An object is acted upon by a constant force which gives it a constant acceleration a. At a certain time t1, having started from rest at t = 0, it has kinetic energy K1. At what time t2 has its kinetic energy doubled?

Answer:12.12 m/s

Explanation:

Given

mass of child [tex]m=49 kg[/tex]

height of hill [tex]h=7.5 m[/tex]

inclination [tex]\theta =26^{\circ}[/tex]

Conserving Energy at top and bottom Point of hill

Potential Energy at Top =Kinetic Energy at bottom

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

[tex]v=\sqrt{2gh}[/tex]

[tex]v=\sqrt{2\times 9.8\times 7.5}[/tex]

[tex]v=\sqrt{147}[/tex]

[tex]v=12.12 m/s[/tex]

Answer:

[tex]v\approx 12.129\,\frac{m}{s}[/tex]

Explanation:

The final speed of the child-sled system is determined by means of the Principle of Energy Conservation:

[tex]U_{1} + K_{1} = U_{2} + K_{2}[/tex]

[tex]K_{2} = (U_{1}-U_{2})+K_{1}[/tex]

[tex]\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot \Delta h[/tex]

[tex]v = \sqrt{2\cdot g \cdot \Delta h}[/tex]

[tex]v = \sqrt{2\cdot\left(9.807\,\frac{m}{s^{2}} \right)\cdot (7.50\,m)}[/tex]

[tex]v\approx 12.129\,\frac{m}{s}[/tex]

Answer:

[tex]v = 0.085 m/s[/tex]

Explanation:

First when man throws the ball then the speed of the man is given as

[tex]m_1v_1 = m_2v_2[/tex]

[tex]97 v = 0.365 \times 11.3[/tex]

[tex]v = 0.042 m/s[/tex]

now ball will rebound after collision with the wall

so speed of the ball will be same as initial speed

now again by momentum conservation we will have

[tex]m_1v_1 + m_2v_2 = (m_1 + m_2) v[/tex]

[tex]97 \times 0.042 + 0.365 \times 11.3 = (97 + 0.365) v[/tex]

[tex]v = 0.085 m/s[/tex]

Answer:

See explanation

Explanation:

To do this, you need to use energy conservation.  The sum of kinetic and potential energies is the same at all points along the path so, you can write the expression like this:

1/2mv1² + mgh1 = (1/2)mv2² + mgh2

Where:

v1 = 0 because it's released from rest

h1 = 18.6 m

v2 = speed we want to solve.

h2 = height at point A. In this case, you are not providing the picture or data, so, I'm going to suppose a theorical data to solve this. Let's say h2 it's 12 m.

Now, let's replace the data in the above expression (assuming h2 = 12 m). Also, remember that we don't have the mass of the bead, but we don't need it to solve it, because it's simplified by the equation, therefore the final expression is:

1/2v1² + gh1 = 1/2v2² + gh2  

Replacing the data we have:

1/2*(0) + 9.8*18.6 = 1/2v² + 9.8*12

182.28 = 117.6 + 1/2v²

182.28 - 117.6 = v²/2

64.68 * 2 = v²

v = √129.36

v = 11.37 m/s

Now, remember that you are not providing the picture to see exactly the value of height at point A. With that picture, just replace the value in this procedure, and you'll get an accurate result.

Answer:

Tension in the string, F = 6.316 N

Explanation:

It is given that,

Mass of the Yo - Yo, m = 0.2 kg

Length of the string, l = 0.8 m

It makes a complete revolution each second, angular velocity, [tex]\omega=2\pi\ rad[/tex]

Let T is the tension exist in the string. The tension acting on it is equal to the centripetal force acting on it. Its expression is given by :

[tex]F=\dfrac{mv^2}{r}[/tex]

[tex]F=m\omega^2 r[/tex]

[tex]F=0.2\times (2\pi )^2 \times 0.8[/tex]

F = 6.316 N

So, the tension must exist in the string is 6.316 N. Hence, this is the required solution.

Answer:

Radius, r = 1.5 meters

Explanation:

It is given that,

When an object is moving in an amusement park it will exert centripetal force. The centripetal force acting on the person, F = 560 N

Mass of the person, m = 83 kg

Speed of the wall, v = 3.2 m/s

The centripetal force acting on the person is given by :

[tex]F=\dfrac{mv^2}{r}[/tex]

[tex]r=\dfrac{mv^2}{F}[/tex]

[tex]r=\dfrac{83\times (3.2)^2}{560}[/tex]

r = 1.51 meters

or

r = 1.5 meters

So, the radius of the chamber is 1.5 meters.

Answer:

Speed of proton = 1.69 x 10⁷ m/s to the left

Speed of carbon = 3.08 x 10⁶ m/s to the right

Explanation:

Elastic collision means momentum and kinetic energy are conserved.

Mass of carbon = 12 x Mass of proton.

Initial velocity of proton = 2 x 10⁷ m/s

Initial velocity of Carbon = 0 m/s

Final velocity of proton = u

Final velocity of carbon = v

Initial momentum = Final momentum.

m x 2 x 10⁷ + 12m x 0 = m x u + 12m x v

               u + 12v = 2 x 10⁷

               u = 2 x 10⁷ - 12v

Initial K.E = Final K.E

0.5 x m x (2 x 10⁷)² + 0.5 x 12m x 0² = 0.5 x m x u² + 0.5 x 12m x (v)²

               u² + 12v² = 4 x 10¹⁴

              (2 x 10⁷ - 12v)² + 12v² = 4 x 10¹⁴

               4 x 10¹⁴ - 48 x 10⁷v + 144v² + 12v² = 4 x 10¹⁴

                                      156v² = 48 x 10⁷v

                                             v = 3.08 x 10⁶ m/s

                   u + 12 x 3.08 x 10⁶ = 2 x 10⁷  

                     u = -1.69 x 10⁷ m/s

Speed of proton = 1.69 x 10⁷ m/s to the left

Speed of carbon = 3.08 x 10⁶ m/s to the right

In a head-on perfectly elastic collision between a proton and a carbon atom, the speed of each particle remains the same after the collision, but their directions are reversed.

An elastic collision between a proton and a carbon atom can be analyzed using the conservation of momentum and kinetic energy.

First, we need to determine the initial momentum of the proton and the carbon atom. The momentum of an object is calculated by multiplying its mass by its velocity.

Given that the mass of the carbon atom is 12 times the mass of the proton, the initial momentum of the proton is equal to the momentum of the carbon atom.

After the collision, the total momentum of the system must still be conserved. Since the collision is head-on and elastic, only the directions of the velocities of both particles will change. The speed of each particle will remain the same.

Therefore, after the collision, the speed of the proton will still be 2.0 * 10^7 m/s, but its direction will be reversed to the left. The carbon atom will also have a speed of 2.0 * 10^7 m/s, but it will be traveling in the right direction.

Answer:[tex]180.86^{\circ}C[/tex]

Explanation:

Properties of Fluid at [tex]100^{\circ}C[/tex]

[tex]P_r=0.711[/tex]

[tex]\rho =0.9458 kg/m^3[/tex]

[tex]c_p=1009 J/kg/k[/tex]

[tex]Flux =6100 W/m^2[/tex]

Drag force [tex]F_d=0.3 N[/tex]

[tex]A=0.2\times 0.5=0.1 m^2[/tex]

drag force is given by

[tex]F_d=c_f\cdot A\rho \frac{v^2}{2}[/tex]

[tex]c_f=\frac{2F_d}{\rho Av^2}[/tex]

[tex]c_f=\frac{2\times 0.3}{0.9458\times 0.1\times 100^2}[/tex]

[tex]c_f=\frac{0.6}{945.8}[/tex]

[tex]c_f=0.000634[/tex]

we know average heat transfer coefficient is

[tex]h=\frac{c_f}{2}\times \frac{\rho vc_p}{P_r^{\frac{2}{3}}}[/tex]

[tex]h=\frac{0.000634}{2}\times \frac{0.9458\times 100\times 1009}{(0.711)^{\frac{2}{3}}}[/tex]

[tex]h=37.92 W/m^2-K[/tex]

Surface Temperature of metal Foil

[tex]\dot{q}=h(T_s-T{\infty })[/tex]

[tex]T_s=\frac{\dot{q}}{h}[/tex]

[tex]T_s[/tex] is the surface temperature and T_{\infty }[/tex] is ambient temperature

[tex]T_s=\frac{6100}{37.92}+20=180.86^{\circ}C[/tex]

The required height of the tin can is given by the height at the minimum value of the surface area function of the tin can

Reason:

The given information on the tin can are;

Volume of the tin can = 16·π in.³

The amount of tin to be used = Minimum amount

The height of the can in inches required

Solution;

Let h, represent the height of the can, we have;

The surface area of the can, S.A. = 2·π·r² + 2·π·r·h

The volume of the can, V = π·r²·h

Where;

r = The radius of the tin can

h = The height of the tin can

Which gives;

16·π = π·r²·h

16 = r²·h

[tex]h = \dfrac{16}{r^2}[/tex]

Which gives;

[tex]S.A. = 2 \cdot \pi \cdot r^2 + 2 \cdot \pi \cdot r \cdot \dfrac{16}{r^2} = 2 \cdot \pi \cdot r^2 + \pi \cdot \dfrac{32}{r}[/tex]

When the minimum amount of tin is used, we have;

[tex]\dfrac{d(S.A.)}{dx} =0 = \dfrac{d}{dx} \left( 2 \cdot \pi \cdot r^2 + \pi \cdot \dfrac{32}{r} \right) = \dfrac{4 \cdot \pi \cdot \left (r^3-8 \right)}{r^2}[/tex]

Therefore;

[tex]\dfrac{4 \cdot \pi \cdot \left (r^3-8 \right)}{r^2} = 0[/tex]

4·π·(r³ - 8) = r² × 0

r³ = 8

r = 2

The radius of the tin can, r = 2 inches

The

[tex]h = \dfrac{16}{2^2} = 4[/tex]

The height of the tin can, h = 4 inches

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The required height of can is 4 inches.

Given data:

The volume of cylindrical tin with top and bottom is, [tex]V = 16 \pi \;\rm in^{3}[/tex].

The volume of tin is,

[tex]V = \pi r^{2}h\\16 \pi = \pi r^{2}h\\[/tex]

here, r is the radius of can and h is the height of can.

[tex]16 =r^{2}h\\\\h=\dfrac{16}{r^{2}}[/tex]

The surface area of tin is,

[tex]SA = 2\pi r(r+h)\\SA = 2\pi r(r+\dfrac{16}{r^{2}})\\SA = 2\pi r^{2}+\dfrac{32 \pi}{r})[/tex]

For minimum amount of tin used, we have,

[tex]\dfrac{d(SA)}{dr} =0\\\dfrac{ d(2\pi r^{2}+\dfrac{32 \pi}{r})}{dr} = 0\\4\pi r-\dfrac{32 \pi}{r^{2}}=0\\4\pi r=\dfrac{32 \pi}{r^{2}}\\r^{3}=8\\r =2[/tex]

So, height of can is,

[tex]h=\dfrac{16}{2^{2}}\\h=4[/tex]

Thus, the required height of can is 4 inches.

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Answer:

-No

Explanation:

Given that

Mass of box = 100 kg

Coefficient of friction ,μ= 0.23

We know that friction force depends on the normal force acts on the box

Fr= μ N

When we pull the box then :

Normal force N= mg

Friction force Fr=  μ mg                  

When we push the box :

Lets take pushing force = F

θ =Angle make by pushing force from the vertical line

Normal force

N = mg + F cosθ

Fr= μ ( mg + F cosθ )

The friction force is more when we push the box.That is why this is not easier to push the box.

Therefore answer is ---No

The mass attached to the spring must be 0.72 kg

Explanation:

The frequency of vibration of a spring-mass system is given by:

[tex]f=\frac{1}{2\pi} \sqrt{\frac{k}{m}}[/tex] (1)

where

k is the spring constant

m is the mass attached to the spring

We can find the spring constant by using Hookes' law:

[tex]F=kx[/tex]

where

F is the force applied on the spring

x is the stretching of the spring

When a mass of m = 2.82 kg is applied to the spring, the force applied is the weight of the mass, so we have

[tex]mg=kx[/tex]

and using [tex]g=9.8 m/s^2[/tex] and [tex]x=0.0177 m[/tex], we find

[tex]k=\frac{mg}{x}=\frac{(2.82)(9.8)}{0.0177}=1561.3 N/m[/tex]

Now we want the frequency of vibration to be

f = 7.42 Hz

So we can rearrange eq.(1) to find the mass m that we need to attach to the spring:

[tex]m=\frac{k}{(2\pi f)^2}=\frac{1561.3}{(2\pi (7.42))^2}=0.72 kg[/tex]

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Answer:

166 666 666.7 years

Explanation:

We start the question by making the units uniform. We are told that the continents move at 3 cm/year = 0.03 m/year.

We are also told that the continents are now 5000 km = 5 000 000 m apart

So to calculate the time it took for them to be this far apart

t = distance/speed

t = 5 000 000 m/(0.03 m/year) = 166 666 666.7 years

Most stars have one or more terrestrial planets orbiting within their habitable zones is the correct answer.  

Explanation:                    

According to the science, the  habitable zones  are the region around the stars where one or more terrestrial planets can orbit. A terrestrial planet orbits inhabitable zone is called potentially habitable zone which have  roughly comparable conditions to those of earth.

This situation is a proof stating that most of the stars have one or more terrestrial planets orbiting within their habitable zone.

Answer:

7.30523 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 2.72 m

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 2.72+0^2}\\\Rightarrow v=7.30523\ m/s[/tex]

The speed which a quarter would have reached before contact with the ground is 7.30523 m/s

Final answer:

The speed at which the quarter would have reached before contact with the ground if dropped from rest from the top of Robert Wadlow's head is approximately 7.3 m/s.

Explanation:

To determine the speed at which a quarter would have reached before contact with the ground if dropped from rest from the top of Robert Wadlow's head, we can use the principle of conservation of mechanical energy.

Let's denote the height of Robert Wadlow's head as h and the mass of the quarter as m. We are given that h = 2.72 meters.

The potential energy (PE) of the quarter at the top of Robert Wadlow's head is:

PE = m * g * h

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

The kinetic energy (KE) of the quarter just before it hits the ground is:

KE = (1/2) * m * v^2

where v is the speed of the quarter just before it hits the ground.

According to the principle of conservation of mechanical energy, the total mechanical energy at the top of Robert Wadlow's head is equal to the total mechanical energy just before the quarter hits the ground:

PE = KE

Substituting the expressions for PE and KE, we get:

m * g * h = (1/2) * m * v^2

Solving for v^2, we get:

v^2 = 2 * g * h

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

v^2 = 2 * 9.8 m/s^2 * 2.72 m

v^2 ≈ 53.2 m^2/s^2

Taking the square root of both sides, we get:

v ≈ 7.3 m/s

So, the speed at which the quarter would have reached before contact with the ground if dropped from rest from the top of Robert Wadlow's head is approximately 7.3 m/s.

Answer:

FALSE

Explanation:

The earth's lithosphere is divided into 2 types, namely the continental and the oceanic crust. The continental crust is comprised of lighter rock minerals such as silicate minerals, alkali and potash feldspar, and some oxide minerals, whereas, the oceanic crust is comprised of olivine, pyroxene and some feldpars that are comparatively denser than the minerals present in the continental crust. Due to this composition, the oceanic lithosphere is denser than the continental lithosphere.

During the convergent plate motion, the oceanic lithosphere (plate) subducts below the continental lithosphere due to its greater density.

Thus, the above given statement is False.

Answer:

Explanation:

The tendency of the automobile running on a circular path at high speed to turn towards left or right around one of its wheels , is due to torque by centripetal force acting at its centre of mass about that wheel .

Suppose the automobile tends to turn in clockwise direction about its wheel on the outer edge . A rotational angular momentum is created . To counter this effect , we can take the help of a rotating wheel or flywheel . We shall have to keep its rotation in anticlockwise direction so that it can create rotational angular momentum in direction opposite to that created by centrifugal force on fast moving automobile. Each of them will nullify the effect of the other . In this way , rotating flywheel will save the automobile from turning upside down .

Final answer:

To counteract the tendency for a car to roll over when rounding curves at high speed, a flywheel should be mounted horizontally and perpendicular to the travel direction, with a sense of rotation that produces gyroscopic precession force downward on the inside wheels.

Explanation:

When an automobile rounds a curve at high speeds, the loading on the wheels changes, with the tendency to lighten the load on the inside wheels, which can lead to the car rolling over. To combat this, a large spinning flywheel can help equalize the loading on the wheels. The flywheel should be mounted such that its axis of rotation is horizontal and perpendicular to the direction of the car's travel. The sense of rotation should be such that when the car turns, the gyroscopic precession of the flywheel produces a force that pushes down on the inside wheels. This force counteracts the effect of the weight transfer to the outside wheels, reducing the risk of the car tipping over.

This concept exploits the conservation of angular momentum and the phenomenon known as gyroscopic precession. In a closed system, any attempt to change the direction of the axis of rotation of the spinning flywheel produces a force perpendicular to the direction of the applied force (precession). If the top of the flywheel is tilted outward, the precession force acts downward on the side of the flywheel closest to the inside of the turn, creating a stabilizing effect on the inside wheels.

Answer:

Angular acceleration will be [tex]1135.5rad/sec^2[/tex]

Explanation:

We have given initial angular speed of a particular disk [tex]\omega _i=646.1rad/sec[/tex]

And it finally comes to rest so final angular speed [tex]\omega _f=0rad/sec[/tex]

Time is given as t = 0.569 sec

From third equation of motion we know that

[tex]\omega _f=\omega _i+\alpha t[/tex]

So angular acceleration [tex]\alpha =\frac{\omega _f-\omega _i}{t}=\frac{0-646.1}{0.569}=1135.5rad/sec^2[/tex]

Answer:

[tex]Re=1992.24[/tex]

Explanation:

Given:

vertical height of oil coming out of pipe, [tex]h=25\ m[/tex]

diameter of pipe, [tex]d=0.1\ m[/tex]

length of pipe, [tex]l=50\ m[/tex]

density of oil, [tex]\rho = 900\ kg.m^{-3}[/tex]

viscosity of oil, [tex]\mu=1\ Pa.s[/tex]

Now, since the oil is being shot verically upwards it will have some initial velocity and will have zero final velocity at the top.

Using the equation of motion:

[tex]v^2=u^2-2gh[/tex]

where:

v = final velocity

u = initial velocity

Putting the respective values:

[tex]0^2=u^2-2\times 9.8\times 25[/tex]

[tex]u=22.136\ m.s^{-1}[/tex]

For Reynold's no. we have the relation as:

[tex]Re=\frac{\rho.u.d}{\mu}[/tex]

[tex]Re=\frac{900\times 22.136\times 0.1}{1}[/tex]

[tex]Re=1992.24[/tex]

Answer:

1. Rotating 2. Revolving

Explanation:

Just did it on edge 2021

Final answer:

Using the given formula and solving for the constant using the temperature 10 minutes after pouring, the temperature of the coffee 30 minutes after it was poured is found to be 60 degrees Fahrenheit.

Explanation:

The question asks us to determine the temperature of coffee 30 minutes after it was poured, given the formula F = 120(2–at) + 60, where F is in degrees Fahrenheit and t is the time in minutes since the coffee was poured. Since the temperature of the coffee 10 minutes after it was poured was 120 degrees Fahrenheit, we first need to find the value of the constant a using the given formula. Substituting F with 120 and t with 10 gives us 120 = 120(2-10a) + 60. Solving for a, we find that a is equal to 0.1. Now, to find the temperature of the coffee 30 minutes after it was poured, we substitute t with 30 in the formula, getting F = 120(2-0.1*30) + 60, which simplifies to F = 60 degrees Fahrenheit.

Answer:

6.5 seconds.

Explanation:

Given: h=-16t²+679

When the object reaches the ground, h=0.

∴ 0=-16t²+679

collecting like terms,

⇒ 16t²=679

Dividing both side of the equation by the coefficient of t² i.e 16

⇒ 16t²/16 = 679/16

⇒ t² = 42.25

taking the square root of both side of the equation.

⇒ √t² =√42.25

⇒ t = 6.5 seconds.

Answer: E.  and Electroweak in 1961 the only other, as

Explanation:

This is more an English grammar question than a physics question, so taking that perspective, one should look for the answer that best completes the sentence.

Based on the word "first" in the sentence, implying the need for a conjunction to join the two theories and the last part of the sentence does not give a reason but further supports the determination of the theory being the first unified theory.

So, to complete the sentence, the best option is;

Maxwell’s theory of Electromagnetism in 1865 was the first "unified field theory" and Electroweak in 1961 the only other, as no further theory has united the electroweak field with either the Strong (Hadronic) force or Gravity.

The equivalent resistance is calculated by taking the inverse of the sum of the reciprocals of each resistance.

The smallest resistance you can obtain by connecting a 36.0-ohm, a 50.0-ohm, and a 700-ohm resistor together can be found by connecting them all in parallel. When resistors are connected in parallel, the equivalent resistance (Requivalent) is smaller than the smallest individual resistance in the network. The formula for the equivalent resistance of parallel resistors is:

1/Requivalent = 1/R1 + 1/R2 + 1/R3

Substituting the given values gives us:

1/Requivalent = 1/36 + 1/50 + 1/700

Calculating the sum of inverses and then taking the inverse of that sum gives us the equivalent resistance for our parallel network. The smallest resistance value, which is Requivalent, can be calculated using this method.

Answer:

The answer is to insulate Earth's surface temperature.

Explanation:

Clouds inhibit the outflow of terrestrial radiation. This acts to insulate Earth's surface temperature, keeping it warmer at night and cooler in the day.

Answer:

Equilibrium temperature will be [tex]T=52.2684^{\circ}C[/tex]

Explanation:

We have given weight of the lead m = 2.61 gram

Let the final temperature is T

Specific heat of the lead c = 0.128

Initial temperature of the lead = 11°C

So heat gain by the lead = 2.61×0.128×(T-11°C)

Mass of the water m = 7.67 gram

Specific heat = 4.184

Temperature of the water = 52.6°C

So heat lost by water = 7.67×4.184×(T-52.6)

We know that heat lost = heat gained

So [tex]2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)[/tex]

[tex]0.334T-3.67=1688-32.031T[/tex]

[tex]T=52.2684^{\circ}C[/tex]

By equating the heat gained by the lead to the heat lost by the water, the final temperature is calculated to be approximately 41.5°C.

To determine the final temperature of the lead weight and water, we can use the principle of calorimetry, which states that the heat gained by one object is equal to the heat lost by another object in a closed system.

Given:

Mass of lead[tex](m_l_e_a_d) = 2.61 g[/tex]

Specific heat capacity of lead [tex](c_l_e_a_d)[/tex] = 0.128 J/g°C

Initial temperature of lead[tex](T_l_e_a_d_i_n_i_t_i_a_l)[/tex]) = 11.1°C

Mass of water [tex](m_w_a_t_e_r)[/tex]= 7.67 g

Specific heat capacity of water[tex](c_w_a_t_e_r)[/tex] = 4.184 J/g°C

Initial temperature of water [tex](T_w_a_t_e_r_i_n_i_t_i_a_l)[/tex] = 52.6°C

Let T_final be the final equilibrium temperature.

Heat gained by lead [tex](Q_lead) = m_lead * c_lead * (T_final - T_lead_initia[/tex]l)

Heat lost by water [tex](Q_water) = m_water * c_water * (T_water_initial - T_final)[/tex]

Since the system is thermally insulated, Q_lead = -Q_water

[tex]m_lead * c_lead * (T_final - T_lead_initial) = - m_water * c_water * (T_water_initial - T_final)[/tex]

Solving for T_final:

[tex]T_final = (m_lead * c_lead * T_lead_initial + m_water * c_water * T_water_initial) / (m_lead * c_lead + m_water * c_water)[/tex]

Plugging in the values:

T_final = (2.61 g * 0.128 J/g°C * 11.1°C + 7.67 g * 4.184 J/g°C * 52.6°C) / (2.61 g * 0.128 J/g°C + 7.67 g * 4.184 J/g°C)

T_final ≈ 41.5°C

Therefore, the final temperature of both the lead weight and the water at thermal equilibrium is approximately 41.5°C.

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Answer: Option (A) is the correct answer.

Explanation:

A corrective action is defined as the action with the help of which a person can avoid a difficulty or problem that he/she was facing earlier.

For example, when the chef checked the temperature of soup using thermometer then it was 120 but his operation's critical limit was 135.

So, to avoid this problem he heated the soup to 165 at 15 seconds following which he got the result as desired.

Therefore, reheating the soup was his corrective action.

Thus, we can conclude that reheating the soup was the corrective action.

In this context, the corrective action taken by the chef was reheating the soup because it was not meeting the operation's critical safe temperature limit.

In the provided scenario, the corrective action referring to steps taken to rectify a situation that does not meet specified standards, is reheating the soup. When the chef figured out the soup was below the operation's critical limit, prompt correction was made to ensure food safety by reheating the soup to 165 for 15 seconds. This measure ensures that any potential harmful microorganisms in the food are eliminated, thus making the food safe to consume.

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Answer:

T= 210.15 N

F= 75.31 N

Explanation:

Let the tension in string be T newton.

According to the question

⇒T×cos21°= mg

⇒T= mg/cos21°

⇒T=20×9.81/cos21

⇒T= 210.15 N

now, the magnitude of horizontal force

F= Tsin21°

⇒F= 210.15×sin21°

=75.31 N

Answer:

[tex]fraction = 0.11[/tex]

Explanation:

Linear kinetic energy of the bicycle is given as

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

[tex]K_1 = \frac{1}{2}(13) v^2[/tex]

[tex]K_1 = 6.5 v^2[/tex]

Now rotational kinetic energy of the wheels

[tex]K_2 = 2(\frac{1}{2}(I)(\omega^2))[/tex]

[tex]K_2 = (mR^2)(\frac{v^2}{R^2})[/tex]

[tex]K_2 = mv^2[/tex]

[tex]K_2 = 3.1 v^2[/tex]

now kinetic energy of the rider is given as

[tex]K_3 = \frac{1}{2}Mv^2[/tex]

[tex]K_3 = \frac{1}{2}(38) v^2 [/tex]

[tex]K_3 = 19 v^2[/tex]

So we have

[tex]fraction = \frac{K_2}{K_1 + K_2 + K_3}[/tex]

[tex]fraction = \frac{3.1 v^2}{6.5 v^2 + 3.1 v^2 + 19 v^2}[/tex]

[tex]fraction = 0.11[/tex]

Answer:[tex]24.70 ^{\circ}C[/tex]

Explanation:

Given

mass of lead piece [tex]m_l=234 gm\approx 0.234 kg[/tex]

mass of water in calorimeter [tex]m_w=611 gm\approx 0.611 kg[/tex]

Initial temperature of water [tex]T_w=24^{\circ}C[/tex]

Initial temperature of lead piece [tex]T_l=24^{\circ}C[/tex]

we know heat capacity of lead and water are [tex]125.604 J/kg-k[/tex] and [tex]4.184 kJ/kg-k[/tex] respectively

Let us take [tex]T ^{\circ}C[/tex] be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

[tex]m_lc_l(T_l-T)=m_wc_w(T-T_w)[/tex]

[tex]0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)[/tex]

[tex]86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)[/tex]

[tex]86-T=86.97T-2087.49[/tex]

[tex]T=\frac{2173.491}{87.97}=24.70^{\circ}C[/tex]

Answer:

Total load = 2999.126 kg

Explanation:

Let the spring constant of the shock absorber be k.

We know that the force applied on a spring is directly proportional to elongated length and the constant of proportionality is called spring constant.

Thus

Force, F = kx

where,

x = elongation = 9.1 cm 0.091 m

mass of the people, m = 127 kg

F = weight of the people = mg = 127 x 9.8 = 1244.6 N

substituting these values in the first equation,

1244.6 = k x 0.091

thus, k = 13,676.923 N/m

Now we know that the time period, T of an oscillating spring with a load of mass m is

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

[tex]\frac{m}{k} = \frac{T^{2} }{4\pi ^{2} }[/tex]

thus,

[tex]m = k\frac{T^{2} }{4\pi ^{2}}[/tex]

T = 1.66s

substituting these values in the equation,

[tex]m =13,676.923\frac{1.66^{2} }{4\pi ^{2}  }[/tex]

m = 2999.126 kg

Answer:

The speed v of the particle at t=5.00 seconds = 43 m/s

Explanation:

Given :

mass m = 5.00 kg

force f(t) = 6.00t2−4.00t+3.00 N

time t between t=0.00 seconds and t=5.00 seconds

From mathematical expression of Newton's second law;

Force = mass (m) x acceleration (a)

F = ma              

[tex]a = \frac{F}{m}[/tex]      ...... (1)

acceleration (a) [tex]= \frac{dv}{dt}[/tex]   ......(2)

substituting (2) into (1)

Hence, F [tex]= \frac{mdv}{dt}[/tex]

[tex]\frac{dv}{dt} = \frac{F}{m}[/tex]

[tex]dv = \frac{F}{m} dt[/tex]

[tex]dv = \frac{1}{m}Fdt[/tex]

Integrating both sides

[tex]\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt[/tex]

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;

[tex]v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt[/tex]     ......(3)

Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):

[tex]v = \frac{1}{5} \int\limits^5_0 {(6t^{2} - 4t + 3)} \, dt[/tex]

[tex]v = \frac{1}{5} |\frac{6t^{3} }{3} - \frac{4t^{2} }{2} + 3t |^{5}_{0}[/tex]

[tex]v = \frac{1}{5} |(\frac{6(5)^{3} }{3} - \frac{4(5)^{2} }{2} + 3(5)) - 0|[/tex]

[tex]v = \frac{1}{5} |\frac{6(125)}{3} - \frac{4(25)}{2} + 15 |[/tex]

[tex]v = \frac{1}{5} |\frac{750}{3} - \frac{100}{2} + 15 |[/tex]

[tex]v = \frac{1}{5} | 250 - 50 + 15 |[/tex]

[tex]v = \frac{215}{5}[/tex]

v = 43 meters per second

The speed v of the particle at t=5.00 seconds = 43 m/s