voltage

   V = I R

   V = 200 x 0.05

   V = 10 V

c) Power

    P = I V

    P = 200 x 10 = 2000 W

d) total resistance = 0.05 + 0.09 = 0.14 Ω

 [tex]I= \dfrac{\epsilon}{R+r}[/tex]

   [tex]I= \dfrac{12}{0.14+0.01}[/tex]

     I = 80 A

     V = 80 x 0.05 = 4 V

     P = 4 x 80 = 320 W

Answer:

Explanation:

Resistance of motor, R = 0.05 ohm

internal resistance of battery, r = 0.01 ohm

Voltage of battery, V = 12 V

(a) Total resistance, R' = R + r = 0.05 + 0.01 = 0.06 ohm

Let the current be i.

use Ohm's law

i = V / R'

i = 12 / 0.06 = 200 A

(b) Voltage across motor, V' = i x R = 200 x 0.05 = 10 V

(c) Power, P = i²R = 200 x 200 x 0.05 = 2000 Watt.

(d) Total resistance, R' = 0.05 + 0.1 + 0.09 = 0.15 ohm

i = V / R' = 12 / 0.15 = 80 A

V' = i x R = 80 x 0.05 = 4 V

P' = i²R = 80 x 80 x 0.05 = 320 Watt

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

Answer:

Angular displacement will be [tex]\frac{1}{4}\Theta[/tex]

So option (b) will be the correct option

Explanation:

We have given that firstly object is at rest

So [tex]\omega _i=0rad/sec[/tex]

From law of motion we know that angular displacement is given by

[tex]\Theta =\omega _it+\frac{1}{2}\alpha t^2=0\times t+\frac{1}{2}\alpha t^2=\frac{1}{2}\alpha t^2[/tex]

Now angular displacement by the object in [tex]\frac{t}{2}sec[/tex]

[tex]\Theta =0\times t+\frac{1}{2}\alpha (\frac{t}{2})^2=\frac{1}{4}(\frac{1}{2}\alpha t^2)=\frac{1}{4}\Theta[/tex]

So option (b) will be the correct option

The angle the object rotate through in the time [tex]\frac{1}{2} t[/tex] is [tex]\frac{1}{4} (\theta)[/tex]

Given the following data:

To determine the angle the object rotate through in the time [tex]\frac{1}{2} t[/tex]:

Mathematically, angular displacement is given by this formula:

[tex]\theta = \omega_i t +\frac{1}{2} \alpha t^2[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]\theta = 0( t )+\frac{1}{2} \alpha t^2\\\\\theta = \frac{1}{2} \alpha t^2[/tex]

when t = [tex]\frac{1}{2} t[/tex]:

[tex]\theta = \frac{1}{2} \alpha (\frac{t}{2} )^2\\\\\theta = \frac{1}{2} \alpha (\frac{t^2}{4} )\\\\\theta =\frac{1}{4} (\frac{1}{2} \alpha t^2)\\\\\theta =\frac{1}{4} (\theta)[/tex]

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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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